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Unformatted text preview: Chapter 2 MATHEMATICAL FOUNDATION
21 (a)
Poles: s = 0, 0, Zeros: s = 1, 10; (b) Poles: s = 2, 2; 1 cancel each other. 2, , , . Zeros: s = 0. The pole and zero at s = (c) Poles: s = 0, 1 + j, 1  j; 2. (d) Poles: s = 0, 1, 2, . Zeros: s = 22 (a) G (s) = (d)
G ( s) (b) 5 (c) G ( s) = (e) ( s + 5)
1 s
2 2 (s
= 4s
2 +4 ) + 1 s+ 2 G (s) = 4 s
2 + 4s +8 = +4 G (s) e
k =0 kT ( s + 5 ) =
1 1 e T ( s+5 ) 23 (a)
g ( t ) = u s ( t )  2u s (t  1) + 2 u s( t  2)  2 u s ( t  3) + L G (s ) = 1 s (1  2e  s + 2e2 s  2e 3s + L ) =
1 s s 1+ e ( 1e s s ) gT (t ) = u s (t )  2us (  1) + us (t  2) t GT (s ) = 0 t 2
2 (1  2e  s + e 2s ) = ( 1  e  s )
1 s 1 g(t ) = k =0 g T (t  2k )us (t  2k ) G (s) = s
k =0 1 (1 e s ) e 2 2 ks = 1 e s s s (1 + e ) (b) g ( t) = 2tu s ( t )  4(t  0.5) u s (t  0.5) + 4(t  1) us (t  1)  4(t  1.5)us (t  1.5) + L G ( s) =
g
T 2 s
2 (1  2e 0.5 s + 2e s  2e 1.5 s (  0.5 s ) + L) = 2 0.5 s s (1 + e )
2 1e
2 (t ) = 2 tu s ( t )  4 ( t  0 . 5) u s ( t  0 . 5) + 2( t  1 ) u s ( t  1 ) 2 s
2 0 t 1 GT ( s ) = (1  2e0.5 s + e s ) = s 2 (1  e0.5 s )
2 g (t ) = k=0 g T ( t  k )us ( t  k ) G(s ) = k=0 s2 (
2 1 e 0.5 s ) 2 e  ks = ( 0.5 s ) 2 0.5s s (1 + e )
2 1e 24
g(t ) = ( t + 1 ) u s ( t )  ( t  1 ) u s ( t  1 )  2 u s ( t  1 )  ( t  2 ) u s ( t  2 ) + ( t  3) u s ( t  3) + u s ( t  3) G ( s) = 1 s
2 (1  e  s  e 2 s + e 3 s ) + s (1 2e  s + e 3 s )
1
1 6( s 1 3( s 1 2( s 25 (a) Taking the Laplace transform of the differential equation, we get 1 1 2 ( s + 5s + 4) F ( s) = F (s) = s +2 ( s + 1)( s + 2 )( s f (t ) +4) = + 4) + + 1)  + 2) = 1 6 e 4 t + 1 3 e t  1 2 e 2 t t 0
sX (s) (b) sX ( s )
1  x1( 0 ) = X (s)
2 x (0)
1 =1 2  x 2 ( 0 ) = 2 X 1 ( s )  3 X 2 ( s ) + 1 s x (0)
2 =0 Solving for X1 (s) and X2 (s), we have X 1 ( s) = = s s( s 2 + 3 s +1 1 + 1 )( s + 2 ) = = 1 2s + 1 s +1
1 s  1 2( s + 2) X (s)
2 1
s (s + 1 )( s + 2 )
t +1 + +2
x (t )
2 Taking the inverse Laplace transform on both sides of the last equation, we get x (t )
1 = 0 .5 + e  0 .5 e 2 t t 0 = e t +e 2 t t 0 2 26 (a)
G (s) = 1 3s  1 2( s + 2) + 1 3( s + 3) g (t ) = 1 3  1 2 e 2 t + 1 3 e 3 t t 0 (b)
G (s) = 2 . 5
s +1
50 s + 5 (s + 1) 2 + 2 .5 s +3 g(t ) = 2 . 5 e t + 5 te t + 2 .5 e 3 t t 0 (c)
G (s ) = (
1 s  20 s +1
s  30s + 20 s +4
1 s
2 )
2 e s g (t ) = 50  20e [  (t 1)  30cos2(t  1)  5sin2(t  1) us (t  1) (d)
G (s) =  1
0.5 t s 2 +s+2 = + 1 s g (t ) = 1 + 1.069e [ sin1.323t + sin (1.323t  69.3o ) ] = 1 + e0.5 t (1.447sin1.323t  cos1.323t )
t +s +2  s s
2 +s+2 Taking the inverse Laplace transform, t0 (e) g(t ) = 0 .5 t 2 e t 0 27 1 2 0 A = 0 2 3 1 3 1 0 0 B = 1 0 0 1 u (t ) = u1( t) u ( t) 2 28 (a) Y (s ) R (s ) (c) Y (s ) R (s ) = s ( s + 2) s + 10 s + 2 s + s + 2
4 3 2 (b) = 3s + 1 s + 2 s +5s + 6
(d)
3 2 Y (s) R (s ) = 5 s + 10 s + s + 5 1+ 2e
2 s 4 2 Y (s ) R (s ) = 2s + s + 5 3 4 5 6 7 8 9 10 11 12 13 Chapter 4 MATHEMATICAL MODELING OF PHYSICAL SYSTEMS 41 (a) Force equations: f ( t) = M 1 d y1 dt
2 2 + B1 dy1 dt + B3 dy1  dy 2 + K ( y  y ) 1 2 dt dt 2 dy1  dy2 + K ( y  y ) + M d y2 + B dy2 B3 1 2 2 2 2 dt dt dt dt Rearrange the equations as follows: d y1 dt
2 2 2 = = (B 1 + B 3 ) dy1 dt  M1 + B3 dy2 M 1 dt  K M1 K (y
1 1  y2 ) + f M1 d y2 dt (i) State diagram:
2 B3 dy1 M 2 dt (B 2 + B3 ) dy2
2 + M dt M2 (y  y2 ) Since y 1  y2 appears as one unit, the minimum number of integrators is three. State equations: Define the state variables as x = y  y ,
1 1 2 x 2 = dy dt 2 , x 3 = dy dt 1 . dx1 dt =  x2 + x3 dx2 dt = K M2 x1 
x (B 2 + B3 ) M2 x2 + =
dy dt B3 M2
2 x3
x dx3 dt = =
x K M1 x1 +
dy dt
1 B3 M1
. x2  (B 1 + B3 ) M1 x3 + 1 M f (ii) State variables:
dx dt dx dt 1 = y ,
2 x 2 , 3 y ,
1 4 = State equations:
1 = x2 =
x dx dt dx 2 = = K M
2 x 1 
B M B 2 + B3
2 x K M
3 2 +
x K M
2 x B 3 + B M 3 x 4 3 4 K M
1 4 x dt 1 + x 2  1 M 3  1 + B3
M
1 2 x 4 + 1 M
1 f 1 State diagram: 14 Transfer functions: Y1 ( s ) F (s ) Y2 ( s ) F ( s) = = s M 1 M 2 s + [( B1 + B 3 ) M 2 + ( B2 + B3 ) M 1 ] s + [K ( M 1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B 1 + B2 ) K
3 2 { { M 2 s + ( B2 + B3 ) s + K
2 } } s M1 M 2 s + [( B1 + B3 ) M 2 + ( B2 + B3 ) M1 ] s + [ K ( M1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B1 + B2 ) K
3 2 B3 s + K (b) Force equations: d y1 dt
2 2 = (B 1 + B2 ) dy1 M dt + B2 dy2 M dt + 1 M f dy2 dt = dy1 dt  K B2 y2 (i) State diagram: Define the outputs of the integrators as state variables, x 1 =
K M x y ,
2 x 2 = dy dt 1 . State equations:
dx dt
1 = K B
2 x 1 + x2 dx dt x 2 =
y ,
2 x 1  B 1 x M 2 + =
B 1 M dy dt x f (ii) State equations: State variables:
dx dt
1 1 = 2 = y ,
1 x 1 3 . = K B
2 x 1 + x3 dx dt 2 = x3 dx dt 3 = K M x 1  1 M 3 + 1 M f Transfer functions: 15 Y1 ( s ) B2 s + K = 2 F (s ) s MB2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K (c) Force equations: dy1 dt = dy2 dt + 1 B1 f d y2 dt
2 2 Y2 ( s) F ( s) = B2 M B2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K
2 = (B 1 + B2 ) dy2 M dt + B1 dy2 M dt + B1 dy1 M dt  K M y2 (i) State diagram: State equations: Define the outputs of integrators as state variables.
dx dt
1 = x2 dx dt 2 =
x K M x 1  B 2 x M x 2 +
y ,
2 1 M x f (ii) State equations: state variables:
dx dt
1 1 = y ,
1 2 = 3 =  dy dt
2 2 . = x3 + 1 B
1 f dx dt 2 = x3 dx dt 3 = K M x B 2 x M 3 + 1 M f State diagram: Transfer functions: Y1 ( s ) F (s ) = Ms + ( B1 + B 2 ) s + K
2 B1 s Ms + B2 s + K
2 ( ) Y2 ( s ) F (s ) = 1 Ms + B 2 s + K
2 42 (a) Force equations: 16 y 1 = 1 K
2 ( f + Mg ) + y 2 d 2 y
2 2 dt = B dy M dt 2  K 1 + K2
M y 2 + K 2 y M 1 State diagram: State equations: Define the state variables as: x = y ,
1 2 x 2 = dy dt 1 2 . dx dt 1 = x2
s
2 dx dt 2 = K 1 x M 1  B M x 2 + ( f M + Mg ) Transfer functions:
Y (s)
1 F ( s) = + Bs + K 1 + K 2
2 Y (s)
2 K ( Ms
2 + Bs + K 1 ) B 1 dy 1 F (s) = 1 Ms
2 + Bs + K 1 (b) Force equations: dy1 dt = 1 B1 [ f ( t) + Mg] + dy2 dt  K1 B1 (y 1  y2 ) d y2 dt
2 2 = dy K B B dy  + ( y  y )  ( y  y ) M dt dt M M M dt
2 1 2 2 1 2 1 2 2 State diagram: (With minimum number of integrators) To obtain the transfer functions Y ( s ) / F ( s ) and Y ( s ) / F ( s ), we need to redefine the state variables as: x
1 = y ,
2 x 2 = dy 2 1 / dt , and x 3 = 2 y .
1 State diagram: 17 Transfer functions: Y1 ( s ) F (s ) = s Ms + ( B1 + B 2 ) s + K 1
2 2 [MBs + ( B B
1 1 2 + MK1 )] Y2 ( s ) F (s ) = s [M B1 s + ( B1 B2 + MK1 ) ]
2 Bs + K 1 43 (a) Torque equation: 2 d B d = + 2
dt J dt State diagram:
1 J T (t ) State equations:
dx dt
1 = x2 dx dt 2 = B J x 2 + 1 J T Transfer function: ( s )
T (s) =
s ( Js 1 + B) d 2 dt (b) Torque equations: d 1
2 dt 2 = K J ( 1  2 ) + 1 J T K ( 1  2 ) = B State diagram: (minimum number of integrators) State equations:
dx dt
1 = K B x 1 + x2 dx dt 2 =
x K J x 1 + 1 J T State equations: Let x = ,
1 2 2 = 1, =
x and x 3 =
3 d dt 1 . dx dt 1 = K B x 1 + K B x dx
2 2 dx
3 dt dt = K J x 1  K J x 2 + 1 J T State diagram: 18 Transfer functions: 1 ( s ) T ( s) = s BJs + JKs + BK
2 ( Bs + K ) 2 (s ) T ( s) = s BJs + JKs + BK
2 ( K ) (c) Torque equations: T ( t ) = J1 d 1
2 dt 2 + K ( 1  2 ) K ( 1  2 ) = J 2 d 2
2 dt 2 State diagram: State equations: state variables:
dx dt
1 x 1 =2,
K J
2 x 2 =
dx d dt
3 2 , x 3 = 1,
dx
4 x 4 =
x d dt 1 .
x = x2 dx dt 2 = K J
2 x 1 + x 3 dt = x 4 dt = K J
1 1  K J
1 3 + 1 J
1 T Transfer functions: 1 ( s ) T ( s) = s 2 J1 J2 s + K ( J1 + J 2 ) 2 J 2s + K
2 2 (s ) T ( s) = K s
2 J1 J 2 s + K ( J1 + J 2 ) 2 (d) Torque equations: d m
2 T ( t) = J m dt 2 + K 1 ( m  1 ) + K 2 ( m  2 ) K 1 ( m  1 ) = J 1 d 1
2 dt 2 K2 (m  2 ) = J2 d 2
2 dt 2 19 State diagram: State equations: x = 1 m  1, x 2 = d dt 1 , x 3 = d dt m , x 4 = m 2 , x 5 = d dt 2 . dx dt 1 = x2 + x3 dx dt 2 = K J 1 x dx
1 3 1 dt = K J 1 x 1  K J 2 x 4 + 1 J
m T dx dt 4 = x 3  x5 dx dt 5 = K J 2 x 4 m m 2 Transfer functions: 1 ( s ) T ( s) 2 ( s) T (s ) = s 2 s 4 + ( K1 J2 J m + K2 J 1 J m + K1 J 1 J 2 + K 2 J 1 J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 ) s 4 + ( K 1 J 2 J m + K 2 J 1 J m + K 1 J 1J 2 + K 2 J 1J 2 ) s 2 + K 1 K 2 ( J m + J 1 + J 2 ) K2 ( J1 s + K1 )
2 K 1 (J 2 s + K 2 )
2 = s 2 (e) Torque equations: d 2 m dt
2 = K1 Jm ( m  1 )  K2 Jm ( m  2 ) + 1 Jm T d 1
2 dt 2 = K1 J1 ( m  1 )  B1 d 1 J1 dt d 2 2 dt
2 = K2 J2 ( m  1 )  B 2 d 2 J 2 dt State diagram: 20 State variables: State equations:
dx 1 dt dx dt K1 J
1 x 1 = m  1, x 2 = d dt 1 , x 3 = d dt m , x 4 = m 2 , x 5 = d dt 2 . = x 2 + x 3 2 = x 1  B1 J
1 x dx 3
2 dt = K1 J
m x 1  K J 2 x 4 + 1 J
m T dx dt 4 = x3  x5 dx dt 5 = K2 J
2 x 4  B2 J
2 x 5 m Transfer functions: 1 ( s ) T ( s) =
2 K1 J 2 s + B2 s + K 2
2 ( )
3 2 2 ( s ) T (s ) ( s )
4 m 1 =
2 K 2 J1 s + B s + K1 1
2 ( )
2 2 1 2 1 2 m ( s )
2 1 m 1 ( s ) = s { J 1 J2 Jm s + J + [( B1K 2 (B + B ) s + [ ( K J + K J ) J + ( K + K ) J J + B B J ] s + B K ) J + B K J + B K J ] s + K K ( J + J + J )}
1 2 1 m 1 2 2 2 1 1 1 2 m 1 2 44 System equations: Tm ( t) = Jm d m
2 dt 2 + Bm
e
o d m dt
E 20
L + K ( m  L ) K ( m  L ) = J L d L
2 dt 2 + Bp d L dt Output equation: State diagram: = Transfer function: L (s ) Tm ( s) Eo ( s ) Tm ( s) = = s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3 2 ( ) K ( ) s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3 2 ( ) KE / 2 0 ( ) 45 (a) Tm ( t) = Jm N1 N3 N2N4 d 1
2 dt 1 2 + T1 T1 = N3 N4 N1 N2 T2
2 T3 = d 3 dt
2 N3 N4 T4 T4 = J L d 1
2 d 3
2 dt 2 T2 = T3 2 = N1 N2 1 3 = T2 = T4 = N3 N4 JL 2 N1 N3 d 21 Tm = J m 2 + T4 = J m + JL 2 dt N 2N 4 N 2 N 4 dt N1 N 3 21 (b) Tm = Jm 2 = N1 N2 d 1
2 dt 1 2 + T1 3 = T2 = J2 N 1N 3 N2N4 1 d 2
2 dt 2 + T3
2 T4 = ( J3 + J L ) d 2 dt
2 d 3
2 dt 2 T1 = + N3 N4 N1 N2 T2 d 3
2 T3 = N3 N4 T4 T2 = J 2 + N3 N4 T4 = J 2 d 2
2 dt 2 (J + JL ) 3 dt 2 2 d 3 d 2 2 N3 Tm ( t) = Jm + J 2 dt 2 + N ( J3 + J 4 ) dt 2 = Jm 2 dt N2 4 d 1
2 N1 2 2 d 21 N1 N1 N3 + J2 + N N ( J3 + J L ) dt 2 N2 2 4 46 (a) Force equations: f ( t) = K h ( y1  y2 ) + Bh (b) State variables:
x dy1  dy 2 dt dt x
2 K h ( y1  y2 ) + Bh 2 2 dy1  dy 2 = M d y 2 + B dy 2 t 2 dt dt dt dt 1 = y 1  y2, = dy dt State equations:
dx dt
1 = K B h h x 1 + 1 B
h f (t ) dx dt 2 = B t x M 2 + 1 M f (t ) 47 (a)
T
m = Jm d 2 m
2 dt
2 +T1
2 T 2 =JL d 2 L
2 dt +TL T 1 = N N 1 T 2 = nT 2 m N 1 = L N 2 nTm  n TL
2 2 Tm = Jm L n d m dt
2 + nJL d L dt
2 + nTL = J m + nJ + nT L L L n 2 Thus, L =
2 Jm + n JL
2 2 Set = 0. (T m  2 nTL ) J m + n J L  2nJL nTm  n J L = 0 ( ) ( ) Or, n + J T
m L n J T  J J m L =0 L m Optimal gear ratio: n = J T
m L 2J T + J mTL 2 2 + 4 J m J LT m
L m 2 where the + sign has been chosen. L m 2J T (b) When T L = 0 , the optimal gear ratio is
n = Jm / J L 48 (a) Torque equation about the motor shaft: Relation between linear and rotational displacements: 22 T m = Jm d 2 m
2 dt + Mr 2 d 2 m
2 dt + Bm d dt m y = r m (b) Taking the Laplace transform of the equations in part (a), with zero initial conditions, we have Tm ( s) = Jm + Mr
Transfer function: ( 2 )s
2 m ( s) + Bm s m (s ) Y ( s) = r m ( s) Y ( s) Tm ( s) = s J m + Mr ( r
r )s + B m 49 (a) Tm = Jm
2 d m
2 dt
2 2 + r ( T1  T2 ) Thus, Tm = J m T1 = K2 r m  r p = K 2 ( r m  y ) d m
2 ( ) T2 = K1 ( y  r m ) d y dt
2 2 T1  T2 = M d y dt dt 2 + r ( K1 + K2 )( r m  y ) M = ( K1 + K2 )( r m  y ) (c) State equations: dx1 = rx3  x2 dt (d) Transfer function: Y ( s) Tm ( s) dx2 dt = K1 + K 2 M x1 dx3 dt =  r ( K1 + K 2 ) Jm x1 + 1 Jm Tm = s 2 Jm Ms + ( K1 + K2 ) ( Jm + rM ) 2 2 r ( K1 + K 2 ) (e) Characteristic equation:
2 s J m Ms + ( K1 + K2 ) ( Jm + rM ) = 0 410 (a) Torque equations: Tm ( t) = Jm d m
2 dt 2 + Bm d m dt + K ( m  L ) K ( m  L ) = J L d L
2 dt 2 + BL d L dt State diagram: 23 (b) Transfer functions: L ( s) Tm ( s) K ( s) m ( s) Tm ( s) J L s2 + BL s + K (s ) = = ( s) = s J mJ L s3 + ( Bm J L + BL J m ) s 2 + ( KJ m + KJ L + Bm BL ) s + Bm K (c) Characteristic equation:
(d) Steady state performance: ( s ) = 0
T (t )
m = T m = consta nt. T (s)
m
2 = T m . s lim m ( t) = lim s m ( s) = lim
t s 0 s 0 J m J L s + ( Bm J L + BL J m ) s + ( KJ m + KJ L + Bm BL ) s + Bm K
3 2 J L s + BL s + K = 1 Bm Thus, in the steady state, m = L. m
and (e) The steadystate values of L do not depend on J m and J .
L 411 (a) State equations: d L =L
dt d dt L = K J 2 m  K J 2 L
1 d dt t = t K1 Jm t + d dt t = K J 1 m  Tm K J 1 t L L d m dt = m d m dt = Bm Jm m  (K + K2 ) Jm t t m + K2 Jm L + 1 Jm (b) State diagram: (c) Transfer functions: 24 L ( s ) Tm ( s) = K 2 J t s + K1
2 ( ) t(s ) T m( s) ( s) = K1 J L s + K 2
2 ( )
4 m (s ) Tm ( s ) ( s)
5 = J t J L s + ( K1 J L + K 2 J t ) s + K1 K 2
4 2 ( s)
3 ( s ) = s [ J mJ Ls + B mJ LJ t s + ( K1 J L J t + K 2J L J t + K 1Jm J L + K 2J m J t ) s + Bm J L ( K1 + K 2 ) s + K1 K2 ( J L + J t + J m ) s + BmK 1K 2 ] = 0
2 (d) Characteristic equation: ( s ) = 0 . 412 (a) m (s ) TL ( s)
Thus,
r = 0 K 1H i (s ) K 1 H i (s ) 1 + K1 H e ( s ) + R + L s B + Js H e ( s ) + R + L s B + Js =0 a a a a = 1 ( s) ( s) H e (s ) =  H i (s ) Ra + La s H i (s) H e (s) K1 K i ( s) + =  ( Ra + L a s ) (b) m (s ) r ( s) ( s ) = 1 + K1 H e ( s ) + = 1+ m (s ) r ( s)
TL =0 TL =0 = (R a + La s ) ( B + Js ) (R a + La s )( B + Js ) + K1 K b K1 H i ( s) R a + La s + (R K1 K i K b H e( s)
a + La s ) ( ( B + Js ) (R
a a + La s )( B + Js ) K1 K b (R a + La s ) ( B + Js ) 1 Kb H e ( s) K 1 Ki = (R + La s ) ( B + Js ) + Ki Kb + K1 Ki Kb H e ( s) K1 K i 413 (a) Torque equation: (About the center of gravity C) 2 d J = T s d 2 sin + F d d 1 2
dt Thus, J d
2 F d
a 1 = J 1 =
J d
2 K F d 1 sin 2 dt = T s d 2 + K F d 1 2 dt  K F d 1 = T s d 2 (b) (c) Js 2 ( s )  K F d 1 ( s ) = T s d 2 ( s ) d 2 With C and P interchanged, the torque equation about C is: Ts ( d1 + d 2 ) + F d 2 = J
2 dt 2 Ts ( d 1 + d 2 ) + K F d 2 = J ( s ) (s ) = d 2 Js (s )  K F d 2 ( s ) = Ts ( d1 + d 2 ) ( s ) Ts ( d1 + d 2 ) Js  K F d 2
2 dt 2 25 414 (a) Causeandeffe ct equations: dia dt
2 e = r o Tm = K ii a nKL Jm e = K s e e a = Ke = Ra La ia + 1 La (e
+ a  eb ) Tm  d m dt
2 2 = = Bm d m J m dt 1 J ( n m  o ) T2 = Tm n 2 = n m d o dt
2 KL JL ( 2  o )
x
1 State variables: State equations:
dx dt dx dt
1 = o,
K J x 2 = o ,
nK J x 3 =m, x 4 = m, x 5 = ia = x2
nK J dx dt 2 =
2 L L x + 1 L L x dx
3 3 dt =
dx x 4 4 = L x 1  n KL J
m x 3  Bm J
m x 4 + Ki J
m x 5 5 m dt = KK L
a s x 1  Kb L
a x 4  Ra L
a x 5 + KK L
a s r (b) State diagram: (c) Forwardpath transfer function: o ( s) KK s K i nK L = 4 3 2 2 e ( s ) s J mJ L La s + J L ( Ra J m + Bm J m + Bm La ) s + n K LL a J L + K L J m La + Bm R a J L s +
( ) (n R K J
2 a L L + R a KL J m + B mK L L a s + K i K bK L + R a B mK L KK s K i nK L
4 ) Closedloop transfer function: o ( s) r ( s ) = J mJ LL a s + J
5 L (R J
a m + Bm J m + B m La ) s + n K LL a J L + K L J m L a + B mR a J L s +
2 3 ( ) (n R K J
2 a L L + Ra KL J m + Bm KL L a s + ( K i K b K L + Ra BmK L ) s + nKK s K i K L
2 ) (d) K L = , o = 2 = n m . J L is re flecte d to m otor s ide so J T = J m +n 2 J .
L State equations: 26 d dt m = B J m T m + K J i T i d
a m dt = m di dt a = R L a a i a + KK L
a s r  KK L
a s n m  K L b a m State diagram: Forwardpath transfer function: o ( s) e ( s ) o ( s) r ( s )
From part (c), when K
L = KKs K i n s J T L a s + ( Ra J T + Bm L a ) s + Ra Bm + K i K b 2 Closedloop transfer function: = JT L a s + ( R a J T + Bm La ) s + ( Ra Bm + Ki Kb ) s + KK s K i n
3 2 KK sK in = , all t he ter ms wit hout K L in o (s ) / e ( s ) and o ( s ) / r ( s ) can b e negl ected. The same results as above are obtained. 415 (a) System equations: dv dt ea = Ra ia + ( La + Las ) dia dt di s dt 0 = Rsis + ( L s + L as ) di s dt di a dt f = K ii a = M T (b) + B Tv  Las + eb  L as Take the Laplace transform on both sides of the last three equations, with zero initial conditions, we have Ki I a ( s ) = ( MT s + BT ) V ( s ) Ea ( s ) = [ Ra + ( La + Las ) s ] I a ( s)  Las sI s ( s ) + K b V( s) 0 =  Las sI a ( s) + [ Rs + s ( Ls + Las ) ] I s ( s )
Rearranging these equations, we get V (s ) = Ia (s ) = Ki M T s + BT 1 I a (s ) Y (s ) = V (s ) s = s ( M T s + BT ) Ki Ia (s) L as s Ra + ( La + L as ) s [E a ( s ) + Las sI s ( s )  KbV ( s ) ] Is (s ) = Ra + ( L a + L as ) s I a ( s) Block diagram: 27 (c) Transfer function: Y ( s) E a (s ) K i [R s + ( L s + L as ) s ] = s [R a + ( L a + L as ) s ][ R s + ( L s + L as ) s ]( M T s + BT ) + K i K b [R s + ( L a + L as ) s]  Lass
2 2 (M T s + BT ) 416 (a) Causeandeffect equations: ea  eb Ra eb = Kb m e = r  L Tm = K i ia e = K s e d m dt = 1 Jm
K K s = 1 V/rad Tm  Bm Jm  KL Jm ea = Ke d L dt
15 . 5 1000 ia = = KL JL ( m  L ) ( m  L ) b = 15 . 5 V / KRPM = 2 / 60 = 0 .148 V / rad / sec State equations: d L dt = L d L dt = KL JL m  KL JL L d m dt = m d m dt = Bm Jm m  KL Jm L+ 1 Ki Jm Ra ( KK s e  K b m ) (b) State diagram: (c) Forwardpath transfer function: G ( s) = K i K Ks KL s J mJ L Ra s + ( Bm Ra + K i K b ) J L s + R a K L ( J L + J 3 2 m )s + K L ( Bm Ra + K i K b ) J m Ra J L = 0 .03 1 .15 0 . 05 = 0 . 001725 Bm Ra J L = 10 1 .15 0 .05 = 0 . 575 Ki K bJ L = 21 0 .148 0 . 05 = 0 .1554 28 R K J
a L L = 1.15 50000 0 . 05 = 2875 R K J
a L K L ( Bm Ra + K i K b ) = 50000(10 1.15 + 21 0.148) = 730400 G ( s) = m = 1 .15 50000 0 . 03 = 1725 K KK K
i s L = 21 1 50000 K = 105000 0K s s + 423.42 s + 2.6667 10 s + 4.2342 10
3 2 6 ( 608.7 10 K
6 8 ) (d) Closedloop transfer function: L ( s) r ( s) G( s) 1 + G ( s)
M (s) M (s ) = = = K i K Ks K L J mJ LR a s + ( B m R a + K i K b ) J L s + R a K L ( J L + J m ) s2 + K L ( Bm Ra + K iK b ) s + K i K Ks K L
4 3 = 6 . 087 s
4 10
2 8 K
8 + 423 .42 s 3 + 2 .6667 10
K s 6 s + 4.2342 10 s + 6 . 087 10 = 5476
405 8 K Characteristic equation roots:
K s s s =1 = 2738
s j 1273 . 5 s K =  1.45 =  159
. 88 j 1614. 6 = j 1000 = 211 . 7 = j1223 .4 s = 617 .22 j 1275 =  131 . 05 417 (a) Block diagram: (b) Transfer function: TAO ( s ) Tr ( s ) = ( 1 + s ) (1 + s ) + K
c s KM KR =
m 3.51 20 s + 12 s + 4.51
2 KR 419 (a) Block diagram: 29 (b) Transfer function: ( s ) ( s) = Js + ( JK L + B ) s + K 2 B + K3 K 4 e
2
 D s K1 K 4 e  D s  Ds (c) Characteristic equation:
2 Js + ( JK L + B ) s + K 2 B + K 3 K 4e ( s ) ( s) K1 K 4 ( 2  D s ) ( s) =0 (d) Transfer function: Characteristic equation: ( s ) J D s + ( 2 J + JK 2 D + B D ) s + ( 2 JK 2 + 2B  D K 2 B  D K 3 K4 ) s + 2 ( K 2 B + K 3 K 4 ) = 0
3 2 419 (a) Transfer function: G ( s) = (b) Block diagram: Ec ( s) E (s ) = 1 + ( R1 + R 2 ) Cs 1 + R2C s (c) Forwardpath transfer function: m ( s) E (s ) (d) Closedloop transfer function: m ( s) Fr ( s ) (e) = = [1 + ( R 1 + R 2 ) Cs ] ( K b Ki + Ra JL s ) K K (1 + R2C s ) K (1 + R2C s ) [1 + ( R
= E (s ) 1 Gc ( s) = E c ( s) (1 + R C s )
2 + R2 ) Cs] ( K b K i + Ra J L s ) + K KK e N (1 + R2C s ) RCs 1 Forwardpath transfer function: m ( s) E (s )
Closedloop transfer function: = RCs ( Kb Ki + Ra JL s ) 1 K (1 + R2C s ) 30 m ( s) Fr ( s )
Ke = K K (1 + R2C s ) R1C s (K b K i + Ra J L s ) + K KKe N (1 + R2C s )
/ 2 pulse s / rad = 36 = 120 =
NK pulse s / rev pulse s / sec
e = 36 = 5 . 73 pul = 200( ses / rad. 2 / 60 ) rad / sec (f) f f r m =
pulse s / sec 200 RPM m = 120 = N ( 36 / 2 ) 200( 2 / 60 ) 120 = 120 N pulse s / sec Thus, N = 1. For m = 1800 RPM, = N ( 36 / 2 ) 1800( 2 / 60 ) = 1080 N. Thus, N = 9. 420 (a) Differential equations: d m  d L dt dt dt dt 2 d d L = J d L + B d L + T K ( m  L ) + B m  L 2 L L dt dt dt dt Ki ia = Jm d m
2 2 + Bm d m + K ( m  L ) + B (b) Take the Laplace transform of the differential equations with zero initial conditions, we get Ki I a ( s ) = J m s + Bm s + Bs + K m ( s ) + ( Bs + K ) L ( s )
2 ( ) ( Bs + K ) Solving for m ( s )  ( Bs + K ) L ( s ) = J L s + BL s L (s ) + TL ( s)
2 ( ) m ( s ) and L ( s ) from the last two equations, we have m (s ) = L (s ) =
Signal flow graph: J m s + ( Bm + B ) s + K
2 Ki I a (s ) + m (s )  Bs + K J m s + ( Bm + B ) s + K
2 L (s ) Bs + K J L s + ( BL + B ) s + K
2 J L s + ( BL + B ) s + K
2 TL ( s ) (c) Transfer matrix: 2 1 K i J L s + ( BL + B ) s + K m ( s ) = (s ) (s ) Ki ( Bs + K ) L o Ia ( s ) 2 Jm s + (B m + B ) s + K TL ( s ) Bs + K 31 o ( s ) = J L Jm s + [ J L
3 (B m + B) + J m ( BL + B ) ] s + [ BL Bm + ( BL + BM
3 ) B + (J m +J L ) K]s 2 + K ( BL + B ) s 421 (a) Nonlinear differential equations:
dx ( t ) dt With R
a = v(t ) e(t ) dv ( t ) dt = k ( v )  g ( x ) + =
K i (t )
f f f (t ) =  Bv (t ) +
a f (t ) = 0, (t ) = K v(t )
b
2 =K f i (t )
f = K f a i (t ) Then, i (t ) = e(t 0 K K
b
2 f v (t ) f (t ) = K ( t ) ia ( t ) = i
a K e (t )
i K 2 b K f v (t ) 2 . Thus, dt dv ( t ) =  Bv (t ) + K K K
b
2 i f v (t ) 2 e (t ) (b) State equations: i ( t ) as input. dx ( t ) dt = v (t ) dv ( t ) dt =  Bv (t ) + K K
i f a i (t ) 2 (c) State equati ons: ( t ) as input.
f (t ) = K iK f a i (t ) 2 ia ( t ) = if (t ) =
K K ( t )
K
i f f dx ( t ) dt = v (t ) dv ( t ) dt =  Bv (t ) + 2 (t ) 422 (a) Force and torque equations:
Broom: vertical direction:
f d
b v
2  M bg =
x (t ) M d
b 2 ( L cos dt ) horizontal direction: rotation about CG: f x = M + L sin 2 dt J d
b
2 2 dt =
2 f L sin
y  f L cos
x 2 Car: horizontal direction: u(t ) = f x + Mc
x
2 d x (t )
2 J x (b) State equations:
Eliminating f
x Define the state variables as x 1 = , = dt d dt b = M L
b 3 x
4 , 3 = x , and = dx . dt and f y from the equations above, and sin x 1 x and
1 cos x
2 1 1. dx1 dt dx3 dt (c) Linearization: = x2 = x4 dx2 dt dx4 dt = = (M c L [ 4 (M b + M c ) / 3  M b ]
2 + M b ) gx1  M b Lx2 x1  u ( t ) u ( t) + M b Lx2 x1  3 M b gx1 / 4 (M b + M c )  3M b / 4 32 f1 x1 f2 x1 f2 x4 f 3 x 1 f4 x1 f4 x3 =0 = f1 x2
b c =1
2 2 f1 x3
b =0 f2 x2 = f1 x 4 =0 2 M b x1 x2 f1 u =0 =0 f2 x3 =0 (M + M ) g  M x = 0 L ( M + M  3M / 4)
b c b L ( M b + M c  3M b / 4 ) =0 =0 = f2 u f 3 x 2
2 = 1 L [ 4 (M b + M c ) / 3  M b ] f 3 x 3 =0 f4 x2 = f 3 x 4 =0 2 M b Lx1 x2 f 3 u =0 =0 (M M b Lx2  3 M b g / 4
b + M c )  3M b / 4 f4 x4 =0 (M
= b + M c )  3M b / 4 1 + M c )  3M b / 4 =0 f4 u (M b Linearized state equations: 0 3( M + M ) g & x 1 b c x L ( M + 4 M ) &2 b c = & x 3 0 &  3M b g x 4 M + 4M b c 0 x 3 1 0 0 0 x 2 L ( M b + 4 M c ) + u 0 0 1 x 3 0 x 4 4 0 0 0 M + 4M b c 1 0 0 423 (a) Differential equations:
e(t ) L( y) = L y =
Ri ( t ) + i ( t ) dL ( y ) dy ( t ) dy di ( t ) At equili brium, dt dy
eq = Ri ( t ) + d L ( y ) i( t ) dt Ki
2 2 dt + L di ( t ) y dt = Ri ( t ) d
2  L y
2 i( t ) dy ( t ) dt + L di ( t ) y dt My ( t ) = Mg 
i (t ) y (t ) = 0,
K Mg dy ( t ) dt = 0, y (t )
2 dt =0 Thus, eq = E eq R x dt =0
x y eq = = E eq R dy . dt x (b) Define the state variables as 1 = i,
x 2 =
E y , and x
eq 3 Then, x 1 eq = E eq R 2 eq = K Mg R 3 eq =0 The differential equations are written in state equation form: dx dt
1 = R L x x
1 2 + x x
1 3 x + x 2 e 2 L = f dx
1 2 dt = x 3 = f dx
2 3 dt = g  K x1 M x 2 2 2 = f 3 (c) Linearization: 33 f 1 x 1 = R L x 2 eq + x x 3 eq 2 eq = E eq K Mg f 1 x 2 f 2 x 1 f 3 x 2 L = R L x 1 eq  x x
1 3 x 2 2 + E eq L =0 f 1 x 3 = x x 1 eq 2 eq = Mg K f 1 e f 3 x 1 = x 2 eq L = 1 L K Mg E eq R 2 Rg E
eq =0 = f 2 x 2
2 K x 1 eq M x
3 2 eq 2 =0 =
2 Rg E
eq f 2 x 3
Mg K =1 f 2 e f 3 e =0 = 2 K x 1 eq M x
2 2 eq = =0 The linearized state equations about the equilibrium point are written as: Eeq K  L Mg A = 0 2 Rg  E eq 424 (a) Differential equations:
M1 d
2 0 0 2 Rg Eeq Mg K Mg K 0 0 & x = A x + B e Eeq K RL Mg B = 0 0 y1 (t )
2 dt d
2 = M 1g B dy 1 ( t ) dt dy 
(t ) Ki 2 2 (t ) y1 ( t ) + Ki
Ki
2 2 (t ) y2 (t ) 1  y1 ( t ) 2 M y 2 (t )
2 2 dt = M 2g 2 B
dy dt 2 dt
1  = (t )
2 y2 (t )  y1 (t ) y ,
2 Define the state variables as x 1 = y ,
1 x 2 = , x 3 x 4 = dy dt 2 . The state equations are: dx1 dt = x2 M1 dx2 dt = M 1 g  Bx2 
dx
1 Ki x1 2 2 + Ki 2 2 dx3 dt
4 (x
dx dt
3 3  x1 ) = 0,
2 2 = x4 M2 =0 dx4 dt = M 2 g  Bx4  = 0. Ki 2 2 (x 3  x1 ) At equilibrium, dt = 0, dx dt 2 = 0,
2 dx dt = 0. Thus , x 2 eq and x
2 4 eq M 1g 
Solving for I, with X KI X1 2 + KI (X 3  X 1) =0 M 2g  KI (X 3  X1) 2 =0 1 = 1 , we have
1/2 M + M2 Y2 = X 3 = 1 + 1 M2 (b) Nonlinear state equations: ( M1 + M 2) g I= K 1/2 34 dx1 dt = x2 dx2 dt = g B M1 x2  K M 1x1
2 i +
2 Ki 2 dx3
2 M 1 ( x 3  x1 ) dt = x4 dx4 dt = g B M2 x4  Ki 2 M 2 ( x3  x1 ) 2 (c) Linearization: f1 =0 x1 f2 x1 = 2 KI
2 3 f1 x2 =0
2 3 f1 x 3 =0 = B M1 f1 x 4 =0 = f1 i 2 KI
2 =0 f2
3 M 1 x1 + 2 KI f2 x2 f2 x3 M1 ( X 3  X 1 ) M1 ( X3  X1 ) f3 x4 x4 =0 f2 i = 2 KI 1 1 2+ 2 M 1 X1 ( X 3  X 1 ) 2 KI
2 3 f3 x1 =0 f3 x2 =0 f3 x3 = 0 = 1 f3 i = 0 f4 x1 = f4 x2 M2 ( X3  X1 ) =0 f4 x3 = 2 KI 2 3 f4 x4 M 2 ( X 3  X1 )
M = B M2 = 1. =1 f4 i = 2 KI M2 ( X3  X1)
2 Linearized state equations:
1/2 M 1 = 2, 2 = 1, g = 32 .2, B = 0 .1, K I = 32.2(1 + 2) X = 96.6 X = 9.8285X 1 1 1 1 X1 = 1 9.8285 X 3 = 1 + 1 + 2 X 1 = 2.732 X 1 = Y2 = 2.732 0 2 1 2 KI 1 + 3 3 M1 X1 ( X 3  X 1 ) A = 0 2 2 KI 3 M 2 ( X 3  X1 ) ( ) X 3  X 1 = 1.732 1 B M1 0 0 0 2 KI
2 3 M 1 ( X 3  X1 ) 0 2 KI
2 M 2 ( X 3  X 1) 3 1 0 0 115.2 0.05 = 1 0 0  37.18 0 B M2 0 18.59 0 0 1 37.18  0.1
0 0 0 1 0 2 KI 1 + M 1 X 12 ( X 3  X 1 ) 2  6.552 = B = 0 0  6.552 2 KI 2 M 2 ( X 3  X 1) 425 (a) System equations: 35 Tm = K i ia = ( J m + JL )
T
D d m dt + B m m
e ea = Ra ia + La
b dia dt + K b m
a y = n m
c y = y ( t  TD ) = d (sec) V = r b = Ksy E ( s) = KG ( s ) E ( s ) Block diagram: (b) Forwardpath transfer function: Y (s ) E (s ) = s {( Ra + La s ) [( Jm + J L ) s + Bm ] + K b Ki }
 TD s  TD s K Kin G c ( s ) e  TD s Closedloop transfer function: Y (s ) R (s ) = s ( Ra + La s ) [( Jm + J L ) s + Bm ] + K bK i s + KG c ( s )K i ne K Kin G c ( s ) e 36 Chapter 5 STATE VARIABLE ANALYSIS OF LINEAR DYNAMIC SYSTEMS
51 (a) State variables:
State equations:
x
1 = y, x 2 = dy dt Output equation: dx1 dt 0 1 x1 0 = + r dx2 1 4 x2 5 dt (b) State variables:
State equations:
x
1 y = [1 0] x1 = x1 x2 = y, x 2 = dy , dt x 3 = d 2 y
2 dt Output equation: dx1 dt 1 0 x1 0 dx2 = 0 0 1 x2 + dt dx 1 2.5 1.5 x3 3 dt (c) State variables:
State equations: 0 0 r 0.5 dx1 dt dy dt x1 y = [1 0 0 ] x2 = x1 x3 d y dt
2 2 x1 = t 0 y ( ) d , x2 = , x3 = , x4 = & x1 Output equaton: x & 2 = & x3 x &4 0 0 0 1 0 x 0 2+ r 0 0 0 1 x3 0  1  1  3 5 x 1 4 x
1 1 0 0 x1 0 x1 x 2 y = [1 0 0 0] = x1 x3 x 4
2 (d) State variables:
State equations: = y, x 2 = dy , dt x 3 = d y
2 , x dt 4 = d y dt
3 3 & x1 Output equation: x & 2 = & x3 x &4 52 0 0 0 1 0 x2 0 + r 0 0 1 x3 0 0 1 2.5 0 1.5 x 1 4 1 0 0 x1 0 x1 x 2 y = [1 0 0 0] = x1 x3 x 4 We shall first show that ( s ) = ( sI  A ) =
1 We multiply both sides of the equation by + 2 + +L 2 s s 2! s ( sI  A ) , and we get I = I. Taking the inverse Laplace transform I A 1 A 2 37 on both sides of the equation gives the desired relationship for ( t ) . 53 (a) Characteristic equation:
Eigenvalues:
s ( s ) =
j 1. 323 , sI  A = s2 + s + 2 = 0
 0 .5 
j 1. 323 = 0 . 5 + State transition matrix: ( t) = cos1.323t + 0.378sin1.323t 1.512sin1.323 t
sI  A 0.5t e 1.069sin (1.323t  69.3 ) 0.756sin1.323t
o (b) Characteristic equation:
State transition matrix: ( s ) = = s 2 + 5s + 4 = 0
t Eigenvalues: s =  4, 1 ( t) = (c) Characteristic equation:
State transition matrix: 1.333 e t  0.333e4 t 1.333 e  1.333 e
2 t 4 t 0.333e + 1.333e 0.333 e  0.333e
t 4 t 4 t ( s ) = ( s + 3) = 0 Eigenvalues: s = 3,  3 e 3 t ( t) = 0
(d) Characteristic equation:
State transition matrix: e 0
3 t ( s ) = s 2  9 = 0 Eigenvalues: s = 3 , 3 e3 t ( t) = 0
(e) Characteristic equation:
State transition matrix:
2 e 0
3 t ( s ) = s + 4 = 0 Eigenvalues: s = j2,  j2 ( t) = (f) Characteristic equation:
State transition matrix: cos2 t  sin2 t 3 2 cos2t s s i n 2t =  1,  2 , 2 ( s ) = s + 5 s + 8 s + 4 = 0 Eigenvalues: e t ( t) = 0 0 (g) Characteristic equation: ( s ) =
s
3 0 e
2 t 0
s
2 te 2t e 0
2 t + 15
5 t + 75 s + 125 = 0
5 t 5 t Eigenvalues: s =  5,  5, 5 e ( t) = 0 0 te e 0 te 5 t e 0
5 t 54 State transition equation: x (t ) = (t )x( t ) + (a) (t  )Br ( )d t 0 (t ) for each part is given in Problem 53. 38 t 0 ( t  ) Br ( ) d = L 1 1 s + 1 1 0 1 1 ( sI  A) 1 BR( s ) = L1 ( s ) 2 s 1 0 1 s s+2 2 s(s + s + 2) 1 + 0.378sin1.323 t  cos1.323t 1 = =L t0 1 + 1.134sin1.323 t + cos1.323t s2 s s2 + s + 2 ) (
1 (b) ( t  ) Br ( ) d = L ( sI  A) t 1 0 1 BR ( s ) = L 1 1 1 s + 5 1 1 1 4 s 1 1 s ( s) s+6 1.5 1.67 0.167 s( s + 1)( s + 2) 1 s  s + 1 + s + 4 1.5  1.67 e t + 0.167e 4 t 1 =L =L = s4 1 1.67 0.667 1 + 1.67 e t  0.667 e  4t +  s( s + 1)( s + 4) s s +1 s + 4 (c) t0 1 t 1 1 1 s + 3 0 ( t  ) Br ( ) d = L ( sI  A) BR( s) = L 0 0 0 1 = =L 1 t 0.333 (1  e 3t ) s ( s + 3) (d) 0 1 1 1 s s + 3
0 0 1 t 1 1 1 s  3 ( t  ) Br ( ) d = L ( sI  A) BR( s) = L 0 0 0 = =L 1 s ( s + 3) 1 0 1 1 1 s s + 3
0 0 0.333 (1  e 3t ) t 0 (e) 39 1 2 t 1 1 1 s + 4 ( t  ) Br ( ) d = L ( sI  A) BR( s) = L 0 2 s 2 + 4 2 s 2 1 =L = t 0 1 0.5sin2t ( s2 + 4 ) (f) 0 1 s 1 s 2 s + 4
2 1 s + 1 t 1 1 1 0 ( t  ) Br ( ) d = L ( sI  A) BR( s) = L 0 0 0 0 1 1 0.5 (1  e 2 t ) = =L t0 s( s + 2) 0 0 (g) 0 1 s+ 2 0 0 1 1 1 2 (s + 2 ) s 0 1 s +2 0 1 s + 5 t 1 1 1 0 ( t  ) Br ( ) d = L ( sI  A) BR (s ) = L 0 0 1 ( s + 5)
1 s +5 0 2 0 1 1 0 2 ( s + 5) s 1 1 s+5 0 0 0 0 1 1 0.04 0.04 0.2 1 5 t 5 t =L =L 2 s  s + 5  ( s + 5 ) 2 = 0.04  0.04 e  0.2 te u s ( t) 5t s ( s + 5) 0.2  0.2 e 1 0.2 0.2  s ( s + 5) s s +5 55 (a) (b)
Not a state transition matrix, since Not a state transition matrix, since ( 0 ) I ( 0 ) I (identity matrix). (identity matrix). (c) ( t ) is a state transition matrix, since ( 0 ) = I and 40 1 [ ( t ) ] =  t 1  e
1 1 = t 1  et e 0
2 t 1 0 e
t = (  t) (d) ( t ) is a state transition matrix, since ( 0 ) = I , and  te e
2t e2 t [ ( t) ]1 = 0 0 56 (a) (1) Eigenvalues of A:
2 .325 , t e / 2
2 2t  te e 2t 0 2t = (  t)  0 .3376 
j 0 . 5623  0 . 3376 + j 0 . 5623 , (2) Transfer function relation: s 1 1 X( s) = ( sI  A ) B U ( s) = 0 (s ) 1 1 s 2
3 1 s + 3 2 0 0 1 s 2 + 3s + 2 0 U ( s) = 1 1 (s ) s 1 1 0 U ( s) = 1 s U (s ) s ( s + 3) s (s ) 2 2 2 s  1 s 1 s s +3 1 0 ( s ) = s + 3 s + 2 s + 1 (3) Output transfer function: 1 1 s = = C ( s ) ( sI  A) B = [1 0 0 ] 3 2 U (s ) ( s ) 2 s + 3 s + 2 s + 1 s Y ( s)
1 1 (b) (1) Eigenvalues of A:  1,  1. (2) Transfer function relation: 1 ( s + 1) 2 1 0 1 s + 1 1 U (s ) X ( s ) = ( sI  A) BU ( s ) = U (s ) = (s ) 0 s + 1 1 1 ( s + 1) (3) Output transfer function: ( s ) = s + 2s + 1
2 1 ( s + 1) 2 Y ( s) 1 1 s+2 1 = = C ( s ) ( sI  A) B = [1 1] + = 2 2 U (s ) 1 ( s + 1) s + 1 ( s + 1) s +1 (c) (1) Eigenvalues of A:
0,  1,  1. (2) Transfer function relation: 1 1 X( s) = ( sI  A) BU ( s ) = (s ) s + 2s = 1
2 0 0 1 0 U ( s) = 1 s U ( s) s ( s + 2) ) s (s ) s s 1 s 2 2 s+2 1 0 ( s ) = s s + 2s + 1
2 ( ) 41 (3) Output transfer function: 1 s +1 1 = C ( s ) ( sI  A) B = [1 1 0 ] s = = 2 s ( s + 1 )2 s ( s + 1 ) U (s ) s Y ( s)
1 57 dy We write dt = dx dt 1 + dx dt 2 = x 2 + x3 d 2 y dt = dx dt 2 + dx dt 3 = x1  2 x2  2 x3 + u dx1 dt d x dy = = dt dt 2 d y dt2 0 1 0 x1 0 0 1 1 x + 0 u 2 1 2 2 x3 1 1 0 0 x =  1 1 0 x 1  1 1 0 0 u 1 (1) x1 1 0 0 x = y = 1 1 0 x & y 0 1 1 Substitute Eq. (2) into Eq. (1), we have (2) dx 1 1 0 = A 1 x + B1u = 0 0 1 x = dt 1 0 2 58 (a)
s sI 2
s 0 0 s  A = 1
1 2
0 = s  3 s + 2 = s + a 2 s + a1 s + a0
3 2 3 2 a 0 = 2, a 1 = 0, a 2 = 3 1 a1 M = a2 1 a2 1 0 1 0 0 0 3 1 =  3 1 0 1 0 0 S = B AB 0 2 4 A B = 1 2 6 1 1  1 2 2 2 0 P = SM = 0 1 1 4 2 1 42 (b)
s sI 2
s 0 0 s  A = 1
1 1 1 a2 1 0 = s 3  3 s + 2 = s + a 2 s + a1 s + a 0
2 3 2 a 0 = 2, a 1 = 0, a 2 = 3 1 1 0 0 a1 M = a2 1 0 3 1 =  3 1 0 1 0 0 1 2 6 S = B AB A B = 1 3 8 0 0 1 0 1 1 P = SM = 1 0 1 1 0 0 2 (c)
s sI +2
0 1 s 1 +2
2 s 0 0 A = = s + 7 s + 16 s + 12 = s + a 2 s + a 1 s + a 0
3 2 3 2 a 0 = 12 , a 1 = 16 , a 2 =7 +3 a1 M = a2 1 a2 1 0 1 0 0 16 7 1 = 7 1 0 1 0 0 1 1 0 S = B AB A B = 1 2 4 1 6 23 9 6 1 P = SM = 6 5 1 3 1 1 2 (d)
s sI +1
0 0 s 1 =1
0 s 0 A = 1 = s + 3 s + 3 s + 1 =
3 2 s 3 + a 2 s + a1 s + a0
2 a 0 = 1, a 1 = 3, a 2 =3 +1 a1 M = a2 1 a2 1 0 1 0 0 3 3 1 = 3 1 0 1 0 0 S = B 2 1 P = SM = 2 3 1 2 (e)
sI s 0 1 1 AB A B = 1 0  1 1 1 1 0 1 1 2 A = 1
2 s 1 +3 = s 2 + 2 s 1 = s + a1 s + a 0
2 a 0 =  1, a 1 =2 43 M= a1 1 1 2 1 0 1 0 = P = SM = S = [B AB ] = 0 1 1 3 1 0 1 1 59 (a) From Problem 58(a), 0 3 1 M =  3 1 0 1 0 0 Then, C V = CA = 2 CA (b)
From Problem 58(b), 1 0 1  1 2 1 1 2 1 0.5 1 3 Q = (MV ) = 0.5 1.5 4 0.5 1 2 1 C V = CA = 2 CA (c)
From Problem 58(c), 16 7 1 M = 7 1 0 1 0 0 1 0 1 1  1 3 1 Q = (MV ) = 2 5 1 0.2308 0.3077 1.0769 0.1538 0.5385 1.3846 0.2308 0.3077 0.0769 C V = CA = 2 CA (d)
From Problem 58(d), 1 0 0  2 1 0 4 4 0 Since V is singular, the OCF transformation cannot be conducted. 3 3 1 M = 3 1 0 1 0 0 Then, C V = CA = 2 CA (e)
From Problem 58(e), 1 0 1  1 1 1 1 2 2 1 1 0 Q = ( MV ) = 0 1  2 1 1 1 1 44 M= 2 1 1 0 1, 2.7321, Then, V= C CA 2 = 1 0 1 1 Q = ( MV ) =
1 0 1 1 3 510 (a) Eigenvalues of A: 0.7321 T = [p1
where p , p , and
1 2 p2 0 0.5591 0.8255 p3 ] = 0 0.7637 0.3022 1 0.3228 0.4766 3 p are the eigenvectors. (b) Eigenvalues of A: 1, 2.7321, 0.7321 T = [ p1
where p , p , and
1 2 p2 0 0.5861 0.7546 p3 ] = 0 0.8007 0.2762 1 0.1239 0.5952 p are the eigenvalues.
3 (c) (d) Eigenvalues of A: Eigenvalues of A: 3, 2, 2. A nonsingular DF transformation matrix T cannot be found. 1, 1, 1 The matrix A is already in Jordan canonical form. Thus, the DF transformation matrix T is the identity matrix I. (e) Eigenvalues of A: 0.4142, 2.4142 T = [p2 p2 ] = 0.8629 0.2811  0.5054 0.9597 S is singular. 511 (a) S = [B (b) S = B (c) S = [B (d) AB] = AB] = 1 2 0 0 1 1 1 2 AB A B = 2 2 2 3 3 3 2 2 2 +2 2 2+ 2 S is singular. S is singular. 45 S = B 512 (a) 1 2 4 AB A B = 0 0 0 1 4 14 2 2 S is singular. Rewrite the differential equations as: d m
2 dt 2 = B d m J dt
2  K J
m m +
x Ki J = ia
d dt
m dia dt
, x
3 = Kb d m La dt  Ra La ia + K a Ks La ( r m ) State variables: x = ,
1 2 = ia
Output equation: State equations: dx1 dt 0 dx2 =  K dt J dx K K 3  a s dt La 1   B J Kb La 0 x1 Ki x2 + 0 r J x K K Ra 3 a s  La La 0 y = 1 0 0 x = x1 (b) Forwardpath transfer function: s m ( s ) K G ( s) = = [1 0 0 ] J E (s ) 0 3 1 s+ B J Kb La
2 0 Ki  J Ra s+ La 1 0 KiK a 0 = K o ( s) a La o ( s ) = J La s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + KRa = 0 Closedloop transfer function:
1 s m ( s ) K M ( s) = = [1 0 0 ] J r ( s ) KaKs La =
3 2 1 s+ B J Kb La K i Ka K s 0 Ki  J R s+ a La 0 K s G( s ) 0 = K K 1 + K s (s ) a s La JLa s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + K i K a K s + KRa 513 (a) 46 A= 0 1 1 0 A =
2  1 0 0 1 A =
3 0  1 1 0 A =
4 1 0 0 1 (1) Infinite series expansion:
3 5 t t t2 t4 1  2 ! + 4 !  L t  3! + 5!  L 1 2 2 ( t) = I + A + A t + L = t 3 5 2 4 2! t + t  t + L 1  t + t L 3! 5 ! 2! 4 ! cos t =  sin t sin t cos t (2) Inverse Laplace transform: 1 s 1 s 1 ( s ) = ( sI  A ) = 1 s = s 2 + 1 1 s 1 1 (t ) = cos t  sin t sin t cos t (b) A= 1 0 0 2 A =
2 1 0 0 4 A =
3 1 0 0  8 A =
4 1 0 0 16 (1) Infinite series expansion:
2 3 4 t t t 0 1  t + 2 !  3! + 4 ! + L 1 2 2 ( t) = I + A + A t + L = t 2 3 2! 4t 8t 0 1  2t +  +L 2! 3! e t = 0 e 0
2 t (2) Inverse Laplace transform: ( s ) = ( sI  A ) =
1 s + 1 0 1 s +1 = s + 2 0 0
1 1 s + 2
0 (t ) = e t 0 2 t e 0 (c) A= 0 1 1 0 A =
2 1 0 0 1 A =
3 0 1 1 0 A =
4 1 0 0 1 (1) Infinite series expansion:
3 5 t t t2 t4 1 + 2 ! + 4 ! + L t + 3! + 5 ! L e t + et 1 2 2 ( t) = I + A + A t + L = t = t t 3 5 2 4 2! t + t + t L 1 + t + t + L e + e 3! 5 ! 2 ! 4! e + e
t t e +e t t (2) Inverse Laplace transform: 1 s 1 s 1 1 ( s ) = ( sI  A ) = 1 s = s 2  1 1 s = 1 0.5  s +1 0.5 + s +1 s  1 s + 1 s  1 0.5 0.5 0.5 + s 1 s +1 s 1 0.5  0.5 + 0.5 47 e t + et ( t) = 0.5  t t e + e
514 (a) e = K s ( r  y ia =
Solve for i e + e
t t e +e t t ) ea = e  es eb = K b
d
y es = Rs ia eu = Kea d y
2 eu  eb Ra + Rs y
and d y dt Tm = K i ia = ( J m + J L ) dt 2 a in terms of , we have dt ia =
Differential equation: KKs (r  y )  Kb Rs + Rs + KRs d y dt d y
2 dt 2 = K ii a Jm + J L
1 =
x d y  KK s y + KK s y K b ( J m + J L ) (R a + R s + KRs ) dt Ki
2 State variables: x = ,
y = d dt y State equations: dx1 0 1 0 dt x1 + =  KK s K i  K b Ki  KK s Ki r x2 dx2 ( J + J ) (R + R + KR ) ( J + J ) (R + R + KR ) ( J + J ) (R + R + KR ) m L a s s m L a s s s dt m L a s 1 x1 0 0 = + r 322.58 80.65 x2 322.58 We can let v ( t ) = 322 .58 r, then the state equations are in the form of CCF. (b) ( sI  A ) =
1 1 1 s s + 80.65 1 = 2 322.58 s + 80.65 s + 80.65 s + 322.58 322.58 s 0.014 0.014 0.06  1.059 s + 76.42 s + 4.22 s + 76.42 + s + 4.22 = 1.0622 0.0587 4.468  4.468 s + 76.42 s + 4.22 s + 76.42  s + 4.22 s 1 For a unitstep function input, u ( t ) =1 / s. 322.2 s ( s + 76.42)( s + 4.22) 1 = ( sI  A)1 B = s 322.2 s ( s + 76.42)( s + 4.22) 1 + 0.0584  1.058 s s + 76.42 s + 4.22  4.479 + 4.479 s + 76.42 s + 4.22 48 76.42t 4.22t 0.014e + 0.01e 0.06e 76.42 t  1.059e 4.22 t x (t ) = x (0) 76.42 t 4.22t 76.42 t 4.22 t  4.468e 1.0622 e  0.0587 e 4.468 e 1 + 0.0584 e 76.42 t  1.058e 4.22 t = t0 76.42 t 4.22t + 4.479e 4.479 e (c) Characteristic equation: (d) ( s ) = s 2 + 80 . 65 s + 322 . 58 From the state equations we see that whenever there is to increase the effective value of Ra there is ( 1 + K ) Rs . =0 Thus, the purpose of R is
s Ra by (1 + K ) Rs . This improves the time constant of the system. 515 (a) State equations: dx1 0 1 0 dt x1 +  KKs Ki  Kb Ki KKs Ki = r x2 dx2 J ( R + R + KR ) J ( R + R + KR ) J ( R + R + KR ) s s s s s s dt 1 x1 0 0 = + r 818.18 90.91 x2 818.18
.18 Let v = 818 r. The equations are in the form of CCF with v as the input.
1 1 (b) s 1 1 s + 90.91 ( sI  A ) = 818.18 s + 90.91 = ( s + 10.128 ) ( s + 80.782) 818.18 10.128t 80.78t 10.128t 80.78t  0.142 e 0.01415 e  0.0141e 1.143 e x1 (0) x (t ) = 10.128 t 80.78 t 10.128 t 80.78t + 0.1433e  0.1433 e + 1.143e  11.58 e x2 (0) 11.58 e10.128 t  11.58 e80.78t + t0 10.128 t 80.78t + 0.1433 e 1  1.1434 e ( s ) = s + 90 . 91 s + 818
2 1 s (c) Characteristic equati on:
Eigenvalues: . 18 =0  10.128, 80.782 (d) Same remark as in part (d) of Problem 514. 516 (a) Forwardpath transfer function: Y ( s) 5 ( K1 + K 2s ) G ( s) = = E ( s ) s [s ( s + 4)( s + 5) + 10 ] M ( s) = Y (s ) R (s ) = G (s) 1 + G(s ) =
4 3 Closedloop transfer function: s + 9 s + 20 s + (10 + 5 K 2 ) s + 5 K1
2 5 ( K1 + K 2s ) (b) State diagram by direct decomposition: 49 & x1 State equations: Output equation: x & 2 = & x3 x &4 0 0 0 5 K 1 0 0  (10 + 5 K 2 ) x1 0 1 0 x2 0 + r 0 1 x3 0 20 9 x4 1 0 0
R( s) y = 5 K1 5K2 0 x (c) Final value: r(t ) = u s ( t ), = 1 . s lim y( t) = lim sY ( s ) = lim
t s 0 s 0 s + 9 s + 20 s + ( 10 + 5 K 2 ) s + 5 K1
4 3 2 5 ( K1 + K 2s ) =1 517 In CCF form, 0 0 A= M 0  a0 1 0 M 0  a1 0 1 M 0  a2 0 0 M 0  a3 1 s M 0 a1 L L O L L 0 1 M 0 a2
n 1 0 M 1  an 1 0 0 0 M 0 a3 L 0 0 B = M 0 1 0 s 0 sI  A = M 0 a 0 sI L 0 O M 0 1 L s + an n2 A = s n + a n 1 s + a n2 s + L + a1 s + a 0 Since B has only one nonzero element which is in the last row, only the last column of going to contribute to the last row of adj (s I  A ) is adj (s I  A ) B . The last column of adj (s I  A )
is 1 is obtained from the cofactors of s s
2 ( sI  A ) . Thus, the last column of adj (s I  A ) B L s n 1 ' . 50 518 (a) State variables: x 1 = y, x 2 = dy , dt x 3 = d 2 y
2 dt State equ ations: & x ( t ) = Ax ( t ) + B r ( t ) 0 1 0 A= 0 0 1 1 3 3 (b) State transition matrix:
1 0 B = 0 1 s2 + 3 s + 3 s + 3 1 s 1 0 1 1 2 ( s ) = ( sI  A ) = 0 s 1 = s + 3s s 1 (s ) 2 s 1 3 s + 3 3 s  1 s 1 1 1 1 2 1 + + + 2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3 s + 1 ( s + 1) ( s + 1) 1 1 1 2 s = +  2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3 2 s 3 2 s + ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3 (1 + t + t 2 / 2 ) e t 2 t ( t) = t e / 2 ( t + t 2 / 2 ) e t 3 2 (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1) 3 (t + t ) e (1 + t  t ) e
2 t 2 t t e 2 t (t  t / 2 ) e (1  2t + t 2 / 2 ) e t t e /2
2 2 t t (d) Characteristic equation: ( s ) = s + 3 s + 3 s + 1 = 0
Eigenvalues:  1, 1, 1
x
1 519 (a) State variables: = y, x 2 = dy dt State equations: dx1 (t ) dt 0 1 x1 ( t) 0 = + r ( t) dx2 ( t)  1  2 x2 ( t) 1 dt State transition matrix: s+ 2 ( s + 1) 2 s 1 1 ( s ) = ( sI  A ) = 1 s + 2 =  1 ( s + 1) 2 1 ( s + 1) s ( s + 1 )2 1
2 (t ) = (1 + t ) et  te
t (1  t ) e te
t t 51 Characteristic equation: (s ) = ( s +1) = 0
2 (b) State variables:
State equations:
dx dt
1 x 1 = y, x 2 = y + dy dt = dy dt = x 2 y= x 2  x1 dx dt 2 = d 2 y
2 dt + dy dt = y  dy dt +r = x2 +r dx1 dt 1 2 x1 0 = + r dx2 0 1 x2 1 dt State transition matrix: 1 s + 1 2 s + 1 (s ) = = 0 s + 1 0 1 2 ( s + 1) 1 s +1 2
2 (t ) = e t 0  te e t t (c) Characteristic equation: (s ) = ( s +1) = 0 which is the same as in part (a). 520 (a) State transition matrix: s  sI  A =  s  ( t) = L (b) Eigenvalues of A:
1 ( sI  A ) 1 = s  ( s) 1 s   ( s ) = s  2 + + 2 2 ( 2 ) cos t  sin t t ( sI  A )1 = sin t cos t e + j ,  j 521 (a)
Y (s)
1 U ( s)
1 = = s 1+ s
1 3 2 + 2s
s +3s + 3s 3 = = 1 s
3 +s +2s+3
2 Y ( s)
2 3 2 3 1 s
3 U (s)
2 1 +s 1 +2s + s + 2s +3
2 = Y ( s)
1 U (s)
1 1 1 (b) State equations [Fig. 521(a)]: & x = A x+B u
1 1 1 Output equation: y = C x 0 1 0 A1 = 0 0 1 3 2 1 State equations [Fig. 521(b)]: 0 B1 = 0 1 & x = A x+B u
2 2 2 C1 = [1 0 0] = C 2x Output equation: y 2 0 0 3 A 2 = 1 0 2 0 1 1 1 B2 = 0 0 C2 = [0 0 1] 52 Thus, A 2 = A1
' 522 (a) State diagram: (b) State diagram: 523 (a)
G (s) X (s) = Y (s) U ( s) = 10 s 1 + 8 .5 s
1 1 3 2 X (s) + 20 . 5 s + 15 s 3 Y ( s) X (s) X (s)
3 = 10 X (s) = U ( s )  8 .5 s X ( s)  20 . 5 s 2 X ( s )  15 s State diagram: State equation: & x ( t ) = Ax ( t ) + B u ( t ) 53 1 0 0 0 A= 0 1 15 20.5 8.5 (b)
G (s) Y ( s) Y (s) U ( s) s
3 0 B = 0 1 s
4 2 A and B are in CCF = = 10 s 3 + 20
1 4 X (s) X (s) X ( s) 1 + 4. 5 s + 3 .5 s
X (s) = 10 X ( s) + 20 s = 4. 5 s 1 X ( s )  3 .5 s 2 X ( s) +U ( s) State diagram: State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 A= 0 0 (c)
G (s) 0 1 0 0 0 1 0 3.5 4.5 1 0 0
Y (s) U ( s) 0 0 B= 0 1 =
5s
2 A and B are in CCF = = 5( s s(s + 1) + 5s 3 X (s)
2 + 2 )( s + 10 )
3 1 + 12 s 1 + 20 s X (s) Y (s) = 5s 2 X (s) + 5s X (s) X (s) = U ( s )  12 s 1 X (s)  20 s 2 X ( s) State equations: & x ( t ) = Ax ( t ) + B u ( t ) 54 0 0 1 0 0 A= 1 0 20 12 (d) G ( s) =
Y ( s) 0 B = 0 1 1
2 A and B are in CCF Y (s ) U (s )
s
4 = s ( s + 5) s + 2s + 2
X (s) ( ) = s
1 4 2 3 X (s ) X (s )
s
3 2 1 + 7 s + 12 s + 10 s
1 = X ( s) = U ( s)  7s X ( s )  12 s X (s)  10 State diagram: State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 0 1 0 0 1 0 A= 0 0 0 1 0 10 12 7 524 (a)
G (s) 0 0 B= 0 1 A and B are in CCF = Y (s) U ( s) = 10 s
3 + 8 . 5 s + 20 . 5 s + 15
2 = 5 . 71 s + 15  6 . 67 s +2 + 0 . 952 s +5 55 State equations: & x ( t ) = Ax ( t ) + B u ( t ) 1.5 0 0 A= 0 2 0 0 0 5 (b)
G (s) 5.71 B = 6.67 0.952 The matrix B is not unique. It depends on how the input and the output branches are allocated. = Y (s) U ( s) = 10( s s (s
2 + 2) = 1 )( s + 3 . 5) = 4. 5
s +
s 0 .49 + 3 .5 +
s 4 +1 + 5 . 71 s
2 State diagram: State equation: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 1 0 0 0 0 0 A= 0 0 1 0 0 0 0 3.5 0 1 B= 1 1 +
0 . 313 s (b)
G (s) = Y ( s) U (s) = 5( s s( s + 1) + 2 )( s + 10 ) = 2 .5 s +2  0 . 563 s + 10 State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 0 0 2 0 A= 0 0 10 (d) 1 B = 1 1 56 G ( s) = Y (s ) U (s ) = s ( s + 5) s + 2s + 2
2 ( 1 ) = 0.1 s  0.0118 s+ 5  0.0882s + 0.235 s + 2s + 2
2 State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 0 2 0 0 0 5 A= 0 0 0 0 525 (a)
G (s) 0 1 2 0 1 1 B= 0 1 = Y (s) U ( s) = 10 (s + 1. 5)( s + 2 )( s + 5) State diagram: State equations: & x ( t ) = Ax ( t ) + B u ( t ) 5 1 A = 0 2 0 0 (b) G ( s) = Y (s ) U (s )
State diagram: 1 1.5 0 10(s + 2) 0 B= 0 10 = = 10 s + 2 1 2 s ( s + 1)( s + 3.5) s s + 1 s + 3.5 2 57 State equations: & x ( t ) = Ax ( t ) + B u ( t ) 0 0 1 0 0 0 1 1 A= 0 0 1 1 0 0 0 3.5 (c) G ( s) = Y (s ) U (s )
State diagram: 0 0 B= 0 10 = = 5 s + 1 1 s( s + 2)( s + 10) s s + 2 s + 10 59 s + 1) State equations: & x ( t ) = Ax ( t ) + B u ( t ) 1 0 0 0  10 1 A= 0 0 2 (d) G ( s) =
State diagram: 0 B = 0 5 Y (s) U (s ) = s ( s + 5) s + 2s + 2
2 ( 1 ) 58 State equations: 0 1 0 0 A= 0 2 0 0 526 (a)
G (s) & x ( t ) = Ax ( t ) + B u ( t ) 0 1 2 0 0 1 5 0 0 0 B= 0 1 = Y (s) E (s) = 10 s(s + 4 )( s + 5) = 10 s 1+ 9 s
1 3 X (s) s
2 + 20 X (s) (b) Dynamic equations: & x1 x = & 2 x3 & 1 0 0 0 10 20 x1 0 1 x2 + 0 r 9 x3 1 0 y = [10 0 0] x (c) State transition equation: s 1 (1 + 9s 1 + 20 s 2 ) s 2 (1 + 9 s 1 ) s 3 x1 (0) s 3 X 1 (s ) 3 1 1 2 X (s ) = 1 1 2 1  10 s s (1 + 9 s ) s 2 x2 (0) + ( s ) s s (s ) 2 2 1 s 1 X 3 (s )  10 s  20 s s x3 (0) 1 2 s s + 9 s + 20 s+9 1 x1 (0) 1 1 = s( s + 9) s x2 (0) + 1  10 c (s ) c (s ) 2 s  10 s  10 ( 2 s + 1 ) s x3 (0) ( s ) = 1 + 9 s
1 + 20 s 2 + 10 s 3 c( s) = s 3 + 9 s + 20 s + 10
2 59 1.612 0.946 0.114 0.706 1.117 0.169 e 0.708t + 1.692 x(t ) = 1.14 0.669 0.081 2.678 4.056 e 2.397t 0.807 0.474 0.057 4.056 6.420 0.972 0.0935 0.171 0.055 + 0.551 1.009  0.325 e 5.895t x(0) 3.249 5.947 1.915 0.1  0.161e0.708t + 0.0706 e2.397t  0.00935e5.895 t 0.708 t  2.397 t 5.895 t +  0.169e + 0.055e 0.114 e 0.087 e0.708t + 0.406 e 2.397t  0.325e 5.895t (d) Output: t 0 y ( t ) = 10 x1 (t ) = 10 1.612e ( 0.708t  0.706e 2.397t + 0.0935e 5.897 t + 10 1.141 e ( 0.708 t  0.169 e 2.397 t + 0.0550 e 5.895 t ) x (0) + 10 ( 0.946e  1.117e + 0.1711e ) x (0) ) x (0) + 1 1.61e 0.708t + 0.706e 2.397 t  0.0935e 5.895t
0.708 t 2.397t 5.895t 1 2 3 t0 527 (a) Closedloop transfer function:
Y (s) R( s) =
s 10
3 + 9 s + 20 s + 10
2 (b) State diagram: (c) State equations: & x1 x = & 2 x3 & 1 0 0 0 10 20 (d) State transition equations:
[Same answers as Problem 526(d)] x1 0 1 x2 + 0 r 9 x3 1 0 (e) Output: [Same answer as Problem 526(e)] 528 (a) State diagram: (b) State equations: 60 & x1 x & 2 = & x3 x &4 (c) Transfer function relations:
From the system block diagram, 2 20 1 0 x1 0 1 0 10 1 0 x2 0 0 u + 0.1 0 20 1 x3 0 0 T D 0 x 30 0 0 0  5 4 Y ( s) = 0.2s 0.3 30 e U ( s) 90U( s) 1 TD ( s) + TD ( s) + + s+2 ( s) ( s + 2)( s + 20) ( s + 2)( s + 5)( s + 20) (s + 5)( s + 20) 1 ( s ) = 1 +
Y ( s) 0 . 1e (s 0 . 2 s + 2 )( s + 20
0 . 2 s =
) T
D (s + 2 )( s + 20 ) + 0 .1 e
(s  0. 2 s + 2 )( s + 20
30 e
0 . 2 s ) s =
(s  ( s + 19 . 7 ) + 2 )( s + 20 ) + 0 .1e ( s + 20
(s )
0 . 2 s (s)+ (s + 90(
0 . 2 s + 2 )U ( s )
0 . 2 s + 5) (s + 2 )( s + 20 ) + 0 . 1e
30 e U ( s) ( s ) = + 2 )( s + 20 ) + 0 .1e T (s)
D +
(s + 5) (s + 2 )( s + 20 ) + 0 .1e 0 . 2 s 529 (a) There should not be any incoming branches to a state variable node other than the s should create a new node as shown in the following state diagram. 1 branch. Thus, we (b) State equations:
dx dt
1 Notice that there is a loop with gain 1 after all the s 1 2 x dx
2 2 1 branches are deleted, so = 2. y = 17 2 x 1 + dt = 15 2 x 1  1 2 x 2 + 1 2 r Output equation: = 6 .5 x 1 + 0 .5 x 2 530 (a) Transfer function: Y (s ) R (s ) = Ks + 5 s + 1
2 (b) Characteristic equation:
2 ( s + 1) s + 11s + 2 ( )
s s
3 2 ( s + 1) ( s 2 + 11s + 2 = 0 ) Roots of characteristic equation: 1, 0.185, 10.82. These are not functions of K. (c) When K = 1: Y (s) R( s) = + 5s +1
s
2 + 12 + 13 s + 2 State diagram: 61 (d) When K = 4: Y (s ) R (s )
State diagram: = ( s + 1) ( s 4s + 5s + 1
2 2 + 11s + 2 ) ( s + 1) (s = (s + 1)(4 s + 1)
2 + 11s + 2 ) = 4s + 1 s + 11s + 2
2 (e)
Y (s) R( s) = Ks (s 2 + 5s =1
s + 1 )( s + 0 .185)( + 10 .82 When K = 4, 2.1914, 0.4536, polezero cancellation occurs. ) 531 (a) Gp ( s) = Y ( s) U (s ) = ( 1 + 0.5s ) (1 + 0.2s + 0.02s 1 2 ) = 100 s + 12 s + 70 s + 100
3 2 State diagram by direct decomposition: State equations: & x1 x = & 2 x3 & 0 0 100 x1 0 0 1 x2 + 0 u 70 12 x3 1 1 0
Roots of characteristic equation: (b) Characteristic equation of closedloop system: 62 s 3 + 12 s 2 + 70 s + 200 = 0  5. 88 ,  3 . 06 + j 4. 965 ,  3 . 06  j 4. 965 532 (a) Gp ( s) = Y ( s) U (s ) (1 + 0.5s ) (1 + 0.133s + 0.0067s 1  0.066 s 2 ) = 20( s  15) s + 22 s + 190 s + 300
3 2 State diagram by direct decomposition: State equations: & x1 x = & 2 x3 & 0 0 300 x1 0 0 1 x2 + 0 190 22 x3 1 1 0
Roots of characteristic equation: Characteristic equation of closedloop system:
s
3 + 22 s 2 + 170
x s + 600 = 0 = m
and x 12, 5 + j5, 5 j5 = D
K KK JR d dt K J K J 533 (a) State variables:
d dt 1 2 State equations:
m = K K
b i + K bRa
a JR + D J D + i a e D = D R m  D R D (b) State diagram: 63 (c) Openloop transfer function: m ( s) E (s ) KK i ( J R s + K D ) = JJ RR a s + ( K b J R Ki + K DR a J R + K D JRa ) s + K DK b Ki
2 Closedloop transfer function: m ( s) r ( s) = JJ R Ra s + ( Kb J R Ki + KD R a J R + KD JRa + Ks K Ki J R ) s + KD K b Ki + Ks K Ki KD
2 Ks KKi ( J R s + KD ) (d) Characteristic equation of closedloop system: ( s ) = JJ R R a s + ( K D J R K i + K D R a J R + K D JRa + K s KK i J R ) s + K D K b K i + K s K Ki KD = 0
2 ( s ) =
Characteristic equation roots: s 2 + 1037 s + 20131 .2 =0 19.8, 1017.2 & 534 (a) State equations: x ( t ) = Ax ( t ) + B r ( t ) A= b d 2 1 c  a = 2  1 B= 0 1 S = [B AB ] = 0 1 1 1 Since S is nonsingular, the system is controllable. (b) S = [B 535 (a) S = B (b) AB AB ] = 0 d 1  a 1 1 1 A B = 1 1 1 1 1 1 2 The system is controllable for d 0. S is singular. The system is uncontrollable. 64 S = B 1  1 1 AB A B = 1  2 4 1 3 9 2 S is nonsingular. The system is controllable. & 536 (a) State equations: x ( t ) = Ax ( t ) + B u ( t ) A= 2 3 1 0 B= 1 1 1 0 x S = [B AB ] = 1 1 1 1 0 S is singular. The system is uncontrollable. Output equation: y = = Cx C = 1 V = C ' AC ' ' 1 2 = 0 3 Y (s) U ( s) V is nonsingular. The system is observable. (b) Transfer function: =
s s
2 +3 +2s3 =
s 1 1 Since there is polezero cancellation in the inputoutput transfer function, the system is either uncontrollable or unobservable or both. In this case, the state variables are already defined, and the system is uncontrollable as found out in part (a). 537 (a) = 1, (b) 2 , or 4 . These values of will cause polezero cancellation in the transfer function. 2 + 2) 4
6( s The transfer function is expanded by partial fraction expansion, Y (s) 1 R( s) = 3( s + 1)  2( s + +4)
output equation: y ( t ) & By parallel decomposition, the state equations are: x ( t ) = Ax ( t ) + B r ( t ) , = C x ( t ). 1 6 1 0 0 A = 0 2 0 0 0 4 The system is uncontrollable for 1 B =  2  4 D= 1  1 3 2 = 1, or = 2, or = 4. (c) Define the state variables so that 1 0 0 A = 0 2 0 0 0  4 1 3 1 B= 2 1 6 D = [  1 2  4] The system is unobservable for = 1, or = 2, or = 4. 65 538 S = [B AB] = b 1 b ab  1 S = ab  1  b 0
2 The boundary of the region of controllability is described by ab  1  b = 0.
2 Regions of controllability: 539 S = [B AB] = b1 b1 + b 2 b b2 2 2 S = 0 when b1 b2  b1 b2  b2 = 0,or b2 = 0
2 The system is completely controllable when b 0. V = 0 when d 1 0 . V = C ' AC ' ' d2 d1 = d d +d 2 1 2
2 The system is completely observable when d 0. 540 (a) State equations: dh 1 K nN K = ( qi  qo ) = I m  o h dt A A A
State variable: State equations:
x
1 d m dt
3 = d m
m = Ki Kb JRa dt m + Ki K a JRa ei J = Jm +n 2 J L = h, x 2 =m, x = d dt m = m & x = Ax + B e i K o A A= 0 0 State diagram: K I nN A 0 0 1 = KK  i b JRa 0 0 1 0.016 0 0 1 0 0 11.767 0 B = 0 = K K i a JRa 0 0 8333.33 66 (b) Characteristic equation of A: s+ sI  A = Ko A  K I nN A s 0
0, 0 1 s+ Ki K b JRa = s s + 0 0 Ko s + Ki K b = s ( s +1)( s + 11.767) A JRa Eigenvalues of A: 1, 11.767. (c) Controllability: S = B (d) Observability: (1) C =
1 0 0 AB 0 133.33 0 0 A B = 8333.33  98058 8333.33 98058 1153848 2 S 0. The system is controllable. :
' ' V = C ' AC 1 1 1 0 0.016 0.016 (A ) C = 0 0 0.016 ' 2 ' V is nonsingular. The system is observable. (2) C = 0 1 0 :
' ' V = C (3) C =
0 0 ' AC 0 0 0 1 0 (A ) C = 0 0 1 11.767 ' 2 ' V is singular. The system is unobservable. 1 :
' ' V = C ' AC 0 0 0 0 (A ) C = 0 0 1 11.767 138.46 ' 2 ' V is singular. The system is unobservable. 541 (a) Characteristic equation: ( s ) = s I  A = s 4  25 . 92 s 2 =0 Roots of characteristic equation: 5.0912, 5.0912, 0, 0 (b) Controllability: 67 S = B AB A B 2  0.0732 0  1.8973 0 0.0732 0 1.8973 0 3 A B = 0 0.0976 0 0.1728 0.0976 0 0.1728 0 B S is nonsingular. Thus, A , is controllable. (c) Observability: (1)
C = 1 0 0 0 V = C ' A C ' ' (A ) C ' 2 ' 1 0 ' 3 ' (A ) C = 0 0 0 1 0 0 25.92 0 0 0 25.92 0 0 0 S is singular. The system is unobservable. (2) C = 0 1 0 0 V = C ' A C ' ' (A ) C ' 2 ' 0 671.85 0 25.92 1 0 25.92 0 ' 3 ' (A ) C = 0 0 0 0 0 0 0 0 S is singular. The system is unobservable. (3) C = 0 0 1 0 V = C ' A C ' ' (A ) C ' 2 ' 0 0 ' 3 ' (A ) C = 1 0 0 0 0 1  2.36 0 0 0  2.36 0 0 0 S is nonsingular. The system is observable. (4) C = 0 0 0 1 68 V = C ' A C ' ' (A ) C ' 2 ' 0 61.17 0 2.36 0 0 2.36 0 ' 3 ' (A ) C = 0 0 0 0 1 0 0 0 S is singular. The system is unobservable. 542 The controllability matrix is 0  384 0 1 0 16 1 0 16 0 384 0 0 0 0 16 0 512 S= 0 512 0 0 0 16 0 1 0 0 0 0 1 0 0 0 0 0 543 (a) Transfer function: v ( s ) R (s ) = J vs
2 Rank of S is 6. The system is controllable. (J KI H
G s + K Ps + K I + K N
2 ) State diagram by direct decomposition: State equations: & x ( t ) = Ax ( t ) + B r ( t ) 1 0 1 0  ( KI + KN JG 0 0 A = 0 0 0 0 0 )
2 (b) Characteristic equation: Jvs 2 (J 0 1  KP JG 0 0 0 B= 0 1 G s + K P s + K I + KN = 0 ) 69 & 544 (a) State equations: x ( t ) = Ax ( t ) + B u 1 ( t ) A= 3 1 0  2 B= 0 1 S = [B AB ] = 0 1 1 2 S is nonsingular. A , B is controllable. C = 1
1 Output equation: y 2 = C x V = C (b)
With feedback, u 2 ' AC ' ' =  kc 2 , & the state equation is: x ( t ) 1 3 = 1 3 =
Ax ( t ) V is singular. The system is unobservable. + B u1 ( t ) .
1 0 1+k S= 1 2 3  2 k A = 1+ g 0 State diagram: 1+ k 2 1 0 B= 1 S is nonsingular for all finite values of k. The system is controllable. Output equation: y2 = Cx V = D' 1 1 + k 1 1 + K ' ' AD= 1 1+ k C= 1+ k 1 (1 + k ) 3 + 2k  2 (1 + k ) 3 + 2k
2 V is singular for any k. The system with feedback is unobservable. 545 (a) S = [B V = C (b) u =  k 1
k2 x
' AB ] = 1 2 2 7 1  1 ' ' AC = 1 2 S is nonsingular. System is controllable. V is nonsingular. System is observable. A c = A  BK = 0 1 k1  1 3 2 k1 1  k2 k1 = 2 k2  1  2 k1  3  2 k 2 k2 70 S = [B
For controllabillity, k 2 A cB ] = 
11 2
' ' 1  k1  2 k2 + 2 2  7  2k  4k 1 2 S =  11  2 k2 0 V = C For observability, V ' A cC  1  1  3 k1 = 1 2  3 k 2 = 1 + 3k 1  3 k 2 0 71 Chapter 6
61 (a)
Poles are at s STABILITY OF LINEAR CONTROL SYSTEMS
= 0 ,  1. 5 +
j 1. 6583 ,  1. 5  j 1. 6583 One poles at s = 0. Marginally stable . Two poles on j axis. Marginally stabl e . Two poles in RHP. Unstable . All poles in the LHP. Stable . Two poles in RHP. Unstable . Two poles in RHP. Unstable . (b) Poles are at s =  5,  j 2 , j 2 (c) Poles are at s =  0 .8688 , 0 .4344 + j 2 . 3593 , 0 .4344  j 2 . 3593 (d) Poles are at s =  5,  1 + j ,  1  j (e) Poles are at s =  1.3387 , 1. 6634 + j 2 . 164, 1. 6634  j 2 .164 (f) Poles are at s =  22 . 8487 j 22 . 6376 , 21 . 3487 j 22 . 6023 62 (a)
s
3 + 25
3 s 2 + 10 s + 450 = 0 Roots:  25 . 31 , 0 .1537 + j 4.214, 0 .1537  4.214 Routh Tabulation:
s s s 1 25 250 10 450 2 1  450
25 = 8 Two sign changes in the first column. Two roots in RHP.
0 s 0 450 s
2 (b) s 3 + 25
3 + 10 s + 50 = 0 Roots:  24. 6769 ,  0 .1616 + j 1.4142 ,  0 .1616  j 1.4142 Routh Tabulation:
s s s 1 25 250 10 50 2 1  50
25 =8 No sign changes in the first column. No roots in RHP.
0 s 0 50 s
2 (c) s 3 + 25
3 + 250 s + 10 = 0 Roots:  0 . 0402 ,  12 .48 + j 9 . 6566 ,  j 9 . 6566 Routh Tabulation:
s s s 1 25 6250 250 10 2 1  10 = 249 No sign changes in the first column. No roots in RHP.
.6 0 25 s
0 10 (d) 2s 4 + 10 s 3 + 5 . 5 s + 5 . 5 s + 10 = 0
2 Roots:  4.466 ,  1.116 , 0 .2888 + j 0 . 9611 , 0 .2888  j 0 . 9611 Routh Tabulation: 71 s s s s 4 2 10 55 5 .5 5 .5 10 3 2  11 = 4.4 =  75 . 8 10 1 10 24.2  100 4.4 s
0 10 Two sign changes in the first column. Two roots in RHP. (e) s 6 + 2 s + 8 s + 15
5 4 6 s 3 + 20 s 2 + 16 s + 16 = 0 Roots:  1.222 j 0 .8169 , 0 . 0447 j 1.153 , 0 .1776 j 2 .352 Routh Tabulation:
s s s 1 2 16 8 15 20 16 16 5 4  15
2 = 0 .5 40  16
2 = 12 s 3  33 396 + 24 33  541 .1 + 528
11 .27  48 = 11 .27 = 1.16
16 s 2 s 1 0 s 0 0 Four sign changes in the first column. Four roots in RHP. (f) s 4 + 2 s + 10
3 4 s 2 + 20 s + 5 = 0 Roots: 0 .29 ,  1. 788 , 0 . 039 + j 3 .105 , 0 . 039  j 3 .105 Routh Tabulation:
s s s 1 2 20 10 20 5 3 2  20
2 =0 5 s 2 20 5 Replac e 0 in last row by s 1  10  10 Two sign changes in first column. Two roots in RHP. s 0 5
3 63 (a) s 4 + 25 s + 15 s 2 + 20 s + K = 0 Routh Tabulation: 72 s s s s 4 1 25 375 15 20 K 3 2  20
25 = 14.2
K K K 1 284  25 = 20  1. 76 20  1. 76 >0 K >0 or K < 11 . 36 14.2 s
0 K K Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable. The auxiliary equation equation is A ( s ) = 14.2 s frequency of oscillation is 0.894 rad/sec.
2 + 11 . 36 = 0 . The solution of A(s) = 0 is s 2 =  0 .8 . The (b) s 4 + Ks + 2 s + ( K + 1) s + 10 = 0
3 2 Routh Tabulation:
s s s
4 1 K 2K K 2 10 K K 3 +1
10 >0 >1
2 2  K 1
K
2 = K 1
K s 1 9 K  1
K 1
9 K 1 > 0 s 0 10
2 2 The conditions for stability are: K > 0, K > 1, and  9 K  1 > 0 . Since K is always positive, the last condition cannot be met by any real value of K. Thus, the system is unstable for all values of K. (c) s 3 + ( K + 2 ) s + 2 Ks + 10 = 0
2 3 Routh Tabulation:
s s 1 K 2K 2K 10 K 2 +2
2 > 2
2 s 1 + 4 K  10
K +2 K + 2 K 5 > 0 s 0 10 The conditions for stability are: K > 2 and K + 2 K  5 > 0 or (K +3.4495)(K  1.4495) > 0, or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is
2 marginally stable. The auxiliary equation is A ( s ) The frequency of oscillation is 1.7026 rad/sec. = 3.4495 s 2 + 10 = 0 . The solution is s 2 = 2 .899 . (d) s 3 + 20 s 2 + 5 s + 10 K =0 Routh Tabulation: 73 s s s 3 1 20 100 5 10 K K 2 1  10
20 = 5  0 .5 K 5  0 .5 K > 0 >0 or K < 10 s 0 10 K K The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system is marginally stable. The auxiliary equation is A ( s ) equation is s
2 = 20 s 2 + 100 = 0 . The solution of the auxiliary = 5 .
2 The frequency of oscillation is 2.236 rad/sec. K (e) s 4 + Ks + 5 s + 10 s + 10
3 =0
10 K K Routh Tabulation:
s s s
4 1 K 5K 5 10 10 K 3 >0  10 > 0
or K 2  10
K 5K >2 50 K s
1  100
K 5K  10 K 2  10
K = 50 K  100  10
5K K 3  10 5K  10  K > 0
3 s 0 10 K K >0
The last condition is written as The conditions for stability are: K > 0, K > 2, and 5 K K + 2 . 9055 2 K 2  2 . 9055 K + 3 .4419  10  K > 0 .
3 <0. The secondorder term is positive for all values of K. Since these are contradictory, the system Thus, the conditions for stability are: K > 2 and K < is unstable for all values of K. 2.9055. (f) s 4 + 12 . 5 s + s + 5 s + K = 0
3 Routh Tabulation:
s s s
4 1 12 . 5 12 . 5 1 5 K 3 2 5 = 0 .6 = 5  20 .83 K 12 . 5 s
1 3  12 . 5 K
0 .6 K 5  20 . 83 K > 0 or K < 0 .24 s K K >0 The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The auxiliary
0 equation is A ( s ) = 0 . 6 s oscillation is 0.632 rad/sec. 2 + 0 .24 = 0 .
2 The solution of the auxiliary equation is s 2 =  0 .4. The frequency of 64 The characteristic equation is Ts 3 + ( 2T +1) s + ( 2 + K ) s + 5 K = 0 Routh Tabulation: 74 s s 3 T 2T ( 2T K +2
5K T T >0 > 1 /
2 2 +1 s 1 + 1 )( K + 2 )  5 KT
2T +1 K (1  3 T ) K + 4T + 2 > 0 s 0 5K >0 The conditions for stability are: T > 0, K > 0, and K < 4T 3T +2 1 . The regions of stability in the TversusK parameter plane is shown below. 65 (a) Characteristic equation: s 5 + 600 s 4 + 50000 s 3 + Ks 2 + 24 Ks + 80 K =0 Routh Tabulation:
s s
5 1 600 3 50000 K
7 24 K 80 K 4 s 3 10 K
00 K
7 14320 K 600 K < 3 10 7 600 s
2 214080 3 K 2 10  K
16 80 K
11 K < 214080
2 00 s 1  7 .2 10 + 3 .113256 10
600( 214080 00 K )  14400 K 2 K K  2 .162 10 7 K + 5 10 12 <0 s 0 80 K K >0 Conditions for stability: 75 From the s From the s 3 2 1 row: row: K
2 From the s row: Thus, From the s
0 < 3 10 7 K < 2 .1408 10 7 12  2 .162 10 K + 5 10 < 0
K
7 or (K  2 . 34 10
7 5 )( K  2 .1386 10 7 ) <0 2 . 34 K>0 10 5 < K < 2 .1386 10 10
5 row: Thus, the final condition for stability is: When K When K 2 . 34 < K < 2 .1386 10 7 = 2 . 34 10 7 = 2 .1386 10
5 = 10 . 6 rad/sec. = 188 . 59 rad/sec.
2 (b) Characteristic equation: s 3 + ( K + 2 ) s + 30 Ks + 200 K =0 Routh tabulation:
s s s
3 1 K 30 K 30 K 200 K K K K 2 +2
2 > 2 > 4. 6667 >0 =0.
The solution is s
2 1  140 +2 K s
0 200 K K > 4.6667 K Stability Condition: When K = 4.6667, the auxiliary equation is A ( s ) The frequency of oscillation is 11.832 rad/sec. = 6 . 6667 s 2 + 933 . 333 =  140 . (c) Characteristic equation: s 3 + 30 s 2 + 200 s +K =0 Routh tabulation:
s s s
3 1 30 6000 30 200 K 2 1 K K < 6000 >0
2 s 0 K 0 K Stabililty Condition: < K < 6000 = 30
s
2 When K = 6000, the auxiliary equation is A ( s ) The frequency of oscillation is 14.142 rad/sec. + 6000 = 0 . The solution is s =  200 . (d) Characteristic equation: s 3 + 2 s + ( K + 3) s + K + 1 = 0
2 Routh tabulation:
s s s
3 1 2 K K +3 2 K +1 K 1 +5
30 > 5 > 1 s 0 K +1 K> K Stability condition: 1. When K = 1 the zero element occurs in the first element of the 76 s row. Thus, there is no auxiliary equation. When K = 1, the system is marginally stable, and one of the three characteristic equation roots is at s = 0. There is no oscillation. The system response would increase monotonically.
0 66 State equation: Openloop system: A= 1 2 10 0 1 10  k 1
k 2 2 s + k2
or k
2 & x ( t ) = Ax ( t ) + B u ( t ) B= Closedloop system: & x ( t ) = ( A  BK ) x ( t ) 2 0 1 A  BK = Characteristic equation of the closedloop system: sI  A + BK =
Stability requirements: s 1 10 + k1
k
2 = s + ( k 2  1 ) s + 20  2 k1  k 2 = 0
2 1 > 0 >1
or k
2 20  2 k1  k 2 > 0 < 20  2 k 1 Parameter plane: 67 Characteristic equation of closedloop system: s 1 0 sI  A + BK = 0 k1
Routh Tabulation: s k2 + 4 1 s + k3 + 3 = s + ( k 3 + 3 ) s + ( k 2 + 4 ) s + k1 = 0
3 2 s s s
s 3 1 k3 + 3 k2 + 4 k1 k3 +3>0 or k3 >  3 2 1 (k 3 + 3) ( k 2 + 4 )  k 1 k3 + 3 (k
k >0
1 3 + 3 )( k + 4)  k > 0
2 1 0 k 1 Stability Requirements: k 3 >  3, 68 (a) k 1 > 0, (k
2 3 + 3) ( k 2 + 4 )  k 1 > 0 Since A is a diagonal matrix with distinct eigenvalues, the states are decoupled from each other. The second row of B is zero; thus, the second state variable, x is uncontrollable. Since the uncontrollable 77 state has the eigenvalue at 3 which is stable, and the unstable state x with the eigenvalue at 2 is
3 controllable, the system is stabilizable. (b) 69 Since the uncontrollable state x has an unstable eigenvalue at 1, the system is no stabilizable.
1 The closedloop transfer function of the sysetm is Y (s) R (s )
The characteristic equation is: = 1000 s + 15.6 s + ( 56 + 100 K t ) s + 1000
3 2 s 3 + 15 . 6 s + ( 56 + 100
2 K )s
t + 1000 = 0 Routh Tabulation:
s s s
3 1 15 . 6 873 . 6 56 + 100
1000 K t 2 1 + 1560 Kt  1000 15 .6 s
0 1560 K t  126 .4 >0 1000 K
t Stability Requirements: > 0 . 081 610 The closedloop transfer function is Y (s) R( s) The characteristic equation: = K(s s
3 + 2 )( s + ) + Ks
3 2 + ( 2 K + K  1 ) s + 2 K
2 s + Ks + ( 2 K + K  1) s + 2 K = 0 Routh Tabulation:
s s
3 1 K (2 2K + K  1
2 K K 2 >0  1  2 > 0 s 1 + ) K  K  2 K
2 (2 + )K K s
0 2 K K >0 > 0,
K Stability Requirements: > 0 ,
K versus Parameter Plane: > 1 + 2 + . 2 78 611 (a) Only the attitude sensor loop is in operation: K t = 0. The system transfer function is: K s
2 ( s) r ( s )
If KK If KK
s s = G (s)
p 1+ K G ( s)
s p =  + KK s >, , the characteristic equation roots are on the imaginary axis, and the missible will oscillate. the characteristic equaton roots are at the origin or in the righthalf plane, and the system is unstable. The missile will tumble end over end. (b) Both loops are in operation: The system transfer function is ( s) r ( s )
For stability: When K
t = G (s)
p 1 + K sG
t p (s) + K sG p ( s ) = K s
2 + KK t s + KK s  KK and KK =0
s t s > 0, >,
t KK s  > 0 . the characteristic equation roots are on the imaginary axis, and the missile will oscillate back and forth. For any KK  if KK < 0, the characteristic equation roots are in the righthalf plane, and the system If KK is unstable. The missile will tumble end over end. > 0 , and KK < , the characteristic equation roots are in the righthalf plane, and the system is
t t unstable. The missile will tumble end over end. 612 Let s 1 = s + ,
1 then when s = , s 1 = 0. This transforms the s =  axis in the splane onto the imaginary axis of the s plane. (a) F (s) Or =
s
1 s
2 2 + 5s +3 = 0 Le t s = s1  1 We get (s 1 1) + 5 ( s1 1) + 3 = 0
2 + 3 s1  1 = 0
s1
2 1 3 1 Routh Tabulation: s s 1 1 0 1 1 Since there is one sign change in the first column of the Routh tabulation, there is one root in the region to the right of s = 1 in the splane. The roots are at 3.3028 and 0.3028. (b) F (s) = s 3 + 3s + 3s +1 = 0
2 Let s = s 1 1 We get ( s1 1)3 + 3( s1 1) + 3 ( s1 1) + 1 = 0
2 79 Or s 3 1 = 0.
3 2 The three roots in the s plane are all at s
1 1 = 0. Thus, F(s) has three roots at s = 1.
2 (c) F (s) Or = s s + 4 s + 3 s + 10 = 0 + s 1  2 s1 + 10 = 0
2 Let s = s 1 1 We get (s 1 1) + 4 ( s1 1) + 3 ( s1 1) +10 = 0
3 3 1 s s 3 1 2 1 1 1 1 2
10 Routh Tabulation: s1 s
1  12
10 0 Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = 1 in the splane. The roots are at 3.8897, 0.0552 + j1.605, and 0.0552  j1.6025.
2 (d) F (s) Or = s 3 +4s +4s +4 = 0
3 Let s = s 1 1 We get (s 1 1) + 4 ( s1 1) + 4 ( s1 1) + 4 = 0
3 2 s 1 + s1  s1 + 3 = 0
2 s s 3 1 2 1 1 1 1 1
3 Routh Tabulation: s1 s
1 4
3 0 Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = 1 in the splane. The roots are at 3.1304, 0.4348 + j1.0434, and 0.4348 j1.04348. 613 (a) Block diagram: (b) Openloop transfer function: G ( s) = H ( s) E (s )
Closedloop transfer function:
H (s) R( s) = s (R a Js + K i K b ) ( As + K o ) K a K i nK I N = 16.667 N s( s + 1)( s + 11.767) =
1 G (s) +G ( s) =
s 16 .667 N
3 + 12 . 767 s 2 + 11 . 767 s + 16 . 667 N (c) Characteristic equation: 80 s 3 + 12 . 767
11 . 767 s 2 + 11 . 767 s + 16 . 667 N =0 Routh Tabulation:
s s s
3 1 12 .767 150 .22 2 16 . 667 N N 1 50.22 1  16 . 667  16.667 N >0 or N <9 12 . 767 s
0 16 .667N 0 N >0 Stability condition: < N <9. 614 (a) The closedloop transfer function:
H (s) R( s) =
s ( 0 .06 s 250 N + 0 . 706 )( As + 50 ) + 250 N The characteristic equation:
0 . 06 As
3 + ( 0 . 706 A + 3) s + 35 . 3 s + 250
2 N =0 Routh Tabulation:
s s s
3 0 . 06 A 0 . 706 A 24. 92 A 35 .3 250 N A >0 +3 > 0 + 105
.9 2 +3  15 NA +3 0 . 706 A 24.92 A 1 + 105 .9 07 . 06 A s
0  15 NA >0 250N
1 N > 0 When A From the s row, N < 1.66 + 7.06/A N max 1. 66 Thus, N max = 1. (b) For A = 50, the characteristic equation is 3 s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 250 N = 0
3 2 Routh tabulation 81 s s 3 3 35 . 3 0706 K K 250 N o 2 + 0 . 06
2 o K o >  588 . 33 s 1 0 . 0424 K o + 24. 92 K o K  750 N 35 . 3 s
0 + 0 . 06 N < 0.0000 5653 K o N 2 + 0 . 03323 K o 250 N >0 (c) N = 10, A = 50. The characteristic equation is s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 50 KI = 0
3 2 Routh Tabulution:
s s
3 1 35 . 3 0 . 706 K K 50 K Ko Ko o 2 + 0 . 06
2 o I K o > 588 . 33 s 1 0 . 04236 K o + 24. 92 + 0 . 06  50 KI 35 .3 s
0 KI < 0 . 000847 >0 2Ko 2 + 0 .498 Ko 50 K I K I 615 (a) Block diagram: (b) Characteristic equation: 2 Ms + K s + K + K = 0 D s P
500 s
2 + K D s + 500 + K P = 0 82 (c) For stability, K D > 0, 0 .5 + K P > 0. Thu s, K P >  0 .5 Stability Region: 616 State diagram: = 1 + s + Ks
Characteristic equation:
s
2 2 +s+K =0 Stability requirement: K>0 83 Chapter 7 TIMEDOMAIN ANALYSIS OF CONTROL SYSTEMS
n 2 (b) 0 . 707 n 2 71 (a) 0 . 707 rad / sec 0 rad / sec (c) 0 . 5 1 n 5 rad / sec (d) 0 .5 0 . 707 n 0 . 5 rad / sec 72 (a) 73 (a) (b) (c) (d) (e) (f) Type 0 (b) Type 0 (c) Type 1 (d) Type 2 (e) Type 3 (f) = lim = lim = lim = lim = lim = lim Type 3 K p = lim = lim = lim = lim = lim = lim s0 G ( s) = 1000 = = = = Kv = lim = lim = lim = lim = lim = lim s 0 sG ( s ) =0 =1 =K = =1 = Ka s0 s G (s) 2 =0 =0 =0 =1 =0 =
K K p s0 G ( s) Kv s 0 sG ( s ) Ka s0 s G (s) 2 K p s0 G ( s) Kv s0 sG ( s ) Ka s0 s G (s) 2 K p s0 G ( s) Kv s 0 sG ( s ) Ka s0 s G ( s) 2 K K p p s0 G ( s) Kv Kv s 0 s 0 sG ( s ) sG ( s ) Ka Ka s0 s0 s G (s) s G (s)
2 2 s0 G ( s) = 84 74 (a) Input Error Constants Steady state Error ________________________________________________________________________________
u (t )
s K K K p v a = 1000 =0 =0 1 1001 tu ( t )
s t u (t ) / 2
s 2 0 (b) Input Error Constants Steady state Error ________________________________________________________________________________
u (t )
s K K K p v a = =1 =0 tu ( t )
s 1 t u (t ) / 2
s 2 0 1/ K (c) Input Error Constants Steady state Error ________________________________________________________________________________
u (t )
s K K K p v a = =K =0 tu ( t )
s t u (t ) / 2
s 2 The above results are valid if the value of K corresponds to a stable closedloop system. (d) (e) The closedloop system is unstable. It is meaningless to conduct a steadystate error analysis. Input Error Constants Steady state Error ________________________________________________________________________________
u (t )
s K K K p v a = 0 1 tu ( t )
s =1 =0 t u (t ) / 2
s 2 0 0 1/ K (f) Input Error Constants Steady state Error ________________________________________________________________________________
u (t )
s K K K p v a = =K tu ( t )
s = t u (t ) / 2
s 2 The closedloop system is stable for all positive values of K. Thus the above results are valid. 75 (a) K H = H (0) =1 M (s) a = G (s) 1+ G ( s)H ( s) a
1 =
2, s s
3 +1
2 0 = 3, 1 = 3, a 2 = b + 2 s + 3s +3 = 1, b 1 = 1. 0 Unitstep Input: ess =
Unitramp input:
a
0 b0 KH 1  a KH 0 2 =3 Thus e
ss  b0 K H = 3  1 = 2 0.  b0 K H = 2 0
and a = .
e Unitparabolic Input:
a
0 1  b1 K H = 1 0. Thus ss = . (b) K H = H (0) =5
M (s) = G (s) 1+ G ( s)H ( s) = 1 s
2 +5s + 5 a 0 = 5, a 1 = 5, b 0 = 1, b 1 = 0. 85 Unitstep Input: ess =
Unitramp Input:
i e b0K H 1  a KH 0
1
a a
1 0 = 1 5 1  = 0 5 5
i = 0:
ss  b0 K H = 0 =
5 25 = 1: a 1  b1 K H = 5 0 =  b1 K H
a K
0 = 1 5 H Unitparabolic Input:
e
ss = (c) K H = H (0) =1/5
M (s) = G (s) 1+ G ( s)H ( s) a
0 = s s
4 3 +5
2 = 1, a 1 = 1, + 15 s + 50 s + s + 1 a = 50 , a = 15 , b = 5 , 2 3 0 The system is stable. b 1 =1 Unitstep Input: ess =
Unitramp Input:
i e b0K H 1  a KH 0
1
a a
1 0 5/5 = 5 1  1 = 0 i = 0:
ss  b0 K H = 0 =
11 / 5 1/5 = 1: a 1  b1 K H = 4 /5 0 =  b1 K H
a K
0 =4 H Unitparabolic Input:
e
ss = (d) K H = H (0) = 10
M (s) =
a G (s) 1+ G ( s)H ( s)
0 = 1 s = 10 , a 1 = 5, + 12 s + 5 s + 10 a = 12 , b = 1, 2 0
3 2 The system is stable. b 1 = 0, b 2 =0 Unitstep Input: ess =
Unitramp Input:
i e b0K H 1  a KH 0
1
a a
1 0 1 10 = 10 1  10 = 0 i = 0:
ss  b0 K H = 0 =
5 100 = 1: a 1  b1 K H = 5 0 =  b1 K H
a K
0 = 0 . 05 H Unitparabolic Input:
e 76 (a) M (s) = s s
4 +4
2 ss =
K a
H + 16 s 3 + 48 s + 4 s + 4 a = 4, a = 4, 0 1 ess = 1 =1 = 48 ,
a
3 The system is stable. 2 = 16 , b 0 = 4, b 1 = 1, b 2 = 0, b 3 =0 Unitstep Input: b0K H 1  a KH 0
a
0 4 = 1  4 = 0 i Unitramp input:
i = 0: b0 KH = 0 = 1: a 1  b1 K H = 4  1 = 3 0 86 e ss = a 1  b1 K H
a K
0 = 4 1
4 = 3 4 H Unitparabolic Input:
e K(s s
3 2 ss =
K a (b) M (s) = + 3) + 3s + ( K + 2)s + 3K a = 3K , a = K + 2, 0 1 ess = H =1 = 3,
b
0 The system is stable for K > 0. 2 =3K, b 1 =K Unitstep Input: b0K H 1  a KH 0
1
a a
1 0 3K = 1  3 K = 0 i Unitramp Input:
i e = 0:
ss  b0 K H = 0 =
K 3K = 1: =
2 a 1  b1 K H = K + 2  K = 2 0 =  b1 K H
a K
0 +2 K H 3K Unitparabolic Input:
e
ss = The above results are valid for K > 0. (c) M (s) = s s
4 +5 + 50 s + 10 s a = 0 , a = 10 , 0 1
2 + 15 s 3 H ( s) a = 10 s 2 = s+5 50 , a K H = lim
b
0 H (s) s b
1 3 = 15 , s 0 =2 =1 = 5, Unitstep Input: ess =
Unitramp Input: a 2  b1K H a KH 1
1
e
ss 1 50  1 2 = 2 10 = 2.4 = =
K Unitparabolic Input:
e K(s s
4 ss (d) M (s) = + 5)
s
2 + 17 s 3 + 60
a
0 + 5 Ks + 5 K
a
1 H =1
a The system is stable for 0 < K < 204. = 5K , = 5K , a 2 = 60 , 3 = 17 , b 0 = 5K, b 1 = K Unitstep Input: ess =
Unitramp Input:
i e b0K H 1  a KH 0
1
a a
1 0 5K = 1  5 K = 0 i = 0:
ss b0 KH = 0 =
e 5K = 1:
4 5 a 1  b1 K H = 5 K  K = 4 K 0 =  b1 K H
a K
0 K H 5K = Unitparabolic Input:
ss = The results are valid for 0 < K < 204. 87 77
G (s) = Y (s) E ( s) = KG 1 p ( s ) 20 s + K tG p( s)
K
p =
K 100 K 20 s ( 1 + 0 .2 s + 100
, K Type1 system. K )
t Error constants: = , v = 5K 1 + 100 K
t a =0 (a) (b) (c) r(t ) = u s ( t ): e ss = = 1 1+ K 1 K
v p =0
1 + 100 K 5K
t r(t ) = tu s ( t ): =t
2 e 1 K
a ss = r(t ) u ( t ) / 2:
s e ss = = 78
G p (s) =
(1 100 + 0 .1 s )( 1 + 0 . 5 s )
100 K G (s) = Y (s) E (s) = KG 20 s 1 p (s) + K tG p ( s ) G (s) =
20 s ( 1 + 0 .1 s )( 1 + 0 . 5 s ) + 100
K Kt 5K Error constants: p = , K v = 1 + 100 K ,
t K a =0 (a) (b) (c) r(t ) = u s ( t ): e ss = = 1 1+ K 1 K
v p =0
1 + 100 K 5K
t r(t ) = tu s ( t ): =t
2 e 1 K
a ss = r(t ) u ( t ) / 2:
s e ss = =
t Since the system is of the third order, the values of K and K stable. The characteristic equation is
3 2 must be constrained so that the system is s + 12 s + ( 20 + 2000 K t ) s + 100 K = 0
20 Routh Tabulation:
s s s
3 1 12 240 + 2000
100 K K t 2 1 + 24000
12 Kt  100 K s 0 100 K Stability Conditions: K>0 12 ( 1+ 100 K t )  5 K > 0 or 1 + 100 K t 5K > 1 12 Thus, the minimum steadystate error that can be obtained with a unitramp input is 1/12. 88 79 (a) From Figure 3P19, o ( s) r ( s ) 1+ = 1+ K1 K 2 Ra + La s + K1 K 2 Ra + La s + (R K i K b + KK1 K i K t
a + La s ) ( Bt + J t s ) + s ( Ra + La s )( Bt + J t s ) KK s K 1K i N (R K i K b + KK1 K i K t
a + La s ) ( Bt + J t s ) o ( s) r ( s ) = L a J t s + ( L a Bt + Ra J t + K 1 K 2 J t ) s + ( Ra Bt + K i K b + K Ki K1K t + K 1 K 2 Bt ) s + KK s K 1K i N
3 2 s [( Ra + La s ) ( Bt + Jt s ) + K1 K2 ( Bt + Jt s ) + Ki Kb + KK1 K i K t ] r ( t ) = u s ( t ), r ( s ) = 1 s
e lim s ( s )
s 0 e =0 Provided that all the poles of s ( s ) are all in the left half splane. (b) For a unitramp input, r ( s ) = 1 /
e
ss s . lim s ( s )
s 0 e 2 = t lim e (t ) = = R B
a t + K 1 K 2 B t + K i K b + KK
KK
s 1 K K
i t K K N
1 i if the limit is valid. 710 (a) Forwardpath transfer function: [n(t) = 0]: K (1 + 0.02s ) G ( s) = Y ( s) E (s ) =
2 K (1 + 0.02s ) s ( s + 25) = 2 K Kt s s s + 25 s + KKt 1+ 2 s ( s + 25) ( ) Type1 system. Error Constants: K p = ,
R( s) K v =
1 , 1 , K
t K a =0 = lim
sE ( s ) For a unitramp input, r ( t ) = tu s ( t ), = s 2 e ss = t lim e ( t ) s 0 = 1 K
v = K t Routh Tabulation: s s s s 3 1 25 25 K ( Kt + 0.02)  K 25 K K >0 =1/ K
s. KK t + 0.02 K K 2 1 0 Stability Conditions: 25 ( Kt + 0.02 )  K > 0 or K t > 0.02 (b) With r(t) = 0, n ( t ) = u s ( t ), N (s) System Transfer Function with N( s) as Input: Y (s ) N (s ) = K s ( s + 25) = 3 2 K (1 + 0.02 s) K Kt s s + 25 s + K ( K t + 0.02 ) s + K 1+ 2 + 2 s ( s + 25) s ( s + 25)
2 89 Steady State O utput due to n ( t):
y ss = t lim y ( t ) = lim s0 sY ( s ) =1 if the limit is valid. 711 (a) n(t ) = 0, r (t ) = tu s ( t ). G ( s) = Y ( s) E (s )
n =0 Forwardpath Transfer function: = K ( s + )( s + 3) s s 1
2 Ramperror constant: Steady state error: Characteristic equation: s + Ks Routh Tabulation:
3 2 Kv e = lim =
1 K
v ( ) Type1 system. s 0 sG ( s ) = 3 K ss = 1 3K
v + [ K ( 3 + )  1] s + 3 K = 0
3K s s s 3 1 K K (3K + K  1
3 K 2 1 + K  1 )  3 K
K 3 K 3K s 0 Stability Conditions: + K  1  3 > 0 or K > 1+ 3K 3 + K > 0 (b)
When r(t) = 0, n ( t ) = u s ( t ), N (s) =1/ s. K ( s + 3)
Transfer Function between n ( t) and y( t): Y (s ) N (s )
r =0 = 1+ 2 Ks ( s + 3) s 1 = K ( s + )( s + 3) s 3 + Ks 2 + [ K ( s + )  1]s + 3 K s s 1
2 ( ) Steady State Output due to n ( t):
y ss = t lim y ( t ) = lim s0 sY ( s ) =0 if the limit is valid. 712
Percen t maxi mum ov Thus ershoo t = 0 .25  =e
2 1  2 Solving for 1 2 =  ln0.25 = 1.386 = 0.404. sec. Thus, = 1.922 1  2 ( 2 )
= 343
2 from the last equation, we have t Peak T ime max = n 1 = 0 .01
2 n =
0 . 01 1  ( 0 .404 ) .4 rad / sec Transfer Function of the Secondorder Prototype System:
Y (s) R( s) = n
2 s 2 + 2 n s + n
2 = 117916 s
2 + 277 .3 s + 117916 713 ClosedLoop Transfer Function: Characteristic equation: 90 Y (s ) R (s ) = 25 K s + ( 5 + 500 Kt ) s + 25 K
2 s + ( 5 + 500 Kt ) s + 25 K = 0
2 For a secondorder prototype system, when the maximum overshoot is 4.3%, = 0 . 707 . n =
Rise Time: [Eq. (7104)]
t
r 25 K , 2 2 n = 5 + 500 = 0 .2 K t = 1.414 25 K = = = 1  0 .4167 + 2 . 917 n
2 = 2 .164 n
5 sec Thu s n = 10 . 82 rad / sec Thus, With K K n
2 = ( 10 . 82 ) 25 and K = 4. 68
t + 500 K 25 4. 68 t = 1.414 n = 15 . 3
r func tion i s Thus K t = 10 . 3 500 = 0 . 0206 = 0 . 0206 , the sy stem t ransfe Y (s) R( s) =
s 117
2 + 15 . 3 s + 117 Unitstep Response:
y = 0.1 at t = 0.047 sec. y = 0.9 at t = 0.244 sec. t = 0 .244  0 . 047 = 0 .197
r sec. y max = 0 . 0432 ( 4. 32% max. overs hoot) 714 Closedloop Transfer Function: Y (s ) R (s ) = 25 K s + ( 5 + 500 Kt ) s + 25 K
2 Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0
2 When Maximum overshoot = 10%, 1
2 =  ln0.1 = 2.3 = 5.3 1  2 2 ( 2 ) Solving for , we get = 0.59. The Natural undamped frequency is n =
2 25 K Thus, 5 + 500 K t = 2 n = 1.18 n n = 17 . 7
rad / sec Rise Time: [Eq. (7114)]
t
r = = 1  0 .4167 + 2 . 917 n = 0 .1 = = 1. 7696 n sec. Th us K n
2 = 12 . 58
t 25 With K = 12.58 and K = = 0 . 0318 500 0 . 0318 , the system transfer function is
Thus K
t 15 . 88 Y (s) R( s) =
s 313
2 + 20 .88 s + 314. 5 Unitstep Response:
y = 0.1 when t = 0.028 sec. y = 0.9 when t = 0.131 sec. 91 t r = 0 . 131  0 . 028 = 0 .103 = 1.1 sec. y max ( 10% max. overs hoot ) 715 ClosedLoop Transfer Function:
Y (s) R( s) Characteristic Equation: = 25 K s
2 + ( 5 + 500 K )s
t + 25 K 1
2 s + ( 5 + 500 Kt ) s + 25 K = 0
2 When Maximum overshoot = 20%, =  ln0.2 = 1.61 = 2.59 1  2 2 ( 2 ) Solving for , we get = 0 .456 . 25 K
2 The Natural undamped frequency n = 5 + 500 K t = 2 n = 0 . 912 n
sec. Thus, Rise Time: [Eq. (7114)]
t
r = = 1  0 .4167 + 2 . 917 n = 0 . 05 = 1.4165 n n =
K 1.4165 0 . 05 = 28 .33 = 32 .1 5 + 500 K = 0 . 912 = 25 . 84 Thus, t n 25 With K = 32.1 and K = 0 . 0417 , the system transfer function is
K
t n
2 t = 0 . 0417 Y (s) R( s) = 802 . 59 s
2 + 25 . 84 s + 802 . 59 Unitstep Response:
y = 0.1 when t = 0.0178 sec. y = 0.9 when t = 0.072 sec. t = 0 . 072  0 . 0178 = 0 . 0542
r sec. y max = 1.2 ( 20% max. overs hoot ) 716 ClosedLoop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K
From Eq. (7102), Delay time t
d Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0
2 1.1 + 0 .125 + 0 .469 n
. t
d 2 = 0 .1 sec. When Maximum overshoot = 4.3%,
2 2 = 0 . 707 = 1.423 n = 0 .1 sec. Thus n = 14.23
K rad/sec. n = 14.23 = 8.1 5 + 500 K = 2 = 1.414 = 20.12 K = t n n 5 5 With K = 20.12 and K
t Thus t = 15 .12 500 = 0 . 0302 = 0 . 0302 , the system transfer function is Y (s) R( s) =
s 202 . 5
2 + 20 .1 s + 202 .5 UnitStep Response: 92 When y = 0.5, t = 0.1005 sec. Thus, t = 0 .1005 sec.
d y max = 1. 043 ( 4. 3% max. overs hoot ) 717 ClosedLoop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K
From Eq. (7102), Delay time t
2 2 Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0
2 d 1.1 + 0 .125 + 0 .469 n
5 2 = 0 . 05 = 1. 337 n Thus, n = 1. 337 0 . 05 = 26 . 74 K = n = 26.74 = 28.6 5 5 t + 500 K t = 2 n = 2 0 . 59 26 . 74 = 31 . 55
r func tion i s Thus K t = 0 . 0531 With K = 28.6 and K = 0 . 0531 , the sy stem t ransfe Y (s) R( s) =
s 715
2 + 31 . 55 s + 715 UnitStep Response:
y = 0.5 when t = 0.0505 sec. Thus, t = 0 . 0505 sec.
d y max = 1.1007 ( 10 . 07% max. overs hoot ) 718 ClosedLoop Transfer Fu nction: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For Maximum overshoot = 0.2, = 0.456
From Eq. (7102), Delay time t
d Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0
2 .
2 = 1.1 + 0 .125 + 0 .469 n = 1.2545 n = 0 . 01 K =
t sec.
2 Natural Undamped Frequency 5 n = 1.2545 0 . 01 = 125
.45 .45 rad/sec. Thus, + 500 K t = 2 n = 2 0 .456 125
t = 114.41 Thus, K = 0.2188 n = 15737.7 = 629.5 25 5 With K = 629.5 and K = 0.2188 , the system transfer function is Y (s) R( s) = 15737 . 7 s
2 + 114.41 s + 15737 .7 Unitstep Response: y = 0.5 when t = 0.0101 sec. 93 Thus, t y d = 0 . 0101 sec. x. overs hoot ) max = 1.2 ( 20% ma 719 ClosedLoop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K = 0 .6 2 = 5 + 500 n
From Eq. (7109), settling time t
s Characteristic Equation: s + ( 5 + 5000 Kt ) s + 25 K = 0
2 K t = 1.2 n
3.2 n 3 .2 n 5 = 0 .6 = 0 .1
n sec. Thus, n = 3 .2 0 . 06 = 53 . 33 rad / sec K t = 1.2 = 0 .118
Y (s) R( s) K = n
2 = 113 . 76 500 25 System Transfer Function: = 2844 s
2 + 64 s + 2844
y(t) reaches 1.00 and never exceeds this value at t = 0.098 sec. Thus, t = 0 . 098 sec.
s Unitstep Response: 720 (a) ClosedLoop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For maximum overshoot = 0.1, = 0 . 59 .
Settling time: t
s Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0
2 5 + 500
sec. K t = 2 n = 2 0 . 59 n = 1.18 n n =
3 .2 0 . 05 = 3 .2 n
t = 3 .2 0 .59 n = 0 . 05 0 . 59 = 108 .47 K = 1.18 n 5
500 = 0 .246
Y (s) R( s) K = n
2 = 470 . 63 25 System Transfer Function: = 11765 . 74 s
2 + 128 s + 11765 . 74 UnitStep Response:
y(t) reaches 1.05 and never exceeds this value at t = 0.048 sec. Thus, t = 0 . 048 sec.
s 94 (b) For maximum overshoot = 0.2, Settling time t
s = 0 .456
3 .2 0 .456 K . 5 + 500 K t = 2 n = 0 . 912 n n =
3 .2 0 .456 = 3 .2 n = n = 0 . 01 n 5 sec. 0 . 01 = 701 . 75 rad / sec t = 0 . 912 = 1.27 500 System Transfer Function:
Y (s) R( s) = 492453 s
2 + 640 s + 492453
y(t) reaches 1.05 and never exceeds this value at t = 0.0074 sec. Thus, t = 0 . 0074 sec. This is less
s UnitStep Response: than the calculated value of 0.01 sec. 721 ClosedLoop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K
Damping ratio Characteristic Equation: s + ( 5 + 500 Kt ) s + 25 K = 0
2 = 0 . 707 . Settling time t s = 4. 5 n = 3 .1815 n = 0 .1 sec. Thus, n = 31 .815 n
2 rad/sec. 5 + 500 K t = 2 n = 44. 986
Y (s) R( s) Thus , K t = 0 . 08 K = 2 = 40 .488 System Transfer Function: = 1012 .2 s
2 + 44. 986 s + 1012 .2
s UnitStep Response: The unitstep response reaches 0.95 at t = 0.092 sec. which is the measured t . 95 722 (a) When = 0 . 5 , the rise time is
t
r 1  0 .4167 + 2 . 917 n
2 2 = 1. 521 n =1 sec. Thus n = 1. 521 rad/sec. The secondorder term of the characteristic equation is written s
2 + 2 n s + n =
s
2 s 2 + 1. 521 s + 2 . 313 = 0
3 The characteristic equation of the system is Dividing the characteristic equation by s + ( a + 30 )s 2 + 30 as +K =0 + 1. 521 s + 2 . 313 , we have For zero remainders, 28 .48 a = 45 . 63 Thus, a = 1. 6
69 . 58 K = 65 .874 + 2 . 313 a = 69 . 58 ForwardPath Transfer Function:
G (s) =
s( s + 1.6 )( s + 30 ) UnitStep Response:
y = 0.1 when t = 0.355 sec. y = 0.9 when t = 1.43 sec. Rise Time: t = 1.43  0 . 355
r = 1. 075 sec. (b) The system is type 1. seconds
ss (i) (ii) For a unitstep input, e For a unitramp input, = 0.
K
v = lim s0 sG ( s ) = K 30 a = 60 . 58 30 1. 6 = 1.45 e ss = 1 K
v = 0 .69 723 (a) Characteristic Equation:
s
3 + 3s + (2 + K ) s  K = 0
2 Apply the RouthHurwitz criterion to find the range of K for stability. Routh Tabulation: 96 s s s 3 1 3 6 2 +K 2 K 1 +4K
3 s 0 K
1.5 < K < 0 This simplifies the search for K for two equal roots. When K = 0.27806, the characteristic equation roots are: 0.347, Stability Condition: 0.347, and 2.3054. (b) UnitStep Response: ( K =  0.27806) (c) UnitStep Response ( K =  1) The step responses in (a) and (b) all have a negative undershoot for small values of t. This is due to the zero of G(s) that lies in the righthalf splane. 724 (a) The state equations of the closedloop system are: = 6 x 1 dt dt The characteristic equation of the closedloop system is
1 2 dx =  x1 + 5 x 2 dx  k1 x 1  k 2 x 2 + r = s +1 6 + k1 5 s + k2 = s + ( 1 + k2 ) s + ( 30 + 5 k1 + k 2 ) = 0
2 97 For n = 10 rad / sec, 30 + 5 k 1 + k 2 = n = 100
2 . k Thu s 5 k 1 + k 2 = 70 (b) For = 0 . 707 , 2 n = 1+ k2. Th us
2 n = 1 + 2 . 1.414 n =
2 (1 + k )
2 2
and = 30 + 5k1 + k 2
, 1 Thus k 2 = 59 + 10 k1
2 (c) For n = 10 rad / sec 5k
1 1 = 0 . 707
an d + k 2 = 100
k
1 Solving for k , we hav e = 11 . 37 . + k 2 = 2 n = 14. 14 Th us k 2 = 13 .14 (d) The closedloop transfer function is Y (s) R (s )
For a unitstep input, = 5 s + ( k 2 + 1) s + ( 30 + 5 k1 + k 2 )
2 = 5 s + 14.14 s + 100
2 lim y ( t )
t = lim sY ( s ) s0 = 5 100 = 0 . 05 (e) For zero steadystate error due to a unitstep input, 30 + 5 k1 + k 2 = 5
1 2 Thus 5k 1 + k 2 = 25 Parameter Plane k versus k : 725 (a) ClosedLoop Transfer Function Y (s) 100 ( K P + K D s ) = 2 R ( s ) s + 100 K D s + 100 K P The system is stable for K > 0 and P (b)
For (b) Characteristic Equation:
s K
2 + 100 K D s + 100 K P =0 D > 0. = 1, 2 n = 100
K
P K D . 2
n n = 10 (c) (d) T hus = 100 KD = 20 KP K D = 0 .2 K P See parameter plane in part (g) . See parameter plane in part (g) . 98 (e) Parabolic error constant
s 0 K
2 a = 1000
s 0 sec 2 K a = lim s G ( s ) = lim100 ( K P + K D s ) = 100 K P = 1000 (f)
Natural undamped frequency Thus K P = 10 n = 50
K
P rad/sec. n = 10 (g)
When K
P = 50 Th us K P = 25 = 0,
G (s) = 100 K s
2 D s = 100 K s D (polezero cancellation) 726 (a) Forwardpath Transfer Function: G ( s) =
When r ( t ) Y ( s) E (s )
K
v = s [ Js(1 + Ts ) + K i K t ]
sG ( s ) KKi
K K
t = s 0.001 s + 0.01 s + 10 K t
2 ( 10 K
K K ) = tu s ( t ), = lim s0 = e ss = 1 K
v = t (b) When r(t) = 0 Y (s ) Td ( s)
For T ( s )
d = 1 + Ts s [ Js(1 + Ts) + Ki Kt ] + KKi
lim y ( t )
t = s 0.001 s + 0.01 s + 10 Kt + 10 K
2 ( 1 + 0.1s ) = 1 s = lim sY ( s )
s 0 = 1 if the system is stable. 10 K (c) The characteristic equation of the closedloop system is 0 . 001 s
3 + 0 . 01 s + 0 .1 s + 10
2 K =0 The system is unstable for K > 0.1. So we can set K to just less than 0.1. Then, the minimum value of the steadystate value of y(t) is 1 + =1  10 K K =0 .1 However, with this value of K, the system response will be very oscillatory. The maximum overshoot will be nearly 100%. (d) For K = 0.1, the characteristic equation is 0 . 001 s
3 + 0 . 01 s + 10
2 K s
t +1 = 0 or s 3 + 10 s 2 + 10 4 K s
t + 1000 = 0 For the two complex roots to have real parts of 2/5. we let the characteristic equation be written as 99 ( s + a) ( s
Then, a The three roots are: s 2 + 5s + b = 0
a ) or s + ( s + 5)s + (5a+ b) s+ ab = 0
3 2 + 5 = 10 =  a = 5 =5 ab = 1000 b = 200 s =  a = 5 s = 2 . 5 j13 . 92 5a + b = 10 4 K t K t = 0 . 0225 727 (a) K t = 10000 oz  in / rad The ForwardPath Transfer Function: G ( s) = s s + 5000 s + 1.067 10 s + 50.5 10 s + 5.724 10
4 3 7 2 9 ( 9 10 K
12 12 ) =
Routh Tabulation:
s s s s
5 9 10 K
12 s ( s + 116)( s + 4883)( s + 41.68 + j3178.3)( s + 41.68  j 3178.3) 10 10
7 1 500 0 5.7 2. 895 16 . 6 1.067 50.5
5 5.7 24 9
9 10
12 12 4 9 10
0 K 3 10
8 5 . 72
7 10
9 12  1. 8 10
12 K 2 10 + 1. 579 10
13 K 10 K s 1 10 + 8 .473 10
29 12 K  2 . 8422 10
K 9 K 2 + 1. 579 10
12 s 0 9
1 K From the s row, the condition of stability is 165710 or K
2 + 8473
(K K  2 . 8422
K  2981 .14 K  58303 .427 <0 or + 19 .43)( >0  3000 . 57 ) < 0
K
2 Stability Condition: 0 < K < 3000.56 The critical value of K for stability is 3000.56. With this value of K, the roots of the characteristic equation are: 4916.9, 41.57 + j3113.3, 41.57 + j3113.3, j752.68, and j752.68 (b) K L = 1000 ozin/rad. The forwardpath transfer function is G ( s) = s s + 5000 s + 1.582 10 s + 5.05 10 s + 5.724 10
4 3 6 2 9 ( 9 10 K
11 11 ) = 9 10 K
11 s (1 + 116.06)( s + 4882.8)( s + 56.248 + j1005)( s + 56.248  j1005) (c) Characteristic Equation of the ClosedLoop System:
s
5 + 5000 s 4 + 1. 582 10 6 s 3 + 5 . 05 10 9 s 2 + 5 . 724 10 11 s + 9 10 10 10
6 11 K =0
5.7 24 9 Routh Tabulation:
s s s s
5 1 500 0 5.72 4.6503 1 .582 5. 05
5 10
11 11 4 9 10
0 K 3 10
7 5 .724
6 10 11  1.8 10
11 8 K 2 10 + 1. 5734 10 K 9 10 K 100 s 1 26 . 618 10 18 + 377 .43
7 10
11 15 K  2 . 832 10
6 14 K 2 4. 6503 s
0 10 + 1. 5734 10
9 K 10 K 26 . 618 or (K From the s row, the condition of stability is Or, K
2 1 10 4  1332 . 73 K  93990 < 0  1400 )( K + 3774. 3 K  2 .832 + 67 .14 ) < 0 K 2 >0 Stability Condition: 0 < K < 1400 The critical value of K for stability is 1400. With this value of K, the characteristic equation root are: 4885.1, 57.465 + j676, 57.465  j676, j748.44, and j748.44 (c) K L = . G ( s) = = nK s K i K s L a J T s + ( Ra J T + R mL a ) s + R a B m + K i K b 2 ForwardPath Transfer Function: JT = J m + n J L
2 s s + 5000 s + 566700
2 ( 891100 K ) = 891100 K s ( s + 116)( s + 4884) Characteristic Equ ation of the ClosedLoop System:
s
3 + 5000 s 2 + 566700 s + 891100 K =0 Routh Tabulation:
s s s s
3 1 5000 5 66700 5667 00 89 1100 K .22 K 2 1  178 0 89 91100 K
1 From the s row, the condition of K for stability is 566700  178.22K > 0. Stability Condition : 0 < K < 3179.78 The critical value of K for stability is 3179.78. With K = 3179.78, the characteristic equation roots are 5000,
When the motor shaft is flexible, K becomes stiffer, K K
L L j752.79, and j752.79. is finite, two of the openloop poles are complex. As the shaft =, L increases, and the imaginary parts of the openloop poles also increase. When the shaft is rigid, the poles of the forwardpath transfer function are all real. Similar effects
L are observed for the roots of the characteristic equation with respect to the value of K . 728 (a)
G (s)
c =1 G (s) = 100( s s
2 + 2) 1 K p = lim G (s) s0 =  200 When d(t) = 0, the steadystate error due to a unitstep input is e ss = 1 1+ K
p =
1 1  200 = 1 199 =  0 . 005025 101 (b) Gc (s ) = (c) s + s G (s ) = 100( s + 2)( s + ) s s 1
2 ( ) Kp = e ss = 0 =5 = 50 = 500 m aximum ma ximum overs hoot oversh oot = 5.6% = 22% = 54.6% max imum o versho ot As the value of increases, the maximum overshoot increases because the damping effect of the zero at s =  becomes less effective. UnitStep Responses: (d) r(t ) =0 and G (s)
c = 1. d (t ) = u s(t ) D (s) = 1 s System Transfer Function: ( r = 0)
Y ( s) D (s)
r =0 = 100( s s
3 + 2) + 100 s 2 + ( 199 + 100 ) s + 200 100( s Output Due to UnitStep Input:
Y ( s) =
s s + 2) 3 + 100 s 2 + ( 199 + 100 ) s + 200 sY ( s ) y ss = lim y ( t )
t = lim s0 = 200 200 = 1 (e) r(t ) = 0, d (t ) = us(t )
G (s)
c = s +
s System Transfer Function [ r( t) = 0]
Y (s) D ( s)
r =0 = 100 s ( s s
3 + 20 + 100 s 2 + ( 199 + 100 ) s + 200 D (s) = 1 s 102 y ss = t lim y ( t ) = lim s0 sY ( s ) =0 (f) =5
Y (s) D ( s ) r =0 Y (s) D ( s) Y ( s) D (s)
r =0 r =0 = 100 s ( s s
3 + 2)
s + 100 s 2 + 699 + 1000 = 50 = 100 s ( s s
3 + 2)
s + 100 s 2 + 5199 + 10000 = 5000 = 100 s ( s s
3 + 2)
s + 100 s 2 + 50199 + 100000 UnitStep Responses: (g) As the value of increases, the output response y(t) due to r(t) becomes more oscillatory, and the overshoot is larger. As the value of increases, the amplitude of the output response y(t) due to d(t) becomes smaller and more oscillatory. 729 (a) ForwardPath Transfer function:
G (s) Characteristic Equation: = H (s) E (s) = 10 N s( s + 1 )( s + 10 ) s N s(s
2 + 1) s 2 +s+N =0 n = 1 n
1 N=1: Characteristic Equation:

1 
2 + s +1 = 0
(16.3%) Peak time t = 0 .5 = rad/sec. Maximu m over shoot = e = 0 .163
s
2 max  = 3 . 628
2 sec. N=10: Characteristic Equation:
 + s = 10 = 0
Peak time t = 0 . 158 = n
1 n = 10 = 1.006
2 rad/sec. Maximum overshoot =e 1  2 = 0 . 605 (60.5%) max sec.  (b) UnitStep Response: N=1 103 Secondorder System Maximum overshoot Peak time
0.163 3.628 sec. Thirdorder System
0.206 3.628 sec. UnitStep Response: N = 10 Secondorder System Maximum overshoot Peak time
0.605 1.006 sec. Thirdorder System
0.926 1.13 sec. 730 UnitStep Responses: When T z is small, the effect is lower overshoot due to improved damping. When T
z z is very large, the overshoot becomes very large due to the derivative effect. T derivative control or a highpass filter. improves the rise time, since 1 +Tz s is a 104 731 UnitStep Responses The effect of adding the pole at s less stable. When T p = 1 T
p to G(s) is to increase the rise time and the overshoot. The system is > 0 . 707 , the closedloop system is stable. 732 (a) N=1 Closedloop Transfer Function:
M
H (s) = Y ( s) R( s) = 10 s
3 + 11 s + 10 s + 10
2 =
1 1 + s + 1. 1 s + 0 .1 s
2 3 SecondOrder Approximating system:
M M M d e f f (s) (s) l
L (s) = 1 1+ d s
1 2 + d2s = 2 H L = 1+ d s
1 + d2s
2 1+ m s
1 +m2s
2 2 1+ s + 1.1 s + 0 .1 s
l
2 3 1+ l s
1 + l2 s + l3s 3 1 = m1 =
f
2 d 2 = m2
2 1 =1
2 = 1.1 l 3 = 0 .1
e f
4 2 = 2m
2 2  m1 = 2 d 2  d1
2 = = f 4 = 2 m 4  2 m 1m 3 + m 2 = m 2 = d 2
2 2 2 4 2 2 = 2 l 2  l1 = 2 1.1  1 = 1.2 = 2 l6  2 l 1 l 5 + 2 l 2 l 4  l 3 =  l3 =  0 . 01
2 4 2l  2 l1 l 3 + l 2 = 2 1 0 .1 + ( 1.1 ) = 1. 01
2 6 Thus, f 2 = 1.2 = 0 . 01
1 e 2 = 1.2 =
d
2 1 2d 2  d1 2 f 4 = 1. 01
1 e 4 = 1. 01 = d 2 2 Thus d 2 = 1. 005 f M (s) 6 = 2 d 2  1.2 = 0 .81 =
0 . 995 s
2 Thus d = 0 .9
G L ( s) L =
1 + 0 . 9 s + 1. 005 s 2 + 0 . 8955 s + 0 . 995 = 0 . 995 s( s + 0 .895) Roots of Characteristic Equation: Thirdorder System Secondorder System 10.11 0.4457 + j0.8892 0.4457  j0.8892 0.4478 + j0.8914 0.4478  j0.8914 105 The real root at 10.11 is dropped by the secondorder approximating system, and the two complex roots are slightly preturbed. UnitStep Responses: (b) N = 2: Closedloop Transfer Function:
M (s) H = 20 s l
2 3 + 11 s + 10 s + 20
2 = = 1 1 + 0 .5 s
2 + 0 . 55
f
2 s 2 + 0 . 05
2l
2 s
2 3 l 1 = 0 .5 = d2 =
2 l f
4 2 = 0 . 55 3 = 0 . 05 e 2 2d 2  d1 = =  l1 = 2 0 . 55  ( 0 . 5)
d
2 2 2 = 0 .85
d
2 e d 4 2 1 = 2 l1 l 3 + l 2 = 0 . 05 + 0 . 3025 = 0 .2525
d 1. 99 s
2 1 Thus = f 4 = 0 .2525 = 0 . 5025 = 2d 2  e 2 = 2 0 . 5025  0 .85 = 0 .155
1 1 = 0 . 3937
G L(s) M L (s) = + 0 . 3937 s + 0 . 5025 s 2 = + 0 . 7834 s + 1. 99 =
s(s 1. 99 + 0 . 7835) Roots of Characteristic Equations: Thirdorder System Secondorder System  10.213 0.3937 + j1.343 0.3937  j1.343 0.3917 + j1.3552 0.3917  j1.3552 The real root at 10.213 is dropped by the secondorder approximating system, and the two complex roots are slightly preturbed. UnitStep Responses: 106 (c) N=3 ClosedLoop Transfer Function: M (s) H = 30 s
3 + 11 s + 10 s + 30
2 =
1 1 + 0 . 333 s + 0 . 3667 s 2 + 0 . 0333 s 3 l e
2 1 = 0 . 3333
2 l f
2 2 = 0 . 3667
2 2 1 l 3 = 0 . 0333
e
4 = 2 d 2  d1 =
2 2 = 2 l 2  l1 = 0 . 7333  0 .1111 = 0 . 6222
d d
1 = d2 =
2 f 4 = 2 l1 l 3 + l 2 = 0 .1122
2 Thus, d = f 4 = 0 . 1122 = 0 .2186 = 2 d2  f 2 = 2 0 . 335  0 . 6222 = 0 . 0477
d
2 = 0 . 335
2 . 985 G L(s) M L (s) =
1 1 + 0 .2186 s + 0 . 335 s 2 =
s 2 + 0 . 6524 s + 2 .985 =
s( s 2 . 985 + 0 . 6525) Roots of Characteristic Equation: Thirdorder System Secondorder System 10.312 0.3438 + j1.6707 0.3438  j1.6707 0.3262 + j1.6966 0.3262  j1.6966 The real root at 10.312 is dropped by the secondorder approximating system, and the complex roots are slightly preturbed. UnitStep Responses: (d) N= 4 ClosedLoop Transfer Function: 107 M H (s) = 40 s
3 + 11 s + 10 s + 40
2 = 1 1 + 0 .25 s + 0 .275
l
4 3 s 2 + 0 . 025 s 3 l e
2 1 = 0 .25
2 l 2 = 0 .275
e
4 2 1 = 0 . 025
f
2 4 = 2d d 2 2  d1 =
2 f 2 = 2 l 2  l 1 = 0 .4875
d 1 1
2 =d2 =
d = 2 l1 l 3 + l 2 = 0 . 06313
2 Thus, 2 =
M f 4 = 0 . 06313
(s) = 0 .2513 = = 2d  f 2 = 0 . 5025  0 .4875 = 0 . 015
G (s) d 3 . 98 1 = 0 .1225 L = 3 . 98 s
2 + 0 .1225 s + 0 .2513 s 2 + 0 .4874 s + 3 .98 L = s( s + 0 .4874 ) Roots of Characteristic Equation: Thirdorder System Secondorder System  10.408 0.2958 + j1.9379 /2958  j1.9379 0.2437 + j1.98 0.2437  j1.98 The real root at 10.408 is dropped by the secondorder approximating system, but the complex roots are preturbed. As the value of N increases, the gain of the system is increased, and the roots are more preturbed. UnitStep Responses: (e) N= 5 Closedloop Transfer Function:
M (s) H = 50 s
3 + 11 s + 10 s + 50
2 =
1 l 1 + 0 .2 s + 0 .22 s 2 + 0 . 02 s 3 l e e
2 1 = 0 .2
2 l
2 2 = 0 .22
2 3 = 0 . 02 = 2 d1  d 2 = = d2 =
2 f = 2 l 2  l1 = 0 .44  0 .04 = 0 .4
2 4 f 4 = 2 l1 l 3 + l 2 = 0 . 008 + 0 . 0484 = 0 . 0404 108 Thus, d d M
L
2 2 2 1 = = f 4 = 0 .0404
2 d 2 = 0 .201
d
1 2d  e 2 = 0 .402  0 .4 = 0 . 002
s
2 = 0 . 04472
G L(s) (s) =
1 1 + 0 . 0447 s + 0 .201 = 1 s
2 + 0 .2225 s + 4. 975 = 4. 975 s(s + 0 .2225) Roots of Characteristic Equation: Thirdorder System Secondorder System  10.501 0.2494 + j2.678 0.2494 j2.678 0.1113 + j2.2277 0.1113  j2.2277 The real root at 10.501 is dropped by the secondorder approximating system, and the complex roots are changed, especially the real parts. UnitStep Responses: 733 (a) K=1 Forwardpath Transfer Function: G ( s) = s s + 5000 s + 566700
2 ( 891100 ) = 891100 s ( s + 116)( s + 4884) ClosedLoop Transfer Function :
M (s) H =
l 891100 s
3 + 5000 s 2 + 566700
l
2 2 s + 891100
3 =
1 1 + 0 . 636
l
3 s + 5. 611 10
6 3 s 2 + 1.1222 10 6 s 3 1 = 0 . 636 = d2 =
2 = 5. 611 10
2 2 2 = 1.122 10
3 6 e e Thus, 2 = 2 d 2  d1 =
f
4 f = 2 l 2  l1 = 1.1222 10  0 .4045 = 0 . 3933 + ( 5 . 611 10
3 4 =  2 l1 l3 + l 2 = 2 0 . 636 1.1222 10 ) 2 = 0 . 000030 06 109 d d 2 2 2 1 = 0 . 000030 = 2 d2 
f
2 06 d 2 = 0 .005482
d
1 = 0 . 01096 + 0 . 3933 = 0 .4042 =
182 .4 s
2 = 0 . 6358
G L ( s) M L (s) =
1 1 + 0 . 6358 s + 0 . 005482 s 2 + 115 . 97 s + 182 = 182 .4 s( s .4 + 115 . 97 ) Roots of Characteristic Equations: 1.595 114.4 4884 Thirdorder System 1.5948 114.38 Secondorder System The real root at 4884 is dropped by the secondorder approximating system. the other two roots are hardly preturbed. UnitStep Responses (b) K = 100 Closedloop Transfer Function: M H (s) = 891100 s
3 00 00 + 5000
l
1 s 2 + 566700 + 891100
l
2 = 1 1 + 0 . 00636 s
5 5 + 5. 611 10 5 8 s 2 + 1.1222 10 8 s 3 = 0 .00636
2 = 5. 611 10
2 2 l 3 = 1. 1222 10
5 8 e e Thus, d d 2 = 2 d 2  d1 = = d2 =
2 f 2 = 2 l 2  l1 = 11 .222 10  4. 045 10 = 0 . 000071 8
5 4 2 2 2 1 f 4 = 2 l 4  2 l1 l 3 + l 2 =  2 0 . 00636 1.1222 10
9 + ( 5 . 611 10 ) 2 = 3 . 0056 10 9 = f 4 = 3 . 0056 10
f
2 d 6 2 = 0 . 000054
8 8 8 d = 2 d2  = 0 . 000109
1  0 . 000071 =
s = 0 . 000054 1 = 0 . 007403
(s) M L (s) =
1 18248
2 + 0 . 007403 s + 0 . 000054 8s 2 + 135 .1 s + 18248 G L = 18248 s( s + 135 . 1) Roots of Characteristic Equations: Thirdorder System Secondorder System 110 4887.8 56.106 + j122.81 56.106  j122.81 67.55 + j114.98 67.55  j114.98 UnitStep Responses (c) K = 1000 Closedloop Transfer Function: M (s) H = 891100 s
3 000 s + 5000
l
1 s 2 + 566700 + 891100
l
2 2 =
000 1 1 + 0 . 000636
6 s + 5 . 611 10 6 9 s 2 + 1.1122 10 9 s 3 = 0 . 000636
2 = 5 . 611 10 l 3 = 1.1222 10
404
9 e e Thus, d d 2 = 2 d 2  d1 = = d2 =
2 f 2 = 2 l 2  l1 = 11 .222 10
2 6  0 .000000 = 0 . 000010 82
6 4 2 2 2 1 f 4 =  2 l1 l3 + l 2 = 2 0 . 000636 1.1222 10
11 + ( 5 . 611 10
482 ) 2 = 3 . 0056 10 11 = = f 4 = 3 . 0055 10
2 d 965 2 = 0 . 000005 = 0 . 000000
182415 .177 2d  f 2 = 0 . 000010
1  0 . 000010 = 818 147 d 1 = 0 . 000382
G L (s) 9 182415 .177 s( s M L (s) =
1 + 0 . 000382 9s + 0 . 000005 482 s 2 s 2 + 69 . 8555 s + 182415 = .177 + 69 . 8555) Roots of Characteristic Equations: Thirdorder System Secondorder System  4921.6 39.178 + j423.7 39.178  j423.7 34.928 + j425.67 34.928  j425.67 111 Unitstep Re sponses 734 Forwardpath Transfer Function
G (s) Closedloop Transfer Function
M (s) = K(s s(s  1) + 1 )( s + 2 )
1 = s +1
s
3 +3s + s +1
2 Secondorder System:
M
L (s)
2 = + c1 s
1 M H (s ) M L (s ) =
l (  s + 1) (1 + d s + d s ) 1 + ( d  1) s + ( d  d ) s  d s = ( s + 3s + s + 1) ( 1+ c s ) 1 + ( c + 1) s + ( c + 3) s + ( 3c + 1) s + c s
2 3 1 2 1 2 1 2 3 2 2 3 1 1 1 1 1 1+ d s + d2s 2 4 1 = 1 + c1
1 l m 2 = 3 + c1 = d 2  d1 l m 3 = 1 + 3 c1 = d 2
2 l 4 = c1 m = d1 1
2 2 3 e2 = f 2 = 2m 2  m1 = 2 ( d 2  d 1 )  ( d 1  1) = 2 d 2 d 1  1
2 = 2l2  l1 = 2 ( 3 + c1 ) ( 1+ c1 ) = 5  c1
2 2 2 2 e4 = f 4 = 2m 4  2m1 m3 + m 2 = 2 ( d 1  1) ( d 2 ) + ( d 2  d1 ) = d 2  2d 2 + d 1
2 2 2 = 2l4  2l1l3 + l2 =  2 ( 1 + c1 ) (1 + 3c1 ) + ( 3 + c1 ) = 7  2c1  5c1
2 2 2 e6 = f 6 = 2m 6  2 m1m 5 + 2 m 2 m4  m 3 = m 3 =  ( d 2 ) = d 2
2 2 2 2 2 2 = 2l6  2l1l5 + 2l 2l4  l3 = 2l2 l4  l3 = 2 ( 3 + c1 ) c1  (1 + 3c1 ) =  1  7c1
2 2 Simultaneous equations to be solved: Solutions: 112 2d 2  d1 = 6 c1
2 2 c1 d d
1 = 1. 0408 = 0 . 971 = 2 . 93
G L(s) d d
2 2 2 2 2 = 1 + 7 c1
2 2  2 d 2 + d 1 = 5  c1 =
1 1 2 M L (s)  1. 0408 s s
2 + 0 . 971 s + 2 . 93 = 0 .3552(
s
2 s  0 . 9608
s ) + 0 . 3314 + 0 .3413 = 0 . 3552(
s( s s  0 . 9608 ) + 0 . 69655) Roots of Characteristic Equations: 2.7693 0.1154 + j0.5897 0.1154  j0.5897
Unitstep Responses Thirdorde r System 0.1657 + j0.5602 0.1657 j0.5602 Secondorder System 735 (a) K = 10 Closedloop Transfer Function: M (s) H = 10 s
4 + 23 s 3 + 62 s 2 + 40 s + 10 = 1 1+ 4 s + 6 .2 s + 2 . 3 s + 0 .1 s
2 3 4 Secondorder System Approximation:
M (s) L =
l 1 1+ d s
1 + d2s 2 l e e e
2 1 =4
2d
2 2 2 2 2 = 6 .2
2 l 3 = 2 .3
2 l 4 = 0 .1 = = = f f f 2 =  d 1 = 2 l 2  l1 = 12 .4  16 = 3 . 6
2 2 2 4 4 = d 2 = 2 l 4  2 l1 l 3 + l 2 = 0 .2  2 4 2 . 3 + ( 6 .2 ) = 20 .24 = 2 d 6  2 d 1 d 5 + 2 d 2 d 4  d 3 =  d 3 = 2 l 2 l 4  l3 =  4. 05
d d
2 2 2 1 6 6 Thus, = = 20 .24 2d d f 2 = 4. 5 = 3 . 55 2  2 = 9 + 3. 6 = 12 . 6 d 1 113 M L (s) =
1 1 + 3 .55 s + 4. 5 s 2 = 0 .2222 s
2 + 0 . 7888 s + 0 .2222 G L(s) = 0 .2222 s( s + 0 . 7888 ) Roots of Characteristic Equations: Fourth order system Secondorder system 2.21 20 0.3957 + j0.264 0.3957 j0.264 (b)
K = 10 Thirdorder System Approximation: 1 1+ d s
1 2 0.3944 + j0.258 0.3944 j0.258 M L (s) = M
2 3 H L (s) (s) + d 2s + d3s M d
2 3 =
1 1 + d1 s + d 2 s + d 3 s
2 2 3 3 + 4 s + 6 .2 s + 2 . 3 s + 0 .1 s
Thus d 4 e e 2 = = f f 2 = 2d 2  d 1 = 3 . 6
2 = f6 =
2 4. 05 d
2 3 = 2 . 0125 4 4 = 2 d 4  2 d 1 d 3 + d 2 =  2 d 1 d 3 + d 2 =  4. 025
d 1 +d2 = f 4 = 20 .24 Thus,
2 1 = 0 . 5 d 1  1.8
2 d , 2 2 Solving for d , d 1 = 3 .9528  3 . 0422 = 0 .25 d 1  1. 8 d 1 + 3 .24 = 20 .24 + 4. 025
4 2 d ,  0 .4553 +
d
1 j 2334, . d
2  0 .4553 
2 1 j 2334. Selecting the positive and real solution, we have 1 1 = 3 . 9528 = = 0 . 5 d 1  1.8 = 6 . 0123
0 .4969 M L (s) = + 3. 9528 s + 6 . 0123 s 2 + 2 . 0125 s 3 s 3 + 2 . 9875 s 2 + 1. 964 s + 0 .4969 GL ( s) =
Roots of Characteristic Equations: Fourth order System s s + 2.9875 s + 1.964
2 ( 0.4969 ) Thirdorder System 2.21 20 0.39565 + j0.264 0.39565  j0.264 2.1963 0.3956 + j0.264 2.1963  j0.264 Unitstep Response 114 (c) K = 40 Closedloop Transfer Function: M (s) H =
l 40 s
4 + 23 s 3 + 62
l
2 s 2 + 40 s + 40
l
3 = 1 1+ s + 1. 55 s l 2 + 0 . 575 = 0 . 025 s 3 + 0 . 025 s 4 1 =1 = 1. 55 = 0 . 575 4 Secondorder System Approximation:
M (s) L = 1 1+ d s
1 + d2s 2 e e Thus, 2 = = f f 2 = 2 d2  d1 =
2 2 2l 2  l 1 = 3 .1  1 =
2 2 2 .1 4 4 = d 2 = 2 l 4  2 l1 l3 + l 2 = 0 . 05  1.15 + 2 .4025 = 1. 3025
d d
2 2 2 1 = 1. 3025 = 2 d2 
f
2 d 2 = 1.1413
1 = 2 1.1413  2 . 1 = 0 .1825 =
0 .8762 s
2 d = 0 .4273
G L ( s) M L (s) =
1 1 + 0 .4273 s + 1.1413 s 2 + 0 . 3744 s + 0 .8762 = 0 .8762 s( s + 0 . 3744 ) Roots of Characteristic Equations: Fourth order System Secondorder System 2.5692 19.994 0.2183 + j0.855 0.2183  j0.855
Thirdorder system Approximation:
M e2 e4 e
6 0.1872 + j0.9172 0.1872  j0.9172 L (s) =
1 1 + d 1s + d 2 s + d 3 s
2 3 = = = f2 f4 f
6 = 2 d2  d1 =
2 2l2
2  l 1 = 3 .1  1 =
2 2 .1
2 = 2 d 1 d 3 + d 2 = 1. 0062
2 2 d1 + d 2 = 2 l 4  2 l1 l 3 + l 2 = 1. 3025
2 =  d 3 = 2 l 2 l 4  l3 =  0 .2531 Equations to be Solved Simultaneously:
2d d
2 2 2  d 1 = 2 .1
2 d
2 2 2 = 1. 00623 d 1 = 1. 3025
d
4 1 Thus d
2 1 d 2 = 0 .5 d 1 + 1. 05
2 1 = 0 .25 d 4 1 + 1. 05 d 1 + 1.1025 Th us + 4.2  4. 0249 d  0 .8 = 0 115 The roots of the last equation are: d 1 =  0 .1688
d
1 , 0 . 9525 , .  0 .392 +
d d j 2 .196 ,  0 . 392  j 2 .196 Selecting the positive real solution, we have d d
2 2 2 3 = 0 .9525 = 1. 00623 = 0 .2531
(s) d 1 + 1.3025 = 2 .261 Th us T hus 2 = 1. 5037 = 0 . 5031
1. 9876 3 M L =
1 1 + 0 . 9525 s + 1. 5037 s 2 + 0 . 5031 s 3 = s 3 + 2 . 9886 s 2 + 1. 8932 s + 1.9876 GL ( s) = s s + 2.9886 s + 1.8932
2 ( 1.9876 ) = 1.9876 s ( s + 2.0772)( s + 0.9114) 2.5692 19.994 0.2183 + j0.855 0.2183  j0.855
Unitstep Responses Roots of Characteristic Equati ons: Fourth order System 2.552 0.2183 j 0 .8551 Thirdorder System 116 Chapter 8
81 (a)
P (s)
4 3 ROOT LOCUS TECHNIQUE
2 = s + 4 s + 4 s +8 s
0, Q (s) = s +1 Finite zeros of P( s): Finite zeros of Q( s): 3.5098, 0.24512 j1.4897
60
o 1
, 180
o Asymptotes: K > 0: Intersect of Asymptotes: , 300 o K < 0: 0 , o 120 o , 240 o 1 = (b)
P( s) 3 . 5  0 .24512  0 .24512  ( 1 )
4 1 = 1 = s + 5s + s
3 2 Q (s) 0, = s +1 Finite zeros of P( s) : Finite zeros of Q( s): 1 4.7912, 0.20871
90
o Asymptotes: K > 0: Intersect of Asymptotes: , 270 o K < 0: 0 , o 180 o 1 = (c)
P( s) 4. 7913  0 .2087  (  1)
3 1 = 2 =s 2 Q (s) = s + 3s + 2s +8
3 2 Finite zeros of P( s): Finite zeros of Q( s): Asymptotes: 0, 0  3.156 , K > 0: 0 . 083156 180
o j 1. 5874 K < 0: 0
o (d) P (s) = s + 2s + 3s
3 2 Q(s ) = s  1
2 ( ) ( s + 3) Finite zeros of P( s): Finite zeros of Q( s): Asymptotes: 1 1, 1, 3
0, j 1.414 There are no asymptotes, since the number of zeros of P( s) and Q( s) are equal.
Q (s) 0, 0, (e) P (s) = s + 2 s + 3s
5 4 3 = s +3s + 5
2 Finite zeros of P( s): Finite zeros of Q( s): 0,
o  1.5 1 , j 1.414
o o j 1. 6583 60 180 , 300 K < 0: 0 ,
o Asymptotes: K > 0: Intersect of Asymptotes: 120 o , 240 o 1 = (f)
P (s) 1  1  ( 1. 5)  ( 1. 5)
5 2 =
1 3 = s + 2 s + 10
4 2 Q (s) = s +5
j 1.4426 ,
o o Finite zeros of P( s): Finite zeros of Q( s):  1.0398 5 1. 0398 o j1.4426 K < 0:
o o o Asymptotes: K > 0: Intersect of Asymptotes: 60 , 180 , 300 0 , 120 , 240 120 1 = 82 (a) Angles of departure and arrival.
K > 0: 1. 0398  1. 0398 + 1. 0398 + 1. 0398  (  5)
4 1 = 5
3  1  2  3 + 4 = 180  1  90  45 + 90 1 = 135
o o o o o o o o o = 180 K < 0:  1  90  45 + 90 1 = 45
o =0 o (b) Angles of departure and arrival.
K > 0: K < 0:  1  2  3 + 4 = 180  1  135
o o  90 + 90
o o o =0 o 1 = 135 (c) Angle of departure:
K > 0:  1  2  3 + 4 = 180  1  135 1 = 90
o o o  90
o o  45 o = 180 (d) Angle of departure
K > 0:  1  2  3  4 = 180  1  135
o o o  135
o o  90 o = 180 1 = 180 121 (e) Angle of arrival
K < 0: 1 + 6  2  3  4  5 = 360 1 + 90  135 1 = 108
.435
o o o o  135 o  45  26 . 565 o = 360 o o 83 (a) (b) (c) (d) 122 84 (a) Breakawaypoint Equation:
Breakaway Points: 2s 5 + 20 s 4 + 74 s 3 + 110 s 2 + 48 s = 0  0 . 7275
6 ,
4  2 . 3887 + 100
s
3 (b) Breakawaypoint Equation: 3 s + 22
Breakaway Points: s 5 + 65 s + 86 s 2 + 44 s + 12 = 0  1,  2 . 5
6 (c) Breakawaypoint Equation: 3 s + 54
Breakaway Points:
6 s 5 + 347 .5 s 4 + 925 s 3 + 867 .2 s 2  781 .25 s  1953 = 0  2 .5 ,
5 1. 09 s
4 (d) Breakawaypoint Equation:  s  8 s  19
Breakaway Points: + 8 s + 94
3 s 2 + 120 s + 48 = 0  0 . 6428 +8) , 2 .1208 85 (a)
G ( s)H ( s) =
o K(s s(s + 5)( s + 6 )
270
o Asymptotes: K > 0: Intersect of Asymptotes: 90 and K < 0: 0 o and 180 o 1 =
Breakawaypoint Equation:
2s
3 0  5  6  ( 8 )
3 s ,
2 1
s =  1.5 + 35 + 176 + 240 = 0  9 . 7098 Breakaway Points: Root Locus Diagram:  2 .2178  5 . 5724, 85 (b)
123 G ( s)H ( s) =
o K s(s , + 1 )( s + 3)( s + 4 )
o Asymptotes: K > 0: Intersect of Asymptotes: 45 135 , 225 0 o , 315 o K < 0: 0 , o 90 o , 180 o , 270 o 1 =
Breakawaypoint Equation: Breakaway Points: Root Locus Diagram:
3 1 3  4
4
2 = 2 4 s + 24 s 0 .4189 , + 38 s + 12 = 0  2 ,  3 . 5811 85 (c)
G ( s)H ( s) = K(s s (s
2 +4) + 2)
o
2 Asymptotes: K > 0: 60 , Intersect of Asymptotes: 180 o , 300 0 o K < 0: 0 , o 120 o , 240 o 1 =
Breakawaypoint Equation: Breakaway Points: Root Locus Diagram:
3s 0,
4 + 0  2  2  ( 4 )
4
3 1
2 =0 + 24 s + 52 s + 32 s = 0  1. 085 ,  2 ,  4. 915 124 85 (d)
G ( s)H ( s) = K(s s( s
2 + 2) + 2s + 2)
o Asymptotes: K > 0: 90 , Intersect of Asymptotes: 270 o K < 0: 0 0 , o 180 o 1 =
Breakawaypoint Equation: Breakaway Points:
2s
3  1  j  1  j  ( 2 )
3
2 1 =0 +8 s +8 s + 4 = 0
The other two solutions are not breakaway points.  2 .8393 Root Locus Diagram 125 85 (e) G ( s) H( s) = s s + 2s + 2
2 ( K ( s + 5)
o )
K < 0: 0 ,
o Asymptotes: K > 0: 90 , Intersect of Asymptotes: 270 o 180 o 1 =
Breakawaypoint Equation: Breakaway Points:
3 0  1  j  1  j  ( 5)
3
2 1 = 1. 5 2 s + 17 s  7.2091 + 20 s + 10 = 0
The other two solutions are not breakaway points. 85 (f) G ( s) H( s) = s ( s + 4 ) s + 2s + 2
2 ( K )
126 Asymptotes: K > 0: 45 , Intersect of Asymptotes: o 135 o , 225 0 o , 315 o K < 0: 0 , o 90 o , 180 o , 270 o 1 =
Breakawaypoint Equation: Breakaway Point:
3  1  j 1 + j  4
4
2 = 1. 5 4 s + 18 s 3 . 0922 + 20 s + 8 = 0
The other solutons are not breakaway points. 85 (g)
G ( s)H ( s) = K(s s (s
2 + 4) + 8)
90
o 2 2 Asymptotes: K > 0: , 270 o K < 0: 0 , o 180 o Intesect of Asymptotes: 1 =
Breakawaypoint Equation: Breakaway Points:
0, s
5 0 + 0  8  8  ( 4 )  ( 4 )
4 s
4 2
3 + 20 + 160 s + 640 s 2 + 1040 s =0 4, 8, 4  j4, 4 + j4 127 85 (h)
G ( s)H ( s) = K s (s
o
2 + 8) 2 Asymptotes: K > 0: 45 , Intersect of Asymptotes: 135 o , 225 o , 315 o K < 0: 0 , o 90 o , 180 o , 270 o 1 =
Breakawaypoint Equation: Breakaway Point:
s 0,
3 8  8
4
2 = 4
s + 12 s + 32 4, 8 =0 85 (i)
128 G ( s) H( s) = K s + 8 s + 20
2 ( )
0 ,
o s ( s + 8)
2 2 Asymptotes: K > 0: 90 , Intersect of Asymptotes: o 270 o K < 0: 180 o 1 =
Breakawaypoint Equation: Breakaway Points:
s
5 8  8  (  4 )  ( 4 )
4 s
4 2
s
3 = 4 + 1280
s + 20 + 128 + 736 s 2 =0  4, 8,  4 + j 4. 9 , 4  j 4. 9 (j) G ( s) H( s) = (s Ks
2 2 4 ) Since the number of finite poles and zeros of G ( s ) H ( s ) are the same, there are no asymptotes. Breakawaypoint Equation: 8 s = 0 Breakaway Points:
s=0 129 85 (k) G ( s) H( s) =
Asymptotes: (s K s 4
2 2 ( +1
90 )( s
o ) 2 +4
270 )
K < 0: 0 ,
o K > 0: , o 180 o Intersect of Asymptotes: 1 = 2 + 2
4 2
s 0,
6 =0  8 s  24
4 Breakawaypoint Equation: Breakaway Points: s 3 .2132 , =0  3 .2132
2 , j 1. 5246 ,  j1. 5246 85 (l)
130 G ( s) H( s) =
Asymptotes: K > 0: (s K ( s  1)
2 2 +1
90 )( s
o 2 +4
270 )
K < 0: 0 ,
o , o 180 o Intersect of Asymptotes: 1 =
Breakawaypoint Equation: Breakaway Points:
s
5 1 + 1
4 2
3 =0  2s  9 s = 0
2 . 07 ,  2 . 07 ,  j 1.47 , j 1.47 (m)
G ( s)H ( s) = K(s + 1 )( s + 2 )( s + 3)
s (s
3  1)
131 Asymptotes: K > 0: 180
6 o K < 0:
5 4 0 s
2 o Breakawaypoint Equation: s + 12 s + 27 s Breakaway Points: + 2 s  18
3 =0
0.683, 0, 0  1.21, 2.4, 9.07, (n)
G ( s)H ( s) = K (s s (s
3 + 5)( s + 40
)( s ) )
o + 250
60
o + 1000
o Asymptotes: K > 0: , 180 , 300 K < 0: 0 , o 120 o , 240 o 132 Intersect of asymptotes: 1 =
Breakawaypoint Equation: 3750 s Breakaway Points:
6 0 + 0 + 0  250  1000  ( 5)  ( 40
5 s 0
5 ) 2
8 =  401 . 67
10 + 335000
0, + 5 .247 10 s 4 + 2 . 9375 10 s 3 + 1. 875 10 11 s 2 =0  7.288, 712.2, 85 (o)
G ( s)H ( s) = K(s s( s  1) + 1 )( s + 2 )
90
o Asymptotes: K > 0: , 270 o K < 0: 0 , o 180 o Intersect of Asymptotes: 133 1 =
Breakawaypoint Equation: Breakaway Points;
s
3 1  2  1
3 1 = 2 3 s 1 = 0
1.879 0.3473, 1.532, 86 (a) Q(s ) = s + 5 P (s ) = s s + 3s + 2 = s (s +1)( s + 2)
2 ( ) Asymptotes: K > 0: 90 , Intersect of Asymptotes: o 270 o K < 0: 0 , o 180 o 1 =
Breakawaypoint Equation:
s
3 1  2  ( 5)
3
2 1 =1 + 9 s + 15 s + 5 = 0
134 Breakaway Points: 0.4475, 1.609, 6.9434 86 (b) Q (s) = s +3 P (s) = s s 2 + s+2
o o Asymptotes: K > 0: 90 , 270 K < 0: 0 , o 180 o Intersect of Asymptotes: 1 =
Breakawaypoint Equation: Breakaway Points:
s
3 1  ( 3)
3
2 1 =1 +5s + 3s + 3 = 0
The other solutions are not breakaway points.  4.4798 135 86 (c) Q (s) = 5s P (s) = s 2 + 10
180
o Asymptotes: K > 0: K < 0: 5s
2 0 o Breakawaypoint Equation: Breakaway Points:  50 = 0
3.162  3.162, 136 86 (d) Q(s ) = s s + s + 2
2 ( ) P( s) = s + 3s + s + 5s +10
4 3 2 Asymptotes: K > 0: 180 o K < 0: s
6 5 4 0 o Breakawaypoint Equation: Breakaway Points: + 2 s + 8 s + 2 s  33
3 s 2  20 s  20 = 0 2, 1.784. The other solutions are not breakaway points. 137 86 (e) Q(s ) = s 1
2 ( ) ( s + 2) P( s) = s s + 2 s + 2
2 ( ) Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes. Breakawaypoint Equation: Breakaway Points: 6s 3 + 12 s 2 + 8s + 4 = 0 1.3848 138 86 (f) Q(s ) = (s + 1)(s + 4)
Asymptotes:
K > 0: P (s ) = s s  2
2 ( )
K < 0: 0 s
2 180 o o Breakawaypoint equations: Breakaway Points: s 4 + 10 s 3 + 14
0.623 8 = 0 8.334, 139 86 (g) Q(s ) = s + 4s + 5
2 P( s) = s
90
o 2 (s
270 2 + 8 s + 16
o )
K < 0: 0 ,
o Asymptotes: K > 0: , 180 o Intersect of Asymptotes: 1 =
Breakawaypoint Equation: Breakaway Points:
s
5 8  ( 4 )
4 s
4 2 + 42  4,
140
s = 2
3 + 10 + 92 2 + s 2 + 80 s =0 2 
j 2 .45 0,  2, j 2 .45 , 86 (h) Q(s ) = s  2
2 ( )( s + 4) P (s ) = s s + 2s + 2
2 ( ) Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes. Breakaway Points:  2, 6.95 141 86 (i) Q (s) = ( s + 2 )( s + 0 . 5)
K > 0: 180
o P( s) =s 
s
2 1 Asymptotes: K < 0: s
4 3 2 0 o Breakawaypoint Equation: Breakaway Points: +5s + 4s 1 = 0 4.0205, 0.40245 The other solutions are not breakaway points. 142 86 (j) Q(s ) = 2 s + 5
Asymptote s:
K > 0: P (s ) = s
60
o 2 (s
, 2 +2s +1 = s
o ) 2 ( s + 1) 2 , 180 o 300 K < 0: 0 , o 120 o , 240 o Intersect of Asymptotes; 1 =
Breakawaypoint Equation:
6s
4 0 + 0  1  1  (  2 . 5)
4 s
3 1
s
2 = =0 0 .5 3 = 0 .167 + 28 + 32
143 + 10 s Breakaway Points: 0, 0.5316, 1, 3.135 87 (a) Asymptotes: K > 0: 45 o , 135 o , 225 o , 315 o Intersect of Asymptotes: 1 =
Breakawaypoint Equation: Breakaway Points:
When 4s
5 2  2  5  6  ( 4 )
5 s
4 1
s
3 =  2 . 75
s
2 + 65 + 396 5.511 + 1100 + 1312 s + 480 = 0 = 0 . 707 0.6325,
K = 13.07 (on the RL) , 144 87 (b) Asymptotes: K > 0: 45 o , 135 o , 225 0 o , 315 o Intersect of Asymptotes: 1 =
Breakawaypoint Equation: When = 0 . 707 , K = 61.5
4s
3  2  5  10
4
2 = 4.25 + 100 = 0 + 51 s + 160 s 145 87 (c) Asymptotes: K > 0: 180 o Breakawaypoint Equation: Breakaway Points:
When s 4 + 4 s + 10
3 s 2 + 300 s + 500 = 0 = 0 . 707 1.727
K = 9.65 (on the RL) , 146 87 (d) K > 0: 90 o , 270 o Intersect of Asymptotes: 1 =
When 2  2  5  6
4 2 = 7 . 5 = 0 . 707 , K = 8.4 147 88 (a) Asymptotes: K > 0: 60 o , 180 o , 300 0 o Intersect of Asymptotes: 1 =
2  10  20
3 = 10
Breakaway Point: (RL) Breakawaypoint Equation: 3 s + 60 s + 200 = 0 4.2265, K = 384.9 148 (b) Asymptotes: K > 0: 45 o , 135 o , 225 0 o , 315 o Intersect of Asymptotes: 1 =
Breakawaypoint Equation: 4 s + 27 s Breakaway Points: (RL) 0.4258
3 2 1  3  5
4 = 2 .25 4.2537
K = 12.95 + 46 s + 15 = 0
K = 2.879 , 149 150 810 P (s) = s + 25
3 s 2 + 2 s + 100
K
t Q (s) 90
o = 100
o s Asymptotes: > 0: , 270 Intersect of Asymptotes: 1 =
Breakawaypoint Equation: Breakaway Points: (RL)
s
3 25  0
3 1
2 = 12 . 5 + 12 . 5 s  50 = 0 12.162 2.2037, 811 Characteristic e quation: s 3 +5s + K ts + K = 0
2 151 (a) Kt = 0 : P (s ) = s 2 ( s + 5)
o Q (s) = 1
180
o Asymptotes: K > 0: 60 , Intersect of Asymptotes: , 300 o 1 =
Breakawaypoint Equation:
3s
2 5  0
3 s = 1. 667
Breakaway Points: + 10 =0 0, 3.333 811 (b) P (s) = s + 5 s + 10 = 0
3 2 Q (s)
o = s Asymptotes: K > 0: 90 , Intersect of Asymptotes: o 270 152 1 =
3 5  0
2 1 =0 Breakawaypoint Equation: 2 s + 5 s  10 = 0 There are no breakaway points on RL. 812 P( s) = s + 116
2 . 84 s + 1843
J
L Q (s) 180
o
4 = 2 . 05 s ( s 2 + 5)
s
2 Asymptotes: = 0: Breakawaypoint Equation:  2 . 05 s Breakaway Points: (RL)  479
0, s 3  12532  37782 s =0 204.18
153 813 (a) P (s ) = s s  1
2 ( ) Q (s ) = ( s+ 5)( s + 3)
K > 0: 180 s
4 Asymptotes: o Breakawaypoint Equation: Breakaway Points: (RL) + 16 s 3 + 46 s 2  15 = 0 0.5239, 12.254 154 813 (b) P (s ) = s s + 10s + 29
2 ( ) Q (s ) = 10(s + 3)
o Asymptote s: K > 0: 90 , Intersect of Asymptotes: 270 o 1 =
3 0  10  ( 3)
3 1
2 = 3 . 5
s Breakawaypoint Equation: 20 s + 190 s There are no breakaway points on the RL. + 600 + 870 = 0 155 814 (a) P( s) = s ( s + 12 . 5)( s + 1 ) Q (s) = 83 .333
60
o Asymptotes: N > 0: Intersect of Asymptotes: , 180 o , 300 0 o 1 =
Breakawaypoint Equation: 3 s Breakaway Point: (RL)  12 . 5  1
3 =  4. 5 + 27 0.4896
2 s _12 .5 =0 156 814 (b) P( s) = s + 12 . 5 s + 833
2 .333 Q (s) = 0 .02 s (s 2 + 12 . 5) A > 0: 180 o
4 Breakawaypoint Equation: 0 . 02 s Breakaway Points: (RL) 0 + 0 . 5 s + 53 .125
3 s 2 + 416 . 67 s =0 157 814 (c) P (s) = s + 12 . 5 s + 1666 . 67 = ( s + 17 . 78 Q ( s ) = 0 . 02 s ( s + 12 . 5)
3 2 )( s  2 . 64 + j 9 . 3)( s  2 . 64  j 9 . 3) Asymptotes: K o > 0: 180 o Breakawaypoint Equation: 0 . 02 s + 0 . 5 s + 3 .125 s Breakaway Point: (RL) 5.797
4 3 2  66 .67 s  416 . 67 =0 158 815 (a) A = K o = 100 : P (s) = s ( s + 12 . 5)( s + 1 )
60
o Q (s)
o = 41 . 67 Asymptotes: N > 0: Intersect of Asymptotes: 180 0 o 300 1 =
Breakawaypoint Equation: Breakaway Points: (RL)
3s
2  1  12 . 5
3 =  4. 5 + 27 s + 12 . 5 = 0 0.4896 159 815 (b) P( s) = s + 12 . 5 + 1666
2 . 67 = ( s + 6 .25 + j 40 . 34 )( s + 6 .25  j 40 . 34 ) Q (s) = 0 . 02 s (s 2 + 12 . 5)
180
o Asymptotes: A > 0: Breakawaypoint Equation: Breakaway Points: (RL) 0 . 02 s 0 4 + 0 . 5 s + 103
3 .13 s 2 + 833 . 33 s =0 160 815 (c) P (s) = s + 12 . 5 s + 833
3 2 . 33 = ( s + 15 . 83)( s  1. 663 + j 7 . 063)( s  1. 663  j 7 .063) Q (s) = 0 . 01 s ( s + 12 . 5)
K
o Asymptotes: > 0: 180 o Breakawaypoint Equation: Breakaway Point: (RL) 0 . 01 s 4 + 0 .15 s 3 + 1. 5625 s 2  16 . 67 s  104. 17 =0 5.37 161 816 (a) P (s) =s 2 (s + 1 )( s + 5) Q (s) =1
45
o Asymptotes: K > 0: Intersect of Asymptotes: , 135 o , 225 0 o , 315 o 1 =
Breakawaypoint Equation:
4s
3 +0 1 5
4 s = 1. 5
Breakaway point: (RL) 0, 3.851 + 18 s 2 + 10 =0 162 (b) P (s) =s 2 (s + 1 )( s + 5) Q (s) = 5s +1
60
o Asymptotes: K > 0: Intersect of Asymptotes: , 0 180 o , 300 o 1 =
4 + 0  1  5  ( 0 .2 )
4
3 1
2 = 5. 8 3 = 1. 93 Breakawaypoint Equation: 15 s + 64 s + 43 s + 10 s = 0 Breakaway Points: (RL) 3.5026 817 P( s) Q (s) =s = 10 2 (s s + 1 )( s + 5) + 10 = ( s + 4. 893)(
T
d s + 1.896
o )( s  0 . 394 + j 0 .96 )( s  0 . 394 + j 0 . 96 ) Asymptotes: > 0: 60 o , 180 o , 300 Intersection of Asymptotes: There are no breakaway points on the RL. 1 = 4. 893  1. 896 + 0 . 3944 + 0 . 3944
4 1 = 2 163 818 (a) K = 1: P (s) =s 3 (s + 117 .23)( s K
L + 4882 .8 ) 90
o Q (s) , 270 .23
o = 1010( s + 1. 5948 )( s + 114.41 )( s + 4884 ) Asymptotes: > 0: Intersect of Asymptotes: 1 =
Breakaway Point: (RL)
0 117  4882 .8 + 1. 5948 + 114.41 + 4884
5 3 = 0 .126 818 (b) K = 1000: P( s) Q (s) =s 3 (s + 117
s s
o
3 .23)( s s + 4882
2 .8 )
5 = 1010( = 1010( + 5000
o + 5 . 6673 10 + 39 .18 + s + 891089 110 ) .7 ) + 4921
270 . 6 )( s j 423 . 7 )( s + 39 .18  423 Asymptotes: K L > 0: 90 , Intersect of Asymptotes: 1 = 117 .23  4882 .8 + 4921
5 .6 + 39 .18 + 39 .18 3 = 0 . 033 164 Breakawaypoint Equation: + 5.279 10 Breakaway points: (RL) 0, 87.576
2020 s
7 7 + 2 .02 10 s 6 10 s 5 + 1. 5977 10 13 s 4 + 1. 8655 10 16 s 3 + 1. 54455 10 18 s 2 =0 819 Characteristic Equation:
P( s)
3 2 s 3 + 5000 s 2 + 572 , 400 s + 900
s ,000 +JL
. 6 )( s 10 s 3 + 50 ,000
.8 ) s 2 = 0 = s + 5000 s + 572 , 400 s + 900 ,000 = ( s + 1. 5945)( Since the pole at 5000 is very close to the zero at 4882.8, P ( s ) and
P( s) + 115 + 4882 Q (s) = 10 s (s 2 + 5000 ) Q ( s ) can be approximated as: ( s + 1. 5945)( s + 115 .6 )
2 Q ( s) 10 .24
s s 2 Breakawaypoint Equation: 1200 s + 3775 =0 Breakaway Points: (RL): 0, 3.146 165 820 (a) = 12 : P ( s) = s (s 2 + 12 )
o Q (s) 270 = s +1
o Asymptotes: K > 0: 90 , Intersect of Asymptotes: K < 0: 0 0 , o 180 o 1 =
Breakawaypoint Equation:
2s
3 + 0  12  ( 1 )
3
2 1
s = 5 . 5
Breakaway Points:
0, + 15 s + 24 =0 2.314, 5.186 820 (b) = 4 : P( s) =s 2 (s +4)
o Q (s) 270
o = s +1
o Asymptotes: K > 0: 90 , Intersect of Asymptotes: K < 0: 0 , 180 o 1 =
Breakawaypoint Equation: 2 s + 7 s
3 2 0 + 0  4  (  10
3 1 =  1. 5 +8 s = 0 Breakaway Points: K > 0 0. None for K < 0. 166 (c) Breakawaypoint Equation:
2 2s 2 + ( + 3) s + 2 s = 0 Solutions: s =
The +3
4 ( + 3)  16 2 , s =0 4 For one nonzero breakaway point, the quantity under the squareroot sign must equal zero. Thus,  10 + 9 = 0 , = 1 or = 9 . The a nswer is cancellation in the equivalent G ( s ). When = 9, = 9. the nonzero breakaway point is at s = 1 solution represents polezero =  3. 1 = 4. 821 (a) P (s) =s 2 (s + 3) Q ( s) = s +
3 Breakawaypoint Equation: 2 s + 3 (1 + ) s + 6 = 0 The roots of the breakawaypoint equation are: s = 3 ( 1 + )
4 9( 1 + )  48 2 4 167 For no breakaway point other than at s = 0 , set 9( 1 + ) 2  48 < 0 or  0 . 333 < < 3 Root Locus Diagram with No Breakaway Point other than at s = 0. 821 (b) One breakaway point other than at s = 0: = 0 . 333 , Breaka way po int at s = 1. 168 821 (d) Two breakaway points : > 3: 169 822 Let the angle of the vector drawn from the zero at s be = j12 to a point s on the root locuss near the zero
1 . 1 = 2 = 3 = 4 = Let angle angle angle angle of th e vect or dra wn fro m the of th e vect or dra wn fro m the pole a t j 10 to s .
1 pole a t 0 to s .
1 of th e vect or dra wn fro m the of th e vect or dra wn fro m the pole a t zero a t   j 10 to s1 . j 12 to s .
1 Then the angle conditions on the root loci are: = 1  2  3 + 4 = 1 = 2 = 3 = 4 = 90
o odd m ultipl es of Thus, 180 = 0
o o The root loci shown in (b) are the correct ones. 170 Chapter 9 FREQUENCY DOMAIN ANALYSIS 91 (a) K=5 n = n = 5 = 2 .24 rad / sec = 4.48 6.54 = 1.46 = 0 . 707 M r =1 =
2 r = 0
1 1 rad / sec (b) K = 21.39 21 . 39 = 4. 62
1 rad / sec = 9.24 6.54 M r =1 
2 r = n (c) 92 (a) (b) (c) (d) (e) (f) (g) (h) 93
Maximu M m over shoot 1 2 1
2 2 = 3 .27
= 20 6.54 rad / sec K = 100 n = 10
( 9 . 38 dB) rad / sec = 0 . 327 M r = 1. 618 r = 9 .45 ra d / sec M M M M M M r = 2 . 944 = 15 . 34 = 4.17 =1 r = 3 rad r =4 / sec BW = 4.495 rad / sec r ( 23 . 71 dB) rad / sec BW = 6.223 rad / sec r ( 12 .4 dB) r = 6 .25 ra
rad / sec d / sec BW = 9.18 r ad / sec r ( 0 dB) r =0 BW = 0.46 r ad / sec r = 1. 57 ( 3 . 918 dB) r = 0 .82 r = 1. 5 rad r = 1.25 ra rad / sec BW = 1.12 r ad / sec r = ( unstab = 3 . 09 = 4.12 le) / sec BW = 2.44 r ad / sec M M r ( 9 . 8 dB) d / sec BW = 2.07 r ad / sec r ( 12 . 3 dB) r = 3 . 5 rad / sec BW = 5.16 r ad / sec = 0.1 t T hus, = 0.59 r = = 1. 05 n = 17 . 7 n r = 1  0 .416 + 2 . 917 n
M 2 = 0 .1 sec Thus, minimum rad / sec Maxi mum r = 1. 05 = 20.56 rad/sec Minimum BW = ( ( 1  2 2 )+ 4  4 + 2
4 2 ) 1/2 94
Maximu m over shoot = 0.2 Thus, 0 .2  =e 1  2 = 0.456 171 M r =
2 1 1
2 = 1.232 t r = 1  0 .416 + 2 . 917 n 2 = 0 .2 Thus, minimum n = 14. 168 rad/sec Maximum M r = 1.232 Minimum BW = ( (1  2
=e 2 )+
2 4  4 + 2
4 2 ) 1/2 = 18.7 rad/sec 95
Maximum overshoot = 0.3 1 2 1
2  Thus, 0 . 3 1  0 .416 1  = 0 . 358
2 M r = = 1.496 t r = + 2 . 917 n = 0 .2 Thus, minimum n = 6 .1246 rad/sec Maximum M r = 1.496 Minimum BW = ( (1  2 2 )+ 4  4 + 2
4 2 )
2 1/2 = 1.4106 rad/sec 96
M 
r = 1.4 =
2 1 Thus, 1
2 = 0.387 Maximum overshoot = e 1  = 0.2675 (26.75%) r = 3 rad
t / sec = n =
2 1  2 2 = 0 .8367 n rad/sec n = 3 0 . 8367 = 3 . 586 r ad/sec max = n
1 3 . 586 1   ( 0 . 387 = 0 .95
)
2 sec At = 0, M = 0 .9 . This indicates that the steadystate value of the unitstep response is 0.9. Unitstep Response: 97
T BW (rad/sec) Mr ________________________________________________________________ 0 0.5 1.0 2.0 1.14 1.17 1.26 1.63 1.54 1.09 1.00 1.09 172 3.0 1.96 4.0 2.26 5.0 2.52 _________________________________________________________________ 1.29 1.46 1.63 98
T BW (rad/sec) Mr _________________________________________________________________ 0 1.14 0.5 1.00 1.0 0.90 2.0 0.74 3.0 0.63 4.0 0.55 5.0 0.50 _________________________________________________________________ 1.54 2.32 2.65 2.91 3.18 3.37 3.62 99 (a)
L( s ) = 20 s ( 1 + 0 . 1 s )( 1 + 0 .5 s ) P = 1, P =0 When = 0: L ( j ) = 90 o
20 L( j ) =
2 When = : L ( j ) = 270 o
2 L( j ) =0
=0 L ( j ) = 0.6 + j 1  0.05
2 2 ( 2 ) = 20 0.6  j 1  0.05 0.36 + 4 2 2 (1  0.05 ) ( ) 2 Setting Im L ( j ) 1  0 .05 = 0
o Thus , = 4.47 rad / sec L ( j 4.47 ) = 1. 667 360 180
o o 11 = 270 = ( Z  0.5 P  P ) 180 = ( Z  0.5 )180
o o Thus, Z = =2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. Nyquist Plot of L ( j ): 173 (b)
L( s ) = 10 s ( 1 + 0 . 1 s )( 1 + 0 .5 s ) Based on the analysis conducted in part (a), the intersect of the negative real axis by the L ( j ) plot is at 0.8333, and the corresponding P is 4.47 rad/sec.
o 11 = 90 o = 
Z 0 . 5 P 180 o = 180 Z  90 Thus, Z = 0 . The closedloop system is stable. Nyquist Plot of L ( j ): (c)
L( s ) = 100( 1 + s ) s ( 1 + 0 . 1 s )( 1 + 0 .2 s )( 1 + 0 . 5 s )
L( j0 )
o P = 1, P = 0. L ( j ) = 270
o When When = 0 : L ( j 0 ) = 90 o = : L ( j ) = 270 = When = :
When L ( j )
o =0
L( j ) L( j ) =0 = : L ( j ) = 270 =0 174 L ( j ) = ( 0.01 100(1 + j )
4  0.8 2 ) + j (1  0.17 )
2 4 = 100(1 + j ) 0.01  0.8 4 ( 0.01 ( 4  0.8 2 )  j (1  0.17 ) ) + ( 1  0.17 )
2 2 2 2 2 2 Setting Im L ( j ) Thus, =0 0 .01  0 .8 2  1 + 0 .17 2 = 0 4  63 2  100 = 0 = 64. 55
2 = 8 . 03 rad/sec 100 ( 0.01 4  0.8 2 ) + 2 (1  0.17 2 ) L ( j 8.03) = =  10 ( 0.01 2  0.8 2 )2 + 2 (1  0.17 2 ) 2 =8.03 11 = 270 = ( Z  0.5 P  P ) 180 = ( Z  0.5 )180
o o o Thus, Z = 2 The closedloop system is unstable.
The characteristic equation has two roots in the righthalf splane. Nyquist Plot of L ( j ): (d)
L( s ) = 10 s ( 1 + 0 .2 s )( 1 + 0 . 5 s )
2 P =2 P =0 When = 0: L ( j ) = 180 o
10
4 L ( j ) =
4 2 When
3 = : L ( j ) = 360 o L ( j ) =0 L ( j ) = ( 0.1  2 )  j 0.7 3 = 10 0.1  + j0.7 ( 0.1 ( 4  2 ) ) 2 + 0.49 6 Setting Im L ( j ) origin where = . = 0 , = . The Nyquist plot of L ( j ) does not intersect the real axis except at the 175 11 = ( Z  0.5 P  P )180 = ( Z  1) 180 o o Thus, Z = 2 . The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. Nyquist Plot of L ( j ): 99 (e) L(s ) = s s + 3s + 1
3 ( 3( s + 2) ) P =1 P=2 When = 0 : L ( j 0 ) = 90 o
3( j + 2)
4 L( j0 ) = When
4 2 = : L ( j ) = 270 o L ( j) =0
= 0, L ( j ) = (  3 2 ) + j
or = 3( j + 2)  3 (4 ( 4  3 2 ) )  j 2 2 + Setting Im L ( j )  3  2 = 0
4 2 = 3 . 56
2
o = 1.89
o rad/sec. L ( j 1.89 )
o =3 11 = ( Z  0.5 P  P )180 = ( Z  2.5 ) 180 =  90 Thus, Z = 2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. Nyquist Plot of L ( j ): 176 99 (f) L(s ) = 0.1 s ( s + 1) s + s + 1
2 ( ) P =1 L( j0 ) P=0 When = 0 : L ( j 0 ) = 90 o
0.1 = When = : L ( j ) = 360 o
2 L ( j) =0
=0 L ( j ) = =
or ( 4  2
2 2 ) + j (1  2 )
2 = 0.1  2 4 (
o ( 4  2 2 )  j (1  2 ) ) + ( 1  2 )
2 2 2 2 2 Settiing Im L ( j ) = 0 .5 = 0 . 707
o rad / sec L ( j 0 . 707 )
o = 0 .1333
The closedloop system is stable. 11 = ( Z  0.5 P  P )180 = ( Z  0.5 ) 180 =  90 Nyquist Plot of L ( j ): Thus, Z = 0 99 (g) L(s ) = 100 s ( s + 1) s + 2
2 ( ) P =3 P=0 177 When = 0 : L ( j 0 ) = 90 o
o L( j0 ) =
rad/sec.
o When = : L ( j ) = 360 o
o o L ( j) =0
o o The phase of L ( j ) is discontinuous at = 1.414 11 = 35.27 + 270  215.27
o ( o ) = 90 11 = ( Z  1.5) 180 = 90 Thus, P 11 = 360 180 = 2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. Nyquist Plot of L ( j ): 99 (h)
L( s ) = 10( s s(s + 10 ) ) + 1 )( s + 100 P =1 P =0 When = 0 : L ( j 0 ) = 90 o
L ( j ) = L( j0 ) =
= When = : L ( j ) = 180 o
2 L ( j) =0 10( j + 10) 101 + j (100  )
2 2 10( j + 10)  101  j 100  10201 + 4 2 (100  )
2 ( 2 ) 2 Setting Im L ( j ) = 0, = 0 is the only solution. Thus, the Nyquist plot of L ( j ) does not intersect the real axis, except at the origin. 11 = ( Z  0.5 P  P )180 = ( Z  0.5 ) 180 =  90 o o o Thus, Z = 0 . The closedloop system is stable. Nyquist Plot of L ( j ): 178 910 (a)
L( s ) =
s(s K + 2 )( s + 10 ) P =1
o P =0
o For stability, Z = 0. 11 = 0 . 5 P 180 = 90 This means that the (1, j0) point must not be enclosed by the Nyquist plot, or 0 < 0 .004167 K Nyquist Plot of L ( j ): < 1. Thus, 0 < K < 240 910 (b)
L( s ) =
s(s K(s + 1) + 2 )( s + 5)( s + 15) 11 = 0 . 5 180
0
o P =1
o P =0 For stability, Z = 0. = 90
9K This means that the (1, j0) must not be enclosed Thus, by the Nyquist plot, or < 0 . 000517 <1 0 < K < 1930.9 Nyquist Plot of L ( j ): 179 910 (c)
L ( j ) = K s (s
2 + 2 )( s + 10 ) P =2 P
o =0 = 180
o For stability, Z = 0. 11 = 0 . 5 P 180 For all K > 0 11 = 180 o , not
o  180
for o . Thus, the system is unstable for all K > 0. For K < 0, the critical point is (1, j0), all K < 0. Thus, the system is unstable for all values of K. 11 = 0 Nyquist Plot of L ( j ): 911 (a)
G (s) = K (s + 5) 2 P =0 P =0 180 G ( j 0 ) = 0
G ( j ) o (K
o > 0) G ( 0 ) = 180 G ( j ) = 0 o (K <0) G ( j0) G ( j ) = K 25 = 180 (K > 0) o (K < 0) =0 11 =  ( 0.5 P + P ) 180 = 0
o For stability, Z = 0. o 0 K < K < 11 = 0 o Stable
o < 25
K 11 = 180 < 0 11 = 0
o Unstable  25 < Stable The system is stable for  25 < K < . 911 (b)
G (s) =
(s K + 5) 3 P =0 P =0
K 125 G ( j 0 ) = 0
G ( j ) o (K
o > 0) G ( 0 ) = 180 o (K
o < 0) G ( j0 ) G ( j ) = =0 = 270 (K > 0) G ( j ) = 270 (K < 0) For stability, Z = 0. 11 =  ( 0.5 P + P ) 180 = 0
o o 0 < K < 1000 K K 11 = 0 o o o Stable Unstable Unstable Stable > 1000 <  125 11 = 360 11 = 180 11 = 0
o 125 < K < 0 The system is stable for  125 < K < 0. 911 (c)
G (s) = K (s + 5) 4 P =P =0
K 625 G ( j 0 ) = 0
G ( j ) o (K > 0) G ( 0 ) = 180 o (K
o < 0) G ( j0 ) G ( j ) = =0 =0 o (K > 0) G ( j ) = 180 (K < 0) 181 For stability, Z = 0. 11 =  ( 0.5 P + P ) 180 = 0
o o 0 K < K < 2500 11 = 0 o o o Stable Unstable Unstable Stable > 2500 <  625 11 = 360 11 = 180 11 = 0
o K 625 < K < 0 The system is stable for  625 < K < 2500. 912 s s + 2s + s +1 + K s + s + 1 = 0
3 2 2 ( Leq ( s) = K s + s +1
2 s s + 2s + s + 1
3 2 ( ( ) ( )
2 ) Leq ( j ) = K (
eq (1  2 4  ) + j = K  ( ) + j (1  2 ) (
2 ) P =1 P =0
6 L eq ( j 0 ) =  90
4 o L eq ( j )
2 = 0 180 o +  4 4 2 )  j (  2 + 1) ) + (1  2 )
2 2 2 2 Setting Im L ( j ) =0 4  2 2 + 1 = 0
Thus, = 1 = K rad/sec are the real solutions. L eq ( j 1) For stability, 11 =  ( 0 . 5 P + P )180 o =  90 o When K = 1 the system is marginally stable. K K >0 <0 11 = 90 11 = +90 o o Stable Unstable Routh Tabulation 182 s s s 4 1 2 K K K +1 +1
K K 3 2 +1
2 K > 1 s 1 K 2  2K +1
K +1 = (K K  1) +1 2 s 0 K
1 K>0
2 When K = 1 the coefficients of the s row are all zero. The auxiliary equation is s + 1 = 0 The solutions are = 1 rad/sec. Thus the Nyquist plot of L eq ( j ) intersects the 1 point when K = 1, when = 1 rad/sec. The system is stable for 0 < K < , except at K = 1.
s 0 913
Parabolic error constant K a = lim s G ( s ) = lim10 ( K P + K D s ) = 10 K P = 100
2 s 0 Thus K P = 10 Characteristic Equation:
10 K s
2 s 2 + 10
P K D s + 100 = 0 G eq ( s) = D s + 100 P =2 =0 For stability, 11 =  ( 0 . 5 P + P )180 o = 180
D o The system is stable for 0 < K < . 914 (a)
1 The characteristic equation is G (s)
2 +G ( s) G ( s)  2 = 1 2
2 G (s) 2 =0
2 G eq ( s ) = 2 G (s) = 2 K
(s + 4) +
G 2 (s + 5)
2 2 P =0 = P =0
2 G eq ( j ) = = ( 400  120
K
2 2 K
2 4 2 K ) + j ( 360  18 )
2 ( 400  120
eq ( 400  120 2 + 2 )  j ( 360  18 2 ) 2 + =0 2 ) + ( 360  18 )
2 2 2 G eq ( j0) 180 o 200 eq ( j ) = 0 180 o Setting Im G ( j ) 183 =0 and = 4.47 r ad / sec G eq ( j 4.47 ) = K 2 800 For stability, 11 =  ( 0.5 P + P ) 180 = 0
o o The system is stable for K or K 2 < 200 < 200 Characteristic Equation:
s
4 + 18 s 3 + 121
s s
4 s 2 + 360 s + 400  2 K 2 =0  2K
2 Routh Tabulation
1 18 101 29160  36 K 101 s
0 2 121 360 400 400 3 s s 2 2K 2 1 29160 + 36 K 2 >0 400 2K 2 K 2 < 200 Thus for stability, K < 200 915 (a)
G (s) = 83 . 33 N s(s Nyquist Plot ) + 2 )( s + 11 . 767 For stability, N < 3.89 Thus N < 3 since N must be an integer. (b)
G (s) =
s ( 0 . 06 s 2500 + 0 . 706 )( As + 100
3 ) Characteristic Equation: 0 . 06 As + ( 6 + 0 . 706 A )s 2 + 70 . 6 s + 2500 = 0 184 G eq (s) = As 6s 2 2 ( 0 . 06 s + 0 . 706 ) + 70 . 6 s + 2500 Since G eq ( s ) has more zeros than poles, we should sketch the Nyquist plot of 1 / G eq ( s ) for stability study. G eq ( 2500  6 ) + j70.6 = ( 2500  6 ) + j70.6 ( 0.706 + j 0.06 ) = ( j ) A ( 0.706  j0.06 ) A ( 0.498 + 0.0036 )
1
2 2 2 3 2 3 4 6 1 / G eq ( j 0 ) =  180
2 o 1 / G eq ( j ) = 0  90
rad/sec o Setting 1 G
eq Im 4.23
A =0 G eq ( j ) 1 100 .156 For stability,  0 . 36 = 0 = 16 . 68 ( j16 . 68 ) = 11 =  ( 0.5 P + P ) 180 = 180
o o For A > 4.23 <
A 11 = 180 o Unstable
o For 0 < 4.23 11 = 180 Stable The system is stable for 0 < A < 4.23. (c)
G (s) = 2500 s ( 0 . 06 s + 0 . 706 )( 50 s +Ko)
s ( 0 . 06 s ) Characteristic Equation:
K s ( 0 . 06 s
o + 0 . 706
P )( 50 s + K o ) + 2500 = 0
G eq ( j 0 ) G eq ( s) = + 0 . 706
2 3s 3 + 35 . 3 s + 2500
3 P =0 =0
2 = 0 90 o G eq ( j ) = 0  90
3 o G eq ( j ) = ( 2500  35.3 )  j3
2 K o ( 0.06 + 0.706 j )
3 = K o  0.06 + 0.706 j ( ) ( 2500  35.5 ) + j 3 ( 2500  35.3 ) + 9
2 2 2 6 Setting Im G G eq ( j 7 .18 ) eq ( j ) =0
Ko + 138
4 .45 2  9805 . 55 =0 = 51 . 6
2 = 7 .18 rad/sec = 0 . 004 185 For stability, 11 =  ( 0.5 P + P ) 180 = 0
o o For stability, 0 < K o < 217 .4 916 (a) K t = 0: G ( s) = Y ( s) E (s ) = s s + 10 s + 10000 Kt
2 ( 10000 K ) = 10000 K s ( s + 10)
2 The (1, j0) point is enclosed for all values of K. The system is unstable for all values of K. (b) K t = 0 . 01 : G ( s) = s s + 10 s + 100
2 ( 10000 K ) G( j ) = 10000 K  10  j 100  2 100 + 4 2 (100  )
2 ( 2 ) 2 186 Setting Im G ( j ) = 0 2 = 100
G ( j 10 ) = 10 rad / sec = 10 K The system is stable for 0 < K < 0.1 (c) K t = 0 .1: G ( s) = s s + 10 s + 1000
2 ( 10000K ) G ( j ) = 10000 K 10  j 1000  2 100 + 4 2 (1000  )
2 ( 2 ) 2 Setting Im G ( j ) = 0 2 = 100 = 31 . 6 rad/sec G ( j 31 . 6 ) = K For stability, 0 < K < 1 917 The characteristic equation for K = 10 is: s G eq ( s )
3 + 10 s 2 + 10 ,000 =0 K s
t + 100 ,000 =0 = 10 ,000 K t s s
3 + 10 s 2 + 100 ,000 P P =2
4 G eq ( j ) = 10,000 K t j 100,000  10  j
2 3 = 10,000 K t  + j 10,000  10 (10,000  10 )
2 ( 2 ) 2 + Setting Im G 6 eq ( j ) =0 187 = 0 , = 10 ,000
2 = 100
For stability, rad / sec G eq ( j 100 ) = Kt 11 =  ( 0.5 P + P ) 180 = 360
o o The system is stable for K > 0.
t 918 (a)
Let G ( s ) = G 1 ( s )e
100
2  Td s Then G1 ( s ) = s s + 10 s + 100 100 ( )
2 Let  10 + j 100  2 ( )
2 =1 or 100 Thus 100 + 4 1 2 The real solution for are = 1 rad/sec. (100  )
2 100 4 + 2 (100  2 )2 6 4 1/2 =1 = 10,000  100 + 10,000  10,000 = 0
2 G1 ( j 1) =  tan Td Equating =1 ( 264.23 =
=
84.23 180 100  2 o 10 = 264.23 =1
o  180 o ) 180 = 1.47 rad Thus the maximum time delay for stability is
T
d = 1.47 sec. (b) T d =1 sec. G ( s) = s s + 10 s + 100
2 ( 100 Ke s ) G( j ) =
rad/sec. 100 Ke
2  j 2  10 + j 100  At the intersect on the negative real axis, = 1.42 ( ) 188 G ( j 1.42 ) = 0 . 7107 K. The system is stable for 0 < K < 1.407 919 (a) K = 0.1 G ( s) = s s + 10 s + 100
2 ( 10 e  Tds ) = G1 (s )e =1 Td s 10
Let
2  10 + j 100  6 4 ( 2 ) or
2 10 100 4 + 2 (100  2 ) 2 The real solutions for 1/2 =1 Thus  100 + 10 ,000  100 = 0
1 is = 0 .1 rad/sec. G1 ( j 0.1) =  tan Td = Equate ( 269.43 100  2 o = 269.43 10 =0.1
o  180 o o ) =0.1 = 1.56 rad We have T 180 d = 15 . 6 sec. We have the maximum time delay for stability is 15.6 sec. 919 (b) T d = 0 .1 sec. G ( s) = s s + 10 s + 100
2 ( 100 Ke 0.1 s ) G( j ) = 100 Ke
2 0.1 j 2  10 + j 100  ( ) At the intersect on the negative real axis, = 6 . 76 rad/sec. G ( j 6 . 76 ) = 0 .1706 K 189 The system is stable for 0 < K < 5.86 920 (a) The transfer function (gain) for the sensoramplifier combination is 10 V/0.1 in = 100 V/in. The velocity of flow of the solution is v = 10 in 3 / sec = 100
d in/sec / v sec. The loop transfer function is 0 .1 in The time delay between the valve and the sensor is T G (s) =D = 100 Ke s
2  Td s + 10 s + 100 190 191 921 (a) The transfer function (gain) for the sensoramplifier combination is 1 V/0.1 in = 10 V/in. The velocity of flow of the solutions is v = 10 in 3 The time delay between the valve and sensor is T G (s) d = 100 in / sec 0 .1 in = D / v sec. The loop transfer function is
T d s / sec = 10 Ke s
2 + 10 s + 100
 j Td (b) K = 10: G ( s) = G1 (s ) e 100
2  Td s G ( j ) = (100  ) + j10
2 2 2 100 e Setting ( 100  ) + j10 =1 (100  ) + 100 = 10,000
2 Thus,  100 = 0
4 2 1 Real s olutio ns: = 0, = 10 rad / sec G1 ( j 10) =  tan
Thus, 10 T Thus, T
d d 10 =  90 o 100  2 =10 o = 20 90 o = 2 rad 180 = = = 0 .157 sec Maximum D vT d = 100 0 .157 = 15 . 7 in (c) D = 10 in. T
d v 100 s + 10 s + 100 The Nyquist plot of G ( j ) intersects the negative real axis at = 12 . 09 rad/sec. G ( j ) For stability, the maximum value of K is 12.94 .
2 = D = 10 = 0 .1 sec G (s) = 10 Ke 0 . 1 s = 0 . 0773 K 922 (a) 192 G H ( s) = s s + 6 s + 12
2 ( 8 )
rad / sec, G L ( s) = 2 . 31 s( s + 2 . 936
M ) M r = 1, r = 0 BW = 1.02 r ad / sec r = 1, r = 0 rad / sec, BW = 1.03 r ad / sec (b) G H ( s) = s 1 + 0.5455 s + 0.0455s
rad / sec, ( 0.909
2 ) G L ( s) = 0 . 995 s ( 1 + 0 .4975 s ) M r = 1, r = 0 BW = 1.4 ra d / sec M r = 1, r = 0 rad / sec, BW = 1.41 r ad / sec (c) G H ( s) = s 1 + 0.75 s + 0.25 s
rad / sec, ( 0.5
2 ) G L ( s) = 0 . 707 s ( 1 + 0 . 3536 s ) M r = 1, r = 0 BW = 0.87 r ad / sec M r = 1, r = 0 rad / sec, BW = 0.91 r ad / sec (d) G H ( s) = s 1 + 0.00283 s + 8.3056 10 s
rad / sec, BW = 119.74 ( 90.3
7 2 ) G L (s) =
s (1 92 . 94 + 0 . 002594 s) M r = 1, r = 0 rad / sec M r = 1, r = 0 rad / sec B W = 118.76 rad / sec (e) G H ( s) = s 1 + 0.00283 s + 8.3056 10 s
rad / sec, BW = 270.55 ( 180.6
7 2 ) G L (s) = 189 . 54 s (1 + 0 . 002644 s) M r = 1, r = 0 rad / sec M r = 1, r = 0 rad / sec, BW = 268.11 rad / sec (f) G H ( s) = s 1 + 0.00283 s + 8.3056 10 s
. 67 rad / sec ( 1245.52
7 2 ) G L (s) = 2617 . 56 s (1 + 0 . 0053 s) M r = 2 . 96 , r = 666 M r = 3 . 74, r = 700 rad / sec BW = 1054.4 1 rad / sec BW = 1128.7 4 rad / sec 923 (a) (b) M r = 2 . 06 , r = 9 . 33 rad / sec, BW = 15.2 r ad / sec 193 M r =
2 1  2 1 1
2 2 = 2 . 06  + 0 . 0589 = 0
4 2 The solution for < 0 . 707 is = 0.25. r = 9 . 33 rad n
2 / sec Thus n = 9 . 33 0 .9354 = 9 . 974 rad / sec GL ( s) = 924 (a) (b)
M s ( s + 2 n ) = 99.48 s( s + 4.987) = 19.94 s(1 + 0.2005 s) BW = 15.21 rad/sec r = 2 . 96 , r = 666 =
2 1 1
2 2 . 67 rad / sec, BW = 1054.4 1 rad / sec M r = 2 . 96  + 0 . 0285 = 0
4 2 The solution for < 0 . 707 is = 0.1715 r 1  2 = 666 n
2 . 67 rad / sec Thus n = 666 . 67 0 . 97 = 687 .19 rad / sec GL ( s) = 925 (a)
G (s) s ( s + 2 n ) = 472227.43 s( s + 235.7) = 2003.5 s(1 + 0.00424 s) BW = 1079.28 rad/sec = 5 s ( 1 + 0 . 5 s )( 1 + 0 .1 s ) 925 (b)
G (s) = 10 s ( 1 + 0 . 5 s )( 1 + 0 .1 s ) 194 (c)
G (s) = 500 (s + 1.2 )( s + 4 )( s + 10 ) (d)
G (s) =
s(s 10( s + 1)
) + 2 )( s + 10 925 (e) 195 G ( s) = s s + s +1
2 ( 0.5 ) (f) G ( s) = 100 e
2 s s s + 10 s + 50 ( ) (g) G ( s) = 100 e
2 s s s + 10 s + 100 ( ) 925 (h) 196 G ( s) = s s + 5s + 5
2 ( 10( s + 5) ) 926 (a)
G (s) = K s ( 1 + 0 .1 s )( 1 + 0 . 5 s ) The Bode plot is done with K = 1. GM = 21.58 dB For GM = 20 dB, K must be reduced by 1.58 dB. Thus K = 0.8337 PM = 60 .42 . For PM = 45 K should be increased by 5.6 dB. Or, K = 1.91
o o (b)
G (s) s ( 1 + 0 .1 s )( 1 + 0 .2 s )( 1 The Bode plot is done with K = 1. GM = 19.98 dB. For GM = 20 dB, K 1. PM = 86 . 9 . For PM = 45 K should be increased by 8.9 dB. Or, K = 2.79.
o o = K(s + 1) + 0 .5 s ) 926 (c)
G ( s) =
(s K + 3) 3 The Bode plot is done with K = 1. GM = 46.69 dB PM = infinity. 197 For GM = 20 dB K can be increased by 26.69 dB or K = 21.6. For PM = 45 deg. K can be increased by 28.71 dB, or K = 27.26. (d)
G (s) = K (s + 3) 4 The Bode plot is done with K = 1. GM = 50.21 dB PM = infinity. For GM = 20 dB K can be increased by 30.21 dB or K = 32.4 For PM = 45 deg. K can be increased by 38.24 dB, or K = 81.66 (e) G ( s) = Ke
s 2 s 1 + 0.1s + 0.01s ( ) The Bode plot is done with K = 1. GM=2.97 dB PM = 26.58 deg For GM = 20 dB K must be decreased by 17.03 dB or K = 0.141. For PM = 45 deg. K must be decreased by 2.92 dB or K = 0.71. 926 (f) G ( s) = K (1 + 0.5s ) s s + s +1
2 ( ) The Bode plot is done with K = 1. GM = 6.26 dB PM = 22.24 deg For GM = 20 dB K must be decreased by  13.74 dB or K = 0.2055. For PM = 45 deg K must be decreased by  3.55 dB or K = 0.665. 927 (a) 198 G (s) = 10 K s ( 1 + 0 .1 s )( 1 + 0 . 5 s ) The gainphase plot is done with K = 1. GM = 1.58 dB PM = 3.95 deg. For GM = 10 dB, K must be decreased by  8.42 dB or K = 0.38. For PM = 45 deg, K must be decreased by 14 dB, or K = 0.2. For M = 1.2 , K must be decreased to 0.16.
r (b)
G (s) = 5K ( s + 1) + 0 .5 s ) s ( 1 + 0 .1 s )( 1 + 0 .2 s )( 1 The Gainphase plot is done with K = 1. GM = 6 dB PM = 22.31 deg. For GM = 10 dB, K must be decreased by 4 dB or K = 0.631. For PM = 45 deg, K must be decrease by 5 dB. For M = 1.2 , K must be decreased to 0.48.
r 927 (c) G ( s) = 10 K
2 s 1 + 0.1s + 0.01s ( ) The gainphase plot is done for K = 1. GM = 0 dB M =
r PM = 0 deg For GM = 10 dB, K must be decreased by 10 dB or K = 0.316. For PM = 45 deg, K must be decreased by  5.3 dB, or K = 0.543. For M = 1.2 , K must be decreased to
r 0.2213. (d) 199 G ( s) = s 1 + 0.1s + 0.01s ( Ke s 2 ) The gainphase plot is done for K = 1. GM = 2.97 dB M = 3 . 09
r PM = 26.58 deg For GM = 10 dB, K must be decreased by  7.03 dB, K = 0.445. For PM = 45 deg, K must be decreased by  2.92 dB, or K = 0.71. For M = 1.2 , K = 0 . 61 .
r 928 (a)
rad/sec Gain crossover frequency = 2.09 PM = 115.85 deg Phase crossover frequency = 20.31 rad/sec Gain crossover frequency = 6.63 rad/sec Phase crossover frequency = 20.31 rad/sec Gain crossover frequency = 19.1 rad/sec Phase crossover frequency = 20.31 rad/sec GM = 21.13 dB PM = 72.08 deg GM = 15.11 dB PM = 4.07 deg GM = 1.13 dB (b) (c) (d) (e) For GM = 40 dB, reduce gain by (40  21.13) dB = 18.7 dB, or gain = 0.116 nominal value. For PM = 45 deg, the magntude curve reads 10 dB. This means that the loop gain can be increased by 10 dB from the nominal value. Or gain = 3.16 nominal value. The system is type 1, since the slope of G ( j ) is GM = 12.7 dB. PM = 109.85 deg. The gain crossover frequency is 2.09 rad/sec. The phase margin is 115.85 deg. Set (f) (g) (h) 20 dB/decade as 0. T d = 2 . 09 T d =
Thus, the maximum time delay is T
d 115 .85 180 o o = 2 .022 rad = 0 .9674 sec. 929 (a) The gain is increased to four times its nominal value. The magnitude curve is raised by 12.04 dB. Gain crossover frequency = 10 rad/sec PM = 46 deg Phase crossover frequency = 20.31 rad/sec GM = 9.09 dB (b) (c) (d) The GM that corresponds to the nominal gain is 21.13 dB. To change the GM to 20 dB we need to increase the gain by 1.13 dB, or 1.139 times the nominal gain. The GM is 21.13 dB. The forwardpath gain for stability is 21.13 dB, or 11.39. The PM for the nominal gain is 115.85 deg. For PM = 60 deg, the gain crossover frequency must be moved to approximately 8.5 rad/sec, at which point the gain is 10 dB. Thus, the gain must be increased by 10 dB, or by a factor of 3.162. With the gain at twice its nominal value, the system is stable. Since the system is type 1, the steadystate error due to a step input is 0. (e) 200 (f) (g) With the gain at 20 times its nominal value, the system is unstable. Thus the steadystate error would be infinite. With a pure time delay of 0.1 sec, the magnitude curve is not changed, but the the phase curve is subject to a negative phase of  0 .1 rad. The PM is .85 PM = 115  0 .1 gain c rossov er fre quency = 115.85  0 .209 = 115 . 64 deg The new phase crossover frequency is approximately 9 rad/sec, where the original phase curve is reduced by 0.9 rad or 51.5 deg. The magnitude of the gain curve at this frequency is 10 dB. Thus, the gain margin is 10 dB. (h) When the gain is set at 10 times its nominal value, the magnitude curve is raised by 20 dB. The new gain crossover frequency is approximately 17 rad/sec. The phase at this frequency is Thus, setting 30 deg.
sec. T d = 17 T d = 30 o o = 0 . 5236 Thus T 180 d = 0 .0308 930 (a) Bode Plot: 201 For stability: 166 (44.4 dB) < K < 7079 (77 dB) Phase crossover frequencies: 7 rad/sec and 85 rad/sec Nyquist Plot: 930 (b) Root Loci. 202 931 (a) Nquist Plot 203 931 Bode Plot 931 (b) Root Loci 932 Bode Diagram 204 When K = 1, GM = 68.75 dB, PM = 90 deg. The critical value of K for stability is 2738. 933 (a) Forwardpath transfer function: 205 G (s) where = L ( s )
E (s) = K G (s)
a p = K K ( Bs
a i +K ) o o = 0.12 s ( s + 0.0325 ) s + 2.5675 s + 6667
2 (
2 ) = s 0.12 s + 0.312 s + 80.05 s + 26
3 ( ) ) G ( s) = (b) Bode Diagram: s s + 2.6 s + 667.12 s + 216.67
3 2 ( 43.33( s + 500) Gain crossover frequency = 5.85 rad/sec Phase crossover frequency = 11.81 rad/sec PM = 2.65 deg. GM = 10.51 dB 933 (c) Closedloop Frequency Response: 206 M r = 17 . 72 , r = 5 . 75 rad / sec, BW = 9.53 r ad / sec 934 (a) When K = 1, the gain crossover frequency is 8 rad/sec. (b) When K = 1, the phase crossover frequency is 20 rad/sec. (c) (d) (e) (f) (g) (h) (i) When K = 1, GM = 10 dB. When K = 1, PM = 57 deg. When K = 1, M When K = 1,
r = 1.2.
rad/sec. r = 3 When K = 1, BW = 15 rad/sec. When K = 10 dB (0.316), GM = 20 dB When K = 10 dB (3.16), the system is marginally stable. The frequency of oscillation is 20 rad/sec. 207 (j) 935 The system is type 1, since the gainphase plot of G ( j ) approaches infinity at 90 deg. Thus, the steadystate error due to a unitstep input is zero. When K = 5 dB, the gainphase plot of G ( j ) is raised by 5 dB. (a) (b) (c) (d) (e) The gain crossover frequency is 10 rad/sec. The phase crossover frequency is 20 rad/sec. GM = 5 dB. PM = 34.5 deg. M
r =2
rad/sec (f) r = 15 (g) (h) 936 (a) BW = 30 rad/sec When K = 30 dB, the GM is 40 dB.
d The phase margin with K = 1 and T produces a phase lag of Thus, =0 sec is approximately 57 deg. For a PM of 40 deg, the time delay 17 deg.
o The gain crossover frequency is 8 rad/sec. T d = 17
T = 17 o o = 0 .2967 rad / sec T hus = 8 rad / sec 180 d = 0 .2967 8 = 0 . 0371 sec (b) With K = 1, for marginal stability, the time delay must produce a phase lag of Thus, at = 8 rad/sec, 57 deg.
T T d = 57 937 (a) =0 o = 57 o o = 0 . 9948 rad 180 d = 0 . 9948 8 = 0 .1244 sec The phase margin with K = 5 dB and T delay must produce a phase lag of d is approximately 34.5 deg. For a PM of 30 deg, the time The gain crossover frequency is 10 rad/sec. Thus, 4.5 deg. T d = 4. 5 = (b) o 4. 5 o o = 0 . 0785 rad Thus T 180 d = 0 . 0785 10 = 0 .00785 sec With K = 5 dB, for marginal stability, the time delay must produce a phase lag of Thus at 34.5 deg. = 10 rad/sec, T d = 34. 938 5 o = 34. 5 180 o o = 0 . 602 rad Thus T d = 0 . 602 10 = 0 .0602 sec For a GM of 5 dB, the time delay must produce a phase lag of 34.5 deg at = 10
rad rad/sec. Thus, T d = 34. 5 o = 34. 5 180 o o = 0 . 602 Thus T d = 0 . 602 10 = 0 .0602 sec 208 939 (a) Forwardpath Transfer Function:
G (s) = Y (s) E ( s) = e 2 s ( 1 + 10 s )( 1 + 25 s) From the Bode diagram, phase crossover frequency = 0.21 rad/sec GM = 21.55 dB gain crossover frequency = 0 rad/sec PM = infinite (b) G ( s) = (1 + 10s )( 1 + 25s ) ( 1 + 2 s + 2 s
gain crossover frequency = 0 rad/sec 1 2 )
PM = infinite From the Bode diagram, phase crossover frequency = 0.26 rad/sec GM = 25 dB (c) G ( s) = 1 s (1 + s ) (1 + 10 s )( 1 + 2s ) From the Bode diagram, phase crossover frequency = 0.26 rad/sec GM = 25.44 dB gain crossover frequency = 0 rad/sec PM = infinite 939 (continued) Bode diagrams for all three parts. 209 940 (a) Forwardpath Transfer Function: 210 ( 1 + 10 s )( 1 + 25 s ) From the Bode diagram, phase crossover frequency = 0.37 rad/sec GM = 31.08 dB gain crossover frequency = 0 rad/sec PM = infinite G (s) = e s (b) G ( s) = (1 + 10s ) ( 1 + 25s ) (1 + s + 0.5s
( 1  0 .5 s ) 1 2 ) From the Bode diagram, phase crossover frequency = 0.367 rad/sec gain crossover frequency = 0 rad/sec GM = 30.72 dB PM = infinite (c)
G (s) ( 1 + 10 s )( 1 + 25 s )( 1 + 0 .5 s ) From the Bode diagram, phase crossover frequency = 0.3731 rad/sec gain crossover frequency = 0 rad/sec = GM = 31.18 dB PM = infinite 941 Sensitivity Plot: 211 S M G
max = 17 .15 max = 5 .75 rad/sec 212 Chapter 10 DESIGN OF CONTROL SYSTEMS 101 Forwardpath Transfer Function:
G (s) =
1 M ( s) M (s) = K s
3 + ( 20 + a ) s + ( 200 + 20
2 a) s + 200 a K For type 1 system, 200 a Ramperror constant: K = 0
K
v Thus sG ( s ) K = 200a = lim s 0 =
200 K + 20 a = 200 a 200 + 20 a =5 Thus a = 10 K = 2000 The forwardpath transfer function is The controller transfer function is G ( s) = s s + 30 s + 400
2 ( 2000 ) Gc ( s) = G ( s) Gp (s ) = 20 s + 10 s + 100
2 (s ( 2 + 30 s + 400 ) ) The maximum overshoot of the unitstep response is 0 percent. 102 Forwardpath Transfer Function:
G (s) =
1 M ( s) M (s) = K s
3 + ( 20 + a ) s + ( 200 + 20
2 a) s + 200 a K For type 1 system, 200 a Ramperror constant: K = 0
K
v Thus sG ( s ) K = 200a = lim s 0 =
200 K + 20 a = 200 a 200 + 20 a =9 Thus a = 90 K = 18000 The forwardpath transfer function is The controller transfer function is G ( s) = s s + 110 s + 2000
2 ( 18000 ) Gc ( s) = G ( s) Gp (s ) = 180 s + 10 s + 100
2 2 ) ( s + 110 s + 2000 ) ( The maximum overshoot of the unitstep response is 4.3 percent. From the expression for the ramperror constant, we see that as a or K goes to infinity, K Thus the maximum value of K
v v approaches 10. that can be realized is 10. The difficulties with very large values of K and a are that a highgain amplifier is needed and unrealistic circuit parameters are needed for the controller. 103 (a) Ramperror Constant: K v = lim s
s 0 1000 ( K P + K D s ) s ( s + 10)
2 = 1000K P 10 = 100 K P = 1000 Thus K P = 10 Characteristic Equation: s + 10 + 1000 K D s + 1000 K P = 0 ( ) n =
Thus 1000 K P =
K 10000 90 1000 = 100 = 0 . 09 rad/sec 2 n = 10 + 1000 K D = 2 0 . 5 100 = 100 D = 213 103 (b) For K v = 1000 and = 0 . 707
K , and from part (a), n = 100 rad/sec, 131 .4 1000 2 n = 10 + 1000 = 1000
and D = 2 0 . 707 100 = 141 .4 Thus K D = = 0 . 1314 (c) For K v K = 1.0 , and from part (a), n = 100 rad/sec, 190 1000 2 n = 10 + 1000 D = 2 1 100 = 200 Thus K D = = 0 . 19 104 The ramperror constant: K v = lim s
s 0 1000 ( K P + K D s ) s ( s + 10) = 100 K P = 10,000 1000 (100 + K D s ) s( s + 10)
Mr 13.5 1.817 1.291 1.226 1.092 1.06 Thus K P = 100 The forwardpath transfer function is: G ( s) = KD 0 0.2 0.4 0.6 0.8 1.0 PM (deg)
1.814 36.58 62.52 75.9 81.92 84.88 GM BW (rad/sec)
493 525 615 753 916 1090
D Max overshoot (%)
46.6 41.1 22 13.3 8.8 6.2 The phase margin increases and the maximum overshoot decreases monotonically as K increases. 105 (a) Forwardpath Transfer Function: G ( s) = G c ( s ) G p ( s) = 4500 K ( K D + K P s ) s ( s + 361.2) = lim
sG ( s ) Ramp Error Constant: K v s0 = 4500 KK 361 .2 P = 12 .458 KK P e ss = 1 K
v = 0 . 0802 KK
P 0 . 001 Thus KK P 80 .2 Let K P =1 and K = 80 .2 Attributes of Unitstep Response:
KD 0 0.0005 0.0010 0.0015 0.0016 0.0017 0.0018 0.0020 Select K 0 . 0017
D tr (sec) 0.00221 0.00242 0.00245 0.0024 0.00239 0.00238 0.00236 0.00233 ts (sec) 0.0166 0.00812 0.00775 0.0065 0.00597 0.00287 0.0029 0.00283 Max Overshoot (%)
37.1 21.5 12.2 6.4 5.6 4.8 4.0 2.8 (b) BW must be less than 850 rad/sec. 214 KD 0.0005 0.0010 0.0015 0.0016 0.0017 0.00175 0.0018 Select K
D GM PM (deg)
48.45 62.04 73.5 75.46 77.33 78.22 79.07
D Mr 1.276 1.105 1.033 1.025 1.018 1.015 1.012 BW (rad/sec)
827 812 827 834 842 847 852 . A larger K 0 . 00175 would yield a BW larger than 850 rad/sec. 106 The forwardpath Transfer Function: N = 20 G ( s) = 200 ( K P + K D s ) s ( s + 1)( s + 10) To stabilize the system, we can reduce the forwardpath gain. Since the system is type 1, reducing the gain does not affect the steadystate liquid level to a step input. Let K = 0 . 05 G ( s) =
Unitstep Response Attributes:
KD 200 ( 0.05 + K D s ) s ( s + 1)( s + 10)
ts P (sec) Max Overshoot (%) When K D 0.01 5.159 12.7 0.02 4.57 7.1 0.03 2.35 3.2 0.04 2.526 0.8 0.05 2.721 0 0.06 3.039 0 0.10 4.317 0 = 0 . 05 the rise time is 2.721 sec, and the step response has no overshoot. 107 (a) For e ss = 1, K v = lim sG ( s ) = lim s
s 0 s 0 200 ( K P + K D s ) s ( s + 1)( s + 10) = 20 K P = 1 Thus K P = 0 . 05 Forwardpath Transfer Function: G ( s) =
Attributes of Frequency Response:
KD 0 0.01 0.02 0.05 0.09 0.10 0.11 0.20 0.30 0.50 200 ( 0.05 + K D s ) s ( s + 1)( s + 10) PM (deg)
47.4 56.11 64.25 84.32 93.80 93.49 92.71 81.49 71.42 58.55
D GM (deg)
20.83 Mr 1.24 1.09 1.02 1.00 1.00 1.00 1.00 1.00 1.00 1.03 BW (rad/sec)
1.32 1.24 1.18 1.12 1.42 1.59 1.80 4.66 7.79 12.36 For maximum phase margin, the value of K is 0.09. PM = 93.80 deg. GM = , M r = 1, 215 and BW = 1.42 rad/sec. (b) Sensitivity Plots:
The PD control reduces the peak value of the sensitivity function S
M G ( j ) 108 (a) Forwardpath Transfer Function: 100 K P +
2 G ( s) =
K KI s s + 10 s + 100
I For K v = 10, K v = lim sG ( s ) = lim s
s 0 s 0 100 ( K P s + K I ) s s + 10 s + 100
2 ( ) = K I = 10 Thus = 10 .
s (b) Let the complex roots of the characteristic equation be written as
2 =  + j 15 an d s
2 =   j 15 . The quadratic portion of the characteristic equation is
3 2 s + 2 s + + 225 = 0 ( ) The characteristic equation of the system is s + 10 s + (100 + 100 KP ) s + 1000 = 0 The quadratic equation must satisfy the characteristic equation. Using long division and solve for zero remainder condition. s + (10  2 ) s + 2 s + + 225 s + 10 s + (100 + 100 K P ) s + 1000
2 2 3 2 s + 2 s + + 225 s
3 2 2 ( ) (10  2 ) s + 100 KP   125 s + 1000
2 2 (10  2 ) s 2 ( ) + ( 20  4 ) s + (10  2 ) ( s
2 2 + 225 ) 216 (100 K
For zero remainder, 2 and The real solution of Eq. (1) is
3 P + 3  20  125 s + 2  10
2 3 ) 2 + 450 1250
(1) (2)  10 2 + 450  1250 = 0
100 K
P + 3  20  125 = 0
2 = 2 .8555
K . From Eq. (2), P =
s 125 + 20  3 100 2 = 1. 5765  2 . 8555 
j 15 , and s The characteristic equation roots are: =  2 .8555 + j 15 , = 10 + 2 = 4.289 (c) Root Contours: G eq ( s ) = 100 K P s s + 10 s + 100 s + 1000
3 2 = ( s + 10) ( s 100 K P s
2 + 100 ) Root Contours: 217 109 (a) Forwardpath Transfer Function: 100 K P +
2 G ( s) = KI s s + 10 s + 100 For K v = 10, K v = lim sG ( s ) = lim s
s 0 s 0 100 ( K P s + K I ) s s + 10 s + 100
2 ( ) = K I = 10 Thus the forwardpath transfer function becomes G ( s) = s 1 + 0.1s + 0.01s ( 10 (1 + 0.1K P s )
2 ) Attributes of the Frequency Response:
KP 0.1 PM (deg)
5.51 GM (dB)
1.21 Mr 10.05 BW (rad/sec)
14.19 218 0.5 0.6 0.7 0.8 0.9 1.0 1.5 1.1 1.2 When K 22.59 25.44 27.70 29.40 30.56 31.25 31.19 31.51 31.40 and K 6.38 8.25 10.77 14.15 20.10 2.24 1.94 1.74 1.88 1.97 2.00 1.81 2.00 1.97 15.81 16.11 16.38 16.62 17.33 18.01 20.43 18.59 19.08 P = 1.1 I = 10 , K v = 10 , the phase margin is 31.51 deg., and is maximum. The corresponding roots of the characteristic equation roots are: 5 .4,  2 .3 + j 13 .41 , and  2 .3  j 13 .41 Referring these roots to the root contours in Problem 108(c), the complex roots corresponds to a relative damping ratio that is near optimal. (b) Sensitivity Function: In the present case, the system with the PI controller has a higher maximum value for the sensitivity function. 1010 (a) Forwardpath Transfer Function: G ( s) =
For K
v 100 ( K P s + K I ) s s + 10 s + 100
2 ( )
100 ( K P s + K I ) s s + 10 s + 100
2 = 100 , K v = lim sG ( s ) = lim s
s 0
3 s 0 ( ) = K I = 100 Thus K I = 100 . (b) The characteristic equation is s + 10 s + (100 + 100 KP ) s + 100 KI = 0
2 Routh Tabulation: 219 s s s s 3 1 10 100 K P 100 + 100 K 10,000 P 2 For stability, 100 K 1  900 P  900 > 0 Thus K P >9 0 0 10,0 00 Root Contours:
G eq ( s ) = 100 K s
3 P s s + 10 s 2 + 100 + 10 ,000 = 100 K P s (s + 23 . 65)( s  6 .825 + j 19 .4 )( s  6 . 825  j19 .4 ) (c) K I = 100 G ( s) = 100 ( K P s + 100 ) s s + 10 s + 100
2 ( )
P The following maximum overshoots of the system are computed for various values of K KP ymax 15 1.794 20 1.779 22 1.7788 24 1.7785 25 1.7756 26 1.779 30 1.782 . 40 1.795 100 1.844 1000 1.859 When K P = 25, minimum ymax = 1.7756 1011 (a) Forwardpath Transfer Function: 220 G ( s) = 100 ( K P s + K I ) s s + 10 s + 100
2 ( ) For K v = 100 K 100 I = 10 , K I = 10 (b) Characteristic Equation:
Routh Tabulation:
s s s s
3 s3 + 10 s2 + 100 K + 1 s + 1000 = 0 P
100 + 100 K 1000 ( ) 1 10 100 K 1000
P P 2 For stability, 1 KP > 0 0 0 Root Contours:
G
eq (s) = 100 K s
3 P s s + 10 s 2 + 100 + 1000 (c)
KP ymax The maximum overshoots of the system for different values of K computed and tabulated below. P ranging from 0.5 to 20 are 0.5 1.393 1.0 1.275 1.6 1.2317 1.7 1.2416 1.8 1.2424 1.9 1.2441 2.0 1.246 3.0 1.28 5.0 1.372 10 1.514 20 1.642 When K P = 1.7, maximum ymax = 1.2416 1012 Gc (s ) = K P + K D s +
where K
P KI s
K = K Ds + K P s + K I
2 s + K D1K I 2 = (1 + K D1 s ) K P2 +
K
D K I2 s I = P2 = K D1K P 2 K = K I2 Forwardpath Transfer Function: 221 G ( s) = G c ( s ) G p ( s) =
Thus K
I 100 ( 1 + K D1 s ) ( K P2 s + K I 2 ) s s + 10 s + 100
2 ( ) Kv = lim s0 sG ( s ) = K I 2 = 100 = K I 2 = 100
D1 Consider only the PI controller, (with K =0)
Characteristic Equation: Forwardpath Transfer Function: G ( s) =
For stability, K 100 ( K P 2 s + 100 ) s s + 10 s + 100
2 ( )
= 10
for fast rise time. s + 10 s + (100 + 100 KP 2 ) s + 10,000 = 0
3 2 P2 > 9. Select K P2 G ( s) =
When K K
D1 1000 ( 1 + K D 1s ) ( s + 10 ) s s + 10 s + 100
2 ( ) = 0 .2 , the rise time and overshoot requirements are satisfied. K D = K D1 K P2 = 0 .2 10 = 2 P = K P2 + K D 1 K I 2 = 10 + 0 .2 100 = 30
G (s)
c = 30 + 2 s + 100 s Unitstep Response 1013 Process Transfer Function: Gp ( s) = (a) PI Controller: KP + G ( s) = G c ( s ) G p ( s) KI 200 ( K P s + K I ) Y ( s) U (s ) = e
0.2s 1 + 0.25 s (1 + 0.25s ) (1 + 0.2s + 0.02s 1 2 ) (1 + 0.25s ) (
= lim
s0 s 2 1 + 0.2 s + 0.02 s
200 K 4
I ) = s ( s + 4 ) s + 10s + 50
2 ( )
K
I For K v = 2, K v sG ( s ) = 50 = K I =2 Thus =2 Thus G ( s) = s ( s + 4 ) s + 10s + 50
2 200 ( 2 + K P s ) ( )
ts ts The following values of the attributes of the unitstep response are computed for the system with various values for K .
P KP Max overshoot (%) (sec) (sec) 222 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0 19.5 13.8 8.8 4.6 1.0 0 0 0 0 0.61 0.617 0.615 0.606 0.5905 0.568 0.541 0.5078 0.44 2.08 1.966 1.733 0.898 0.878 0.851 1.464 1.603 1.618 223 224 225 Unitstep Response. 226 The unitstep response shows a maximum overshoot of 26%. Although the relative damping ratio of the complex roots is 0.707, the real pole of the thirdorder system transfer function is at 0.667 which adds to the overshoot. (c) G ( s) = G c ( s ) G p ( s) = = 100 = 0.00667 (1 + K D1 s ) ( K P2 s + K I 2 ) s s + 0.00667
2 ( ) For K v , K I2 K I = 100 . Let us select K P2 = 50 . Then G ( s) =
For a small overshoot, K 0.00667 (1 + K D1 s ) ( 50 s + 100 ) s s + 0.00667
2 ( ) D1 must be relatively large. When K D1 = 100 , the maximum overshoot is approximately 4.5%. Thus, KP K K
D I = = K K P2 D1 + K D 1 K I 2 = 50 + 100 100 = 10050
K
P2 = 100 50 = 5000 = 100
3 2 System Characteristic Equation: s + 33 . 35 s + 67 . 04 s + 0 .667 = 0 Roots:  0.01,  2.138,  31.2 Unitstep Response. 227 1016 (a)
G (s)
p = Z (s) F (s) = 1 Ms
2 +Ks = 1 150 s
2 +1 = 0 . 00667 s
2 + 0 . 00667 The transfer function G p ( s ) has poles on the j axis. The natural undamped frequency is n = 0 . 0816 (b) PID Controller: rad/sec. G ( s) = G c ( s ) G p ( s) =
Kv 0.00667 K D s + K P s + K I
2 s s + 0.00667
2 ( ( ) ) = lim s 0 sG ( s ) = KI = 100 T hus KI = 100 Characteristic Equation:
For s + 0.00667 K D s + 0.00667 (1 + K P ) s + 0.00667 K I = 0
3 2 = 1 and n = 1 rad/sec, the secondorder term of the characteristic equation is s 2 + 2 s + 1. Dividing the characteristic equation by the seondorder term. s + ( 0.00667 K D  2 ) s + 2s + 1 s + 0.00667 K D s + ( 0.00667 + 0.00667 K P ) s + 0.00667 K I
2 3 2 s + 2s + s
3 2 ( 0.00667 K ( 0.00667 K ( 0.00667 K
For zero remainder, 0 . 00667 K D  2 ) s + ( 0.00667 KP  0.99333 ) s + 0.00667 K I
2 D  2 ) s + ( 0.01334 K D  4 ) s + 0.00667 K D  2
2 P  0.01334 K D + 3.00667 ) s + 0.00667 K I  0.00667 K D + 2  0 . 01334 + 0 . 00667
K K 0 . 00667
From Eq. (2), P D D I + 3. 00667 = 0 +2 =0 (1) (2) K 228 0 . 00667 K From Eq. (1), 0 . 00667 K D = 0 . 00667 = 0 . 01334 K K I + 2 = 2 . 667 Thus K D = 399
K . 85 P D  3 . 00667 = 2 .3273 Th us P = 348 . 93 Forwardpath Transfer Function: G ( s) =
Characteristic Equation: 0.00667 399.85 s + 348.93 s + 100
2 ( s s + 0.00667
2 ( ) )
( s + 0.667 ) = 0 s 3 + 2.667 s 2 + 2.334 s + 0.667 = ( s + 1)
Roots: Unitstep Response. 2  1,  1,  0.667 The maximum overshoot is 20%. 1017 (a) Process Transfer Function:
G ( s) Forwardpath Transfer Function p = 4 s
2 G ( s) = G c ( s ) G p ( s) =
s
2 4 ( KP + KD s) s +1 = 0
2 Characteristic Equation:
K
P + 4K Ds + 4 K P = s 2 + 1.414 s for = 0 .707 , n =1 rad/sec = 0 .25 a nd K D = 0 . 3535 Unitstep Response. 229 Maximum overshoot = 20.8% (b) Select a relatively large value for K The closedloop zero at s = K P D and a small value for K / K
D P so that the closedloop poles are real. is very close to one of the closedloop poles, and the system
D dynamics are governed by the other closedloop poles. Let K The following results show that the value of K Kp 0.1 0.05 0.2
P = 10 and use small values of K P . is not critical as long as it is small. tr ts Max overshoot (%)
0 0 0 (sec)
0.0549 0.0549 0.0549 (sec)
0.0707 0.0707 0.0707 (c) For BW 40 rad/sec and M r = 1, we can again select K D = 10 and a small value for K P . The following frequencydomain results substantiate the design. Kp 0.1 0.05 0.2 Mr 1 1 1 PM (deg)
89.99 89.99 89.99 BW (rad/sec)
40 40 40 1018 (a) Forwardpath Transfer Function: G ( s) = G c ( s ) G p ( s) =
Routh Tabulation: Characteristic Equation:
s
3 10,000 ( K P + K D s ) s
2 ( s + 10) + 10 s 2 + 10 ,000 K D s + 10 ,000 K P =0 230 s s s s 3 1 10 10,000 K 10 ,000 K 10, 000 K D 2 P The system is stable for K 1 D  1000
K
P K P >0 and K D > 0 .1 K P P 0 0 10,000 (b) Root Locus Diagram: G ( s) = 10,000 K P s
2 ( s + 10 ) Root Contours: 0 K D < , = K P = 0 . 001 ,
D 0 . 002 , 0 .005 , 0 . 01 . s K
P G eq (s) 10 ,000 K s
3 + 10 s 2 + 10 ,000 231 (c) The root contours show that for small values of K means that if we choose K
P P the design is insensitive to the variation ofK P . This to be between 0.001 and 0.005, the value of K
P for a relative damping ratio of 0.707. Let K The forwardpath transfer function becomes = 0 . 001 and K D = 0 . 005 D can be chosen to be 0.005 . G (s)
c = 0 . 001 + 0 . 005 s. G ( s) =
Since the zero of G ( s ) is at s = 10 (1 + 5 s ) s
2 ( s + 10 ) 0.2, which is very close to s = 0, G(s) can be approximated as: G (s) 50 s ( s + 10 )
. Using Eq. (7104), the rise time is obtained as For the secondorder system, = 0 . 707
t r = 1  0 .4167 + 2 . 917 n 2 = 0 . 306 sec Unitstep Response: 232 (d) Frequencydomain Characteristics: G ( s) = 10 (1 + 5 s ) s
PM (deg)
63
2 ( s + 10 )
Mr 1.041 GM (dB) BW (rad/sec)
7.156 1019 0 25.92 A = 0 2.36 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 s 25.92 s 0 0 sI  A = 0 0 s  1 2.36 0 0 s s 0 0 25.92 0 2.36 s
2 0 s 0 0 0 1 = s s
2 = sI  A = s 0 s  1 +
0 0 s (s
0 2  25.92 ) s3 2 1 25.92 s 1 ( sI  A ) =  2.36 s 2 2.36 s s 0 0 3 2 2.36 s  25.92 s s  25.92 3 2.36 s 0 s  25.92s 3 233 ( sI  A ) 1 0.0732 s 2 0 0.0732 1 0.0732 s 3 1 = B = ( sI  A ) 0 0.0976 s 2  2.357 0.0976 3 0.0976 s  2.357 s Y ( s) = D sI  A
Characteristic Equation: ( )
4 1 B = [0 0
3 1 0 ] sI  A ( ) 1 B=
2 0.0976 s  24.15
2 s 2 (s
D (
2  25.92 ) ) s + 0.0976 s + ( 0.0976 KP  25.92 ) s  2.357 K D s  2.357 KP = 0
3 1 The system cannot be stabilized by the PD controller, since the s and the s terms involve K require opposite signs for K
D which . 1020 Let us first attempt to compensate the system with a PI controller. G (s)
c = K P + K s I Then G ( s) = Gc ( s ) G ( s) = p 100 ( K P s + K I ) s s + 10 s + 100
2 ( ) Since the system with the PI controller is now a type 1 system, the steadystate error of the system due to a step input will be zero as long as the values of K and K are chosen so that the system is stable. Let us choose the ramperror constant K
v = 100
I P I . Then, K performance characteristics are obtained with K KP 10 20 30 40 50 100 = 100 I = 100 . The following frequencydomain
P and various value of K ranging from 10 to 100. Mr 29.70 7.62 7.41 8.28 8.45 11.04
P PM (deg)
1.60 6.76 7.15 6.90 6.56 5.18 GM (dB) BW (rad/sec)
50.13 69.90 85.40 98.50 106.56 160.00 The maximum phase margin that can be achieved with the PI controller is only 7.15 deg when K = 30 . Thus, the overshoot requirement cannot be satisfied with the PI controller alone. Next, we try a PID controller. Gc (s ) = K P + K Ds + KI s = (1 + K s ) (K
D1 P2 s + KI2 ) = (1 + K s ) (K
D1 P2 s + 100 ) Based on the PIcontroller design, let us select K becomes P2 = 30 . s s Then the forwardpath transfer function G ( s) = 100 ( 30 s + 100 )( 1 + K D1s ) s s + 10 s + 100
2 ( ) The following attributes of the frequencydomain performance of the system with the PID controller are obtained for various values of K ranging from 0.05 to 0.4.
D1 K D1 0.05 0.10 0.20 0.30 PM (deg)
85.0 89.4 90.2 90.2 GM (dB) Mr 1.04 1.00 1.00 1.00 BW (rad/sec)
164.3 303.8 598.6 897.0 234 0.40 We see that for values of K
D1 90.2 D1 1.00 1201.0 greater than 0.2, the phase margin no longer increases, but the . Thus we choose bandwidth increases with the increase in K K K D1 = 0 .2,
K
P2 K I = K I2 = 100 , K D = K D1 K P2 = 0 .2 30 = 6 , P = + K D 1 K I 2 = 30 + 0 .2 100 = 50
G (s)
c The transfer function of the PID controller is s The unitstep response is show below. The maximum overshoot is zero, and the rise time is 0.0172 sec. = 50 + 6 s + 100 1021 (a) Gp (s ) = 4 s
2 G ( s) = Gc ( s ) G p ( s ) = 4 (1 + aTs ) s (1 + Ts )
2 G eq (s) = 4 aTs Ts
3 + s +4
2 Root Contours: ( T is fixed and a varies) 235 Select a value for a. Let T = 0.02 small value for T and a large and a = 100. G (s)
c = 1+ 2 s 1 + 0 . 02 s s
3 G ( s) = + 50
2 400 ( s + 0.5) s
2 ( s + 50) The characteristic equation is The roots are: The system transfer function is s + 400 s + 200 = 0 0.5355, 9.3, 40.17 Y (s) R (s ) = ( s + 0.5355) ( s + 9.3) ( s + 40.17 ) 400 ( s + 0.5 ) Since the zero at 0.5 is very close to the pole at 0.5355, the system can be approximated by a secondorder system, Y (s) R (s ) = 373.48 ( s + 9.3)( s + 40.17 )
= 0 . 6225
sec t The unitstep response is shown below. The attributes of the response are: Maximum overshoot = 5% t
s r = 0 .2173 sec Unitstep Response. 236 The following attributes of the frequencydomain performance are obtained for the system with the phaselead controller. PM = 77.4 deg GM = infinite Mr = 1.05 BW = 9.976 rad/sec 1021 (b) The Bode plot of the uncompensated forwardpath transfer function is shown below. The diagram shows that the uncompensated system is marginally stable. The phase of G ( j ) is frequencies. For the phaselead controller we need to place 180 deg at all m at the new gain crossover frequency to realize the desired phase margin which has a theoretical maximum of 90 deg. For a desired phase margin of 80 deg, a = 1 + sin 80 1  sin 80 o o = 130 The gain of the controller is 20 log 10 a = 42 dB. The new gain crossover frequency is at G ( j ) G ( j ) = 42 2 = 21
2 dB Or = 4 m 2 = 0 . 0877 Thus = 45 . 61 = 6 . 75 = 0 . 013 rad/sec 1 T 1 aT = a = 130 6 . 75 = 77 = 1. 69 Thus T = 0 . 592 1 + aTs 1 + Ts Thus aT Gc (s ) =
Bode Plot. = 1 + 1.702 s 1 + 0.0131s G( s) = s 4 (1 + 1.702s )
2 (1 + 0.0131s ) 237 1022 (a) Forwardpath Transfer Functi on: aT G ( s) = G c ( s ) GP ( s) = = s ( s + 10 ) (1 + Ts ) 1 s ( s + 10 ) s + T 1000 (1 + aTs ) 1000a s + 1 238 Set 1/aT = 10 so that the pole of G ( s ) at s = becomes s
2 10 is cancelled.
1 T s The characteristic equation of the system + + 1000 a =0
Thus a = 40 and T = 0.0025 n = 1000 a 2 n = 1 T =2 1000 a Controller Transfer Function:
G (s)
c Forwardpath Transfer Function: = 1 + 0 . 01 s 1 + 0 . 0025 s G ( s) = 40,000 s ( s + 400 ) The attributes of the unitstep response of the compensated system are: Maximum overshoot = 0 t
r = 0 . 0168 sec t s = 0 . 02367 sec (b) Frequencydomain Design
The Bode plot of the uncompensated forwardpath transfer function is made below. G (s) = 1000 s ( s + 10 )
The attributes of the system are PM = 17.96 deg, GM = infinite. Mr = 3.117, and BW = 48.53 rad/sec. To realize a phase margin of 75 deg, we need more than 57 deg of additional phase. Let us add an additional 10 deg for safety. Thus, the value of for the phaselead controller is chosen to be
m 67 deg. The value of a is calculated from a = 1 + sin 67 1  sin 67 log
10 o o = 24. 16 = 27 . 66
dB. The new gain crossover frequency The gain of the controller is 20 log is at 10 a = 20 24. 16 G ( j ' m ) = 27 . 66 2 =  13 . 83 dB From the Bode plot 1 T m
' is found to be 70 rad/sec. Thus, = aT = 24. 16 70 = 344 =
1 + aTs or T = 0 . 0029
s aT = 0 . 0702 1 + Ts 1 + 0 . 0029 s The compensated system has the following frequencydomain attributes: PM = 75.19 deg GM = infinite Mr = 1.024 BW = 91.85 rad/sec Thus G (s)
c = 1 + 0 . 0702 The attributes of the unitstep response are: Rise time tr = 0.02278 sec Settling time ts = 0.02828 sec Maximum overshoot = 3.3% 239 1023 (a) Forwardpath Transfer Function: ( N = 10) G ( s) = G c ( s ) G p ( s) = 200 ( 1 + aTs ) s ( s + 1) ( s + 10 ) ( 1 + Ts ) Starting with a = 1000, we vary T first to stabilize the system. The following timedomain attributes are obtained by varying the value of T. 240 T 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010 Max Overshoot (%)
59.4 41.5 29.9 22.7 18.5 16.3 15.4 15.4 15.5 16.7 tr ts 5.205 2.911 1.83 1.178 1.013 0.844 0.699 0.620 0.533 0.525 0.370
0.293 0.315 0.282 0.254 0.230 0.210 0.192 0.182 0.163 The maximum overshoot is at a minimum when T = 0.0007 or T = 0.0008. The maximum overshoot is 15.4%. Unitstep Response. ( T = 0.0008 sec a = 1000) 1023 (b) Frequencydomain Design.
Similar to the design in part (a), we set a = 1000, and vary the value of T between 0.0001 and 0.001. The attributes of the frequencydomain characteristics are given below. T 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 PM (deg)
17.95 31.99 42.77 49.78 53.39 54.69 54.62 GM Mr 3.194 1.854 1.448 1.272 1.183 1.138 1.121 (dB)
60.00 63.53 58.62 54.53 51.16 48.32 45.87 BW (rad/sec)
4.849 5.285 5.941 6.821 7.817 8.869 9.913 241 0.0008 0.0009 0.0010 53.83 52.68 51.38 43.72 41.81 40.09 1.125 1.140 1.162 10.92 11.88 12.79 The phase margin is at a maximum of 54.69 deg when T = 0.0006. The performance worsens if the value of a is less than 1000. 1024 (a) Bode Plot. The attributes of the frequency response are: PM = 4.07 deg GM = 1.34 dB Mr = 23.24 BW = 4.4 rad/sec 1024 (b) Singlestage Phaselead Controller. G ( s) = 6 (1 + aTs ) s (1 + 0.2s ) (1 + 0.5s ) (1 + Ts ) We first set a = 1000, and vary T. The following attributes of the frequencydomain characteristics are obtained. T 0.0050 0.0010 0.0007 0.0006 PM (deg)
17.77 43.70 47.53 48.27 Mr 3.21 1.34 1.24 1.22 242 0.0005 0.0004 0.0002 0.0001 48.06 46.01 32.08 19.57 1.23 1.29 1.81 2.97 The phase margin is maximum at 48.27 deg when T = 0.0006. Next, we set T = 0.0006 and reduce a from 1000. We can show that the phase margin is not very sensitive to the variation of a when a is near 1000. The optimal value of a is around 980, and the corresponding phase margin is 48.34 deg. With a = 980 and T = 0.0006, the attributes of the unitstep response are: Maximum overshoot = 18.8% tr = 0.262 sec T = 0.0006) ts = 0.851 sec (c) Twostage Phaselead Controller. ( a = 980, G( s) = 6 ( 1 + 0.588s ) 1 + bT s 2 s (1 + 0.2s )(1 + 0.5 s ) (1 + 0.0006 s ) 1 + T s 2 ( ) ( ) Again, let b = 1000, and vary T . The following results are obtained in the frequency domain.
2 T2 PM (deg) Mr 0.0010 93.81 1.00 0.0009 94.89 1.00 0.0008 96.02 1.00 0.0007 97.21 1.00 0.0006 98.43 1.00 0.0005 99.61 1.00 0.0004 100.40 1.00 0.0003 99.34 1.00 0.0002 91.98 1.00 0.0001 73.86 1.00 Reducing the value of b from 1000 reduces the phase margin. Thus, the maximum phase margin of 100.4 deg is obtained with b = 1000 and T 2 = 0.0004. The transfer function of the twostage phaselead controller is Gc ( s) = (c) Unitstep Responses. ( 1 + 0.588s ) (1 + 0.4s ) (1 + 0.0006s ) (1 + 0.0004s ) 243 1025 (a) The loop transfer function of the system is G ( s) H( s) = 10 K pK aK e Ns (1 + 0.05 s )
s
3 1 + R2C s RCs 1
2 = 68.76 s (1 + 0.05s )
4 1 + R2 10 s 2s
.6 6 The characteristic equation is For root locus plot with R
2 + 20 s + 6 .876 10 R s
2 + 687 =0 as the variable parameter, we have
4 G eq ( s ) = 6.876 10 R2 s s + 20 s + 687.6
3 2 = ( s + 21.5) ( s  0.745 + 6.876 10 R2 s 4 j 5.61) ( s  0.745  j 5.61 ) Root Locus Plot. 244 When R 2 = 2 . 65 10 5 , the roots are at 6 . 02 j 7 . 08 , and the relative damping ratio is 0.65 which is maximum. The unitstep response is plotted at the end together with those of parts (b) and (c). (b) Phaselead Controller. G ( s) H( s) = 68.76 (1 + aTs ) s (1 + 0.05 s ) (1 + Ts ) Characteristic Equation: Ts + 1 + 20 T s + 20 + 1375.2 aT s + 1375.2 = 0
3 2 ( ) ( ) With T = 0.01, the characteristic equation becomes 3 2 s + 120 s + ( 2000 + 1375.2 a ) s + 137520 = 0
The last equation is conditioned for a root contour plot with a as the variable parameter. Thus 1375 .2 as G eq ( s ) = 3 2 s + 120 s + 2000 s + 137 ,520 From the root contour plot on the next page we see that when a = 3.4 the characteristic equation roots are at  39 .2 ,  40 .4 + j 43 . 3, and  40 .4  j 43 . 3 , and the relative damping ratio is maximum and is 0.682. Root Contour Plot ( a varies). 245 Unitstep Responses. 1025 (c) Frequencydomain Design of Phaselead Controller.
For a phase margin of 60 deg, a = 4.373 and T = 0.00923. The transfer function of the controller is G (s)
c = 1 + aTs 1 + Ts = 1 + 0 . 04036 s 1 + 0 . 00923 s 246 1026 (a) Timedomain Design of Phaselag Controller.
Process Transfer Function: Gp ( s) = 200 s ( s + 1 )( s + 10 ) =  0 .475 +
j 0 .471 For the uncompensated system, the two complex characteristic equation roots are at s and 0 .475  j 0 .471 which correspond to a relative damping ratio of 0.707, when the forward path gain is 4.5 (as against 200). Thus, the value of a of the phaselag controller is chosen to be a Then G (s)
c = 4. 5 200 = 0 . 0225
1 + aTs 1 + Ts 1 Select T = 1000 which is a large number. = = + 22 . 5 s 1 + 1000 s G ( s) = Gc ( s ) G ( s) = p s ( s + 1 )( s + 10 )( s + 0.001) 4.5 ( s + 0.0889) ) Unitstep Response. Maximum overshoot = 13.6 Bode Plot (with phaselag controller, a = 0.0225, T = 1000) tr = 3.238 sec ts = 18.86 sec 247 PM = 59 deg. GM = 27.34 dB Mr = 1.1 BW = 0.6414 rad/sec 1026 (b) Frequencydomain Design of Phaselag Controller.
For PM = 60 deg, we choose a = 0.02178 and T = 1130.55. The transfer function of the phaselag controller is 1 + 24. 62 s G (s) = GM = 27.66 dB Mr = 1.093 BW = 0.619 rad/sec c 1 + 1130 . 55 s Unitstep Response. Max overshoot = 12.6%, tr = 3.297 sec ts = 18.18 sec 1027 (a) Timedomain Design of Phaselead Controller
Forwardpath Transfer Function. G ( s) = Gc ( s ) G ( s) = p K (1 + aTs ) s ( s + 5)
2 (1 + Ts ) K v = lim sG ( s ) s 0 = K 25 = 10 Thus K = 250 248 With K = 250, the system without compensation is marginally stable. For a > 1, select a small value for T and a large value for a. Let a = 1000. The following results are obtained for various values of T ranging from 0.0001 to 0.001. When T = 0.0004, the maximum overshoot is near minimum at 23%. T 0.0010 0.0005 0.0004 0.0003 0.0002 0.0001 Max Overshoot (%)
33.5 23.8 23.0 24.4 30.6 47.8 (sec)
0.0905 0.1295 0.1471 0.1689 0.1981 0.2326 tr (sec)
0.808 0.6869 0.7711 0.8765 1.096 2.399 ts As it turns out a = 1000 is near optimal. A higher or lower value for a will give larger overshoot. Unitstep Response. (b) Frequencydomain Design of Phaselead Controller G ( s) = s 250 ( 1 + aTs )
2 ( s + 5 ) (1 + Ts )
2 Setting a = 1000, and varying T, the following attributes are obtained. T 0.00050 0.00040 0.00035 0.00030 0.00020 PM (deg)
41.15 42.85 43.30 43.10 38.60 Mr 1.418 1.369 1.355 1.361 1.513 BW (rad/sec)
16.05 14.15 13.16 12.12 10.04 When a = 1000, the best value of T for a maximum phase margin is 0.00035, and PM = 43.3 deg. As it turns out varying the value of a from 1000 does not improve the phase margin. Thus the transfer function of the controller is 249 G (s)
c = 1 + aTs 1 + Ts =
1 1 + 0 . 35 s and s + 0 . 00035 G ( s) = 250 ( 1+ 0.35s ) s ( s + 5)
2 (1 + 0.00035s ) 1027 (c) Timedomain Design of Phaselag Controller
Without compensation, the relative damping is critical when K = 18.5. Then, the value of a is chosen to be 18 . 5 a = = 0 . 074 250 We can use this value of a as a reference, and conduct the design around this point. The value of T is preferrably to be large. However, if T is too large, rise and settling times will suffer. The following performance attributes of the unitstep response are obtained for various values of a and T. a 0.105 0.100 0.095 0.090 0.090 0.090 0.090 0.090 0.090 0.090 T 500 500 500 500 600 700 800 1000 2000 3000 Max Overshoot (%)
2.6 2.9 2.6 2.5 2.1 1.9 1.7 1.4 0.9 0.7 tr 1.272 1.348 1.422 1.522 1.532 1.538 1.543 1.550 1.560 1.566 ts 1.82 1.82 1.82 2.02 2.02 2.02 2.02 2.22 2.22 2.22 As seen from the results, when a = 0.09 and for T 2000, the maximum overshoot is less than 1% and the settling time is less than 2.5 sec. We choose T = 2000 and a = 0.09. The corresponding frequencydomain characteristics are: PM = 69.84 deg GM = 20.9 dB Mr = 1.004 BW = 1.363 rad/sec (d) Frequencydomain Design of Phaselag Controller G ( s) = 250 ( 1 + aTs ) s ( s + 5)
2 (1 + Ts ) a <1 The Bode plot of the uncompensated system is shown below. Let us add a safety factor by requiring that the desired phase margin is 75 degrees. We see that a phase margin of 75 degrees can be realized if the gain crossover is moved to 0.64 rad/sec. The magnitude of G ( j ) at this frequency is 23.7 dB. Thus the phaselag controller must provide an attenuation of the new gain crossover frequency. Setting 20 log a 23.7 dB at 10 = 23 .7 dB we have a = 0.065 We can set the value of 1/aT to be at least one decade below 0.64 rad/sec, or 0.064 rad/sec. Thus, we get T = 236. Let us choose T = 300. The transfer function of the phaselag controller becomes G (s)
c = 1 + aTs 1 + Ts = 1 + 19 . 5 s 1 + 300 s 250 The attributes of the frequency response of the compensated system are: 251 PM = 71 deg GM = 23.6 dB Mr = 1.065 BW = 0.937 rad/sec The attributes of the unitstep response are: Maximum overshoot = 6% tr = 2.437 sec ts = 11.11 sec Comparing with the phaselag controller designed in part (a) which has a = 0.09 and T = 2000, the time response attributes are: Maximum overshoot = 0.9% tr = 1.56 sec ts = 2.22 sec The main difference is in the large value of T used in part (c) which resulted in less overshoot, rise and settling times. 1028 Forwardpath Transfer Function (No compensation) G ( s) = G p ( s) = 6.087 10
3 2 7 6 8 s s + 423.42 s + 2.6667 10 s + 4.2342 10 ( ) The uncompensated system has a maximum overshoot of 14.6%. The unitstep response is shown below. (a) Phaselead Controller
G (s)
c = 1 + aTs 1 + Ts (a > 1) By selecting a small value for T, the value of a becomes the critical design parameter in this case. If a is too small, the overshoot will be excessive. If the value of a is too large, the oscillation in the step response will be objectionable. By trial and error, the best value of a is selected to be 6, and T = 0.001. The following performance attributes are obtained for the unitstep response. Maximum overshoot = 0% tr = 0.01262 sec ts = 0.1818 sec However, the step response still has oscillations due to the compliance in the motor shaft. The unitstep response of the phaselead compensated system is shown below, together with that of the uncompensated system. (b) Phaselead and Secondorder Controller
The poles of the process G p ( s ) are at 161.3, order term is s
2 131+ j1614.7 and 131  j1614.7. The second + 262 s + 2 ,624,
G ( s) 417 .1 . Let the secondorder controller transfer function be s
2 c1 = + 262
s
2 s + 2 ,624, 417 .1
2 + 2 p n s + n
rad/sec, so that the steadystate error is not affected. The value of n is set to 2 ,624, 417 .1 = 1620
and
2 Let the two poles of G c1 ( s ) be at s =  1620 =
s s  1620
s s . Then, p = 405 . G c1 ( s ) + 262 + 3240 + 2 ,624, + 2 ,624, 417 .1 417 .1
10 2 G ( s) = G c ( s ) G c ( s ) G p ( s) = 1 s ( s + 161.3 ) s + 3240 s + 2,624,417.1 ( 1 + 0.001s )
2 ( 6.087 10 (1 + 0.006 s ) ) The unitstep response is shown below, and the attributes are: 252 Maximum overshoot = 0.2 The step response does not have any ripples. tr = 0.01012 sec ts = 0.01414 sec Unitstep Responses 1029 (a) System Equations.
e a = R a i a + K b m T m = K i ia d L dt Tm = Jm d m dt + Bmm + KL ( m  L ) + BL ( m  L ) K L ( m  L ) + BL ( m  L ) = J L State Equations in Vectormatrix Form: d L dt 0 K d L  L dt J L d = 0 m dt K L d m J m dt 1  BL JL 0 BL Jm  0 KL JL 0 KL Jm 0 BL L 0 JL L + 0 ea 1 m Ka B + Bl Ki K b m  m  Ra J m Jm J m Ra 0 253 State Diagram: Transfer Functions: m ( s) Ea ( s ) L ( s) E a (s ) m ( s) Ea ( s ) L ( s) E a (s ) = Ki s + BL s + KL / Ra
2 J m J L s + ( K e J L + BL J L + BL J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e
2 2 ( ) = J m J L s + ( K e J L + BL J L + B L J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e
3 2 K i ( B Ls + K L ) / R a = 133.33 s + 10 s + 3000
2 3 2 ( ) s + 318.15 s + 60694.13 s + 58240 = 133.33 s + 10s + 3000
2 ( s + 0.9644) ( s + 158.59 + ( j187.71) ( s + 158.59  j187.71) ) = ( s + 0.9644 ) ( s + 158.59 + 1333.33 ( s + 300 ) j187.71 ) ( s + 158.59  j187.71) (b) Design of PI Controller. 1333.33 K P s + = G ( s) = L (s ) E (s ) s ( s + 0.9644 ) s + 317.186s + 60388.23
2 ( ( s + 300 ) KP KI )
Thus Kv = lim s0 sG ( s ) = 1333 .33 0 . 9644 300 K I 60388 . 23 = 6 . 87 KI = 100 KI = 14 .56 With K I = 14 .56 ,
KP 20 18 we study the effects of varying K P . The following results are obtained. tr ts (sec)
0.00932 0.01041 (sec)
0.02778 0.01263 Max Overshoot (%)
4.2 0.7 254 17 16 15 10 With K I 0.01113 0.01184 0.01303 0.02756 0.01515 0.01515 0.01768 0.04040 0 0 0 0.6 = 14 .56 and K P ranging from 15 to 17, the design specifications are satisfied. Unitstep Response: (c) Frequencydomain Design of PI Controller ( KI = 14.56) G ( s) = s s + 318.15 s + 60694.13s + 58240
3 2 ( 1333.33 ( K P s + 14.56 )( s + 300 ) )
tr ts The following results are obtained by setting K I KP 20 18 17 16 15 10 8 7 6 5 = 14 .56 and varying the value of K P . PM (deg)
65.93 69.76 71.54 73.26 74.89 81.11 82.66 83.14 83.29 82.88 GM (dB) Mr 1.000 1.000 1.000 1.000 1.000 1.005 1.012 1.017 1.025 1.038 BW (rad/sec)
266.1 243 229 211.6 190.3 84.92 63.33 54.19 45.81 38.12 Max Overshoot (%)
4.2 0.7 0 0 0 0.6 1.3 1.9 2.7 4.1 (sec)
0.00932 0.01041 0.01113 0.01184 0.01313 0.0294 0.04848 0.03952 0.04697 0.05457 (sec)
0.02778 0.01263 0.01515 0.01515 0.01768 0.0404 0.03492 0.05253 0.0606 0.0606 255 From these results we see that the phase margin is at a maximum of 83.29 degrees when K P = 6 . However, the maximum overshoot of the unitstep response is 2.7%, and M r is slightly greater than one. In part (b), the optimal value of K P from the standpoint of minimum value of the maximum overshoot is between 15 and 17. Thus, the phase margin criterion is not a good indicator in the present case. 1030 (a) Forwardpath Transfer Function Gp ( s) = K m ( s) Tm ( s ) = 100 K s + 10 s + 100
2 3 2 s s + 20 s ( ( ( = + 2100 s + 10,000 ) s ( s + 4.937 ) ( s
ts = 0.4949 sec ) 10,000 s + 10 s + 100
2 2 ) + 15.06 s + 2025.6 ) The unitstep response is plotted as shown below. The attributes of the response are: Maximum overshoot = 57% tr = 0.01345 sec (b) Design of the Secondorder Notch Controller
The complex zeros of the notch controller are to cancel the complex poles of the process transfer function. Thus s s
2 2 G (s)
c = + 15 . 06 s + 2025 + 90 p s + 2025 .6 and .6 ( G ( s) = G ( s ) G ( s) = s ( s + 4.937 ) ( s
c p 10,000 s + 10 s + 100
2 2 ) + 90 p s + 2025.6 ) The following results are obtained for the unitstep response when various values of The maximum overshoot is at a minimum of 4.1% when p p are used. = 1.222 . The unitstep response is plotted below, along with that of the uncompensated system. p
2.444 2.333 2.222 1.667 1.333 1.222 1.111 1.000 2 200 210 200 150 120 110 100 90 n Max Overshoot (%)
7.3 6.9 6.5 4.9 4.3 4.1 5.8 9.8 Unitstep Response 256 1030 (c) Frequencydomain Design of the Notch Controller
The forwardpath transfer function of the uncompensated system is ( G ( s) = s ( s + 4.937 ) ( s 10000 s + 10 s + 100
2 2 ) + 15.06 s + 2025.6 )
22 dB or 0.0794 The Bode plot of G ( j ) is constructed in the following. We see that the peak value of G ( j ) is approximately 22 dB. Thus, the notch controller should provide an attentuation of at the resonant frequency of 45 rad/sec. Using Eq. (10155), we have G ( j 45)
c = z p = 0 .167 p = 0 . 0794 Thus p = 2 .1024 Notch Controller Transfer Function
s s
2 2 Forwardpath Transfer Function G (s)
c = + 15 .06 s + 2025
.216 s .6 .6 + 189 + 2025 G (s ) = ( s ( s + 4.937 ) (s 10,000 s + 10 s + 100
2 2 ) + 189.22s + 2025.6 ) Bode Plots 257 Attributes of the frequency response: PM = 80.37 deg GM = infinite M r = 1. 097 BW = 66.4 rad/sec Attributes of the frequency response of the system designed in part (b): PM = 59.64 deg GM = infinite M
r = 1. 048 BW = 126.5 rad/sec 1031 (a) Process Transfer Function Gp ( s) = s s + 10 s + 1000
2 ( 500 ( s + 10 ) )
M The Bode plot is constructed below. The frequencydomain attributes of the uncompensated system are: PM = 30 deg GM = infinite
r = 1.86 and BW = 3.95 rad/sec The unitstep response is oscillatory. (b) Design of the Notch Controller 258 For the uncompensated process, the complex poles have the following constants: n = 1000 = 31 .6 rad / sec 2 n = 10 Thus = 0.158 The transfer function of the notch controller is s
2 G (s)
c = + 2 z n s + n
2 2 s + 2 p s + n
2 For the zeros of G ( s ) to cancel the complex poles of G p ( s ) ,
c z = = 0 .158 . From the Bode plot, we see that to bring down the peak resonance of G ( j n ) in order to smooth out the magnitude curve, the notch controller should provide approximately 26 dB of attenuation. Thus, using Eq. (10155), z p 26 = 10 20 = 0 . 05 Thu s p = 0 .158 0 . 05 = 3 .1525 The transfer function of the notch controller is G (s)
c = s s
2 2 + 10 s + 1000
. 08 s + 199 + 1000 G ( s) = G c ( s ) G p ( s) = s s + 199.08 s + 1000
2 ( 500 ( s + 10 ) ) The attributes of the compensated system are: PM = 72.38 deg GM = infinite t M
r =1
t
s BW = 5.44 rad/sec Maximum overshoot = 3.4% r = 0 . 3868 sec = 0 .4848 sec Bode Plots 259 Step Responses 1031 (c) Timedomain design of the Notch Controller 260 With z = 0 .158 and n = 31 . 6 , the forwardpath transfer function of the compensated system is G ( s) = G c ( s ) G p ( s) = s s + 63.2 p s + 1000
2 ( 500 ( s + 10 ) ) p.
(sec)
0.5859 0.5657 0.5455 0.5253 0.5152 0.4840 0.4848 ts tr The following attributes of the unitstep response are obtained by varying the value of p
1.582 1.741 1.899 2.057 2.215 2.500 3.318 When 2 n Max Overshoot (%)
0 0 0 0 0.2 0.9 4.1 (sec)
0.4292 0.4172 0.4074 0.3998 0.3941 0.3879 0.3884 100 110 120 130 140 158.25 209.7 p = 2 .5 the maximum overshoot is 0.9%, the rise time is 0.3879 sec and the setting time is 0.4840 sec. These performance attributes are within the required specifications. 1032 Let the transfer function of the controller be Gc ( s) = 20,000 s + 10 s + 50
2 ( ) ( s + 1000 ) 2 Then, the forwardpath transfer function becomes G ( s) = Gc ( s ) G ( s) = p
For G (s) 20,000 K s + 10 s + 50
2 s s + 10 s + 100
2 6 ( ( ) ( s + 1000) ) 2 cf = 1, K v = lim
K sG ( s ) s0 = 10 K
8 = 50 = 6000 Thus t he nom inal K = 5000 10 and For 20% variation in K, min = 4000 K max . To cancel the complex closedloop poles, we let G cf ( s ) = 50 ( s + 1 ) s + 10 s + 50
2 where the (s + 1) term is added to reduce the rise time. Closedloop Transfer Function: Y (s) R (s ) = s s + 10 s + 100
2 ( )( s + 1000 ) + 20,000K ( s
2 7 2 8 10 K ( s + 1)
6 2 + 10 s + 50 ) Characteristic Equation:
K = 4000: s
5 Roots:  97 . 7 ,
5 + 2010 s + 1,020 , 100 s + 9 . 02 10 s + 9 10  648 . 9 ,  1252 . 7 ,  5 . 35 + j 4. 6635 ,
4 3 Max overshoot
K = 5000: s 6.7% + 4 10 = 0  5 . 35  j 4. 6635
9 9 Rise time < 0.04 sec
3 8 2 9 Roots: 132 + 2010
.46 , s 4 + 1,02010 587 .44, Max overshoot 4% 0 s + 1.1 10 s + 1.1 10 s + 5 10  1279 . 6 ,  5 .272 + j 4. 7353 ,  5 .272 =0  j 4. 7353 Rise time < 0.04 sec 261 K = 6000 Roots: 176 s 5 + 2010
. 77 , s , 100 s + 1. 3 10 s + 1.3 10 s + 6 10 = 0  519 . 37 ,  1303 .4,  5 .223 + j 4. 7818 ,  5 .223  j 4. 7818
4 3 8 2 9 9 + 1,020 Max overshoot 2.5% Rise time < 0.04 sec Thus all the required specifications stay within the required tolerances when the value of K varies by plus and minus 20%. Unitstep Responses 1033 Let the transfer function of the controller be Gc ( s) =
The forwardpath transfer function becomes 200 s + 10 s + 50
2 ( ) ( s + 100 ) 2 G ( s) = G c ( s ) G p ( s) = 200,000 K s + 10 s + 50
2 ( ) s ( s + a ) ( s + 100 ) 2 For a = 10, K
v = lim sG ( s ) s 0 = 10 7 K
5 = 100 K = 100 Thus K =1 10 Characteristic Equations: ( K = 1)
a = 10: Roots:
a = 8: + 210 s + 2 .12 10 s + 2 .1 10 s + 10 = 0 4. 978 + j 4. 78 ,  4. 978  j 4. 78 ,  100 + j 447
s
4 3 5 2 6 7 4 3 5 2 6 7 .16  100  j 447 .16 s + 208 s + 2 .116 10 s + 2 . 08 10 s + 10 = 0 Roots:  4. 939 + j 4. 828 ,  4. 939  j 4. 828 ,  99 . 06 + j 446 .97 ,  99 . 06 s
4  j 446 . 97 a = 12: + 212 s 3 + 2 .124 10 + 2 .12 10
5 6 s + 10 7 =0 262 Roots:  5. 017 + j 4. 73 ,  5 . 017  j 4. 73 ,  100 . 98 + j 447 . 36 ,  100 . 98  j 447 . 36 Unitstep Responses: All three responses for a = 8, a = 10, and 12 are similar. 1034 Forwardpath Transfer Function: G ( s) = Y ( s) E (s )
Characteristic Equation: = K s ( s + 1 )( s + 10 ) + K Kt s
3 2 Kv = lim sG( s ) =
s 0
3 2 K =1 10 + KKt s + 11s + (10 + KKt ) s + K = s + 11 s + Ks+ K = 0 K ( s + 1) s
2 For root loci, G eq ( s ) = ( s + 11) 263 Root Locus Plot ( K varies) The root loci show that a relative damping ratio of 0.707 can be realized by two values of K. K = 22 and 59.3. As stipulated by the problem, we select K = 59.3. 1035 Forwardpath Transfer Function: G ( s) = 10 K s ( s + 1) ( s + 10 ) + 10 K t s
K v = lim s 0 sG ( s ) =
10 10 K + 10 K =
t K 1+ K
t =1 Thus K t = K 1 264 Characteristic Equation:
When K = 5.93 and K s
3 2 s ( s + 1 )( s + 10 ) + 10 Kt + 10 K = s + 11 s +10 Ks + 10 K = 0
3 2 t = K  1 = 4. 93
s , the characteristic equation becomes + 11 s + 10 . 046  10 . 046
, + 4. 6 = 0
j 0 .47976 , The roots are:  0 .47723 +  0 .47723  j 0 .47976 1036 Forwardpath Transfer Function: G ( s) = K (1 + aTs ) s ( (1 + Ts ) s + 10 s + KK t
2 ( )
2 K v = lim s 0 sG ( s ) = 1 K
t = 100 Thu s K t = 0 . 01 Let T = 0.01 and a = 100. The characteristic equation of the system is written: s + 110 s + 1000s + K 0.001s + 101s +100 = 0
4 3 2 ( ) To construct the root contours as K varies, we form the following equivalent forwardpath transfer function: G eq ( s ) = 0.001 K s + 101,000 s + 100,000
2 ( s 2 ( s + 10) ( s + 100 ) ) = 0.001K ( s + 1)( s + 50499 )
s
2 ( s + 10) ( s + 100) From the root contour diagram we see that two sets of solutions exist for a damping ratio of 0.707. These are: K = 20: Complex roots:  1.158 + j 1.155 ,  1.158  j1 .155 K = 44.6: Complex roots:  4. 0957 + j 4. 0957 ,  4. 0957  j 4. 0957 The unitstep responses of the system for K = 20 and 44.6 are shown below. Unitstep Responses: 265 1037 Forwardpath Transfer Function: G ( s) = K s K K Ki N 1 s J t L a s + ( Ra J t + L a Bt + K 1 K 2 J t ) s + R a Bt + K 1 K 2 Bt + K b K i + K K Ki Kt 1 2 G ( s) = s s + 3408.33s + 1,204,000 + 1.5 10 KK t
2 8 ( 1.5 10 K
7 ) Ramp Error Constant: K v = lim s 0 sG ( s ) =
1.204 15 K + 150 KK = 100
t Thus 1.204 + 150 KK t = 0 .15 K The forwardpath transfer function becomes G ( s) =
Characteristic Equation: s s + 3408.33s + 150,000 K
2 ( 1.5 10 K
7 3 )
+ 1. 5 10
7 s + 3408 . 33 s + 150 ,000 Ks K =0 When K = 38.667 the roots of the characteristic equation are at  0 .1065 ,  1. 651 + j 1. 65 ,  1.651  j1. 65 ( 0 . 707 for the complex roots) The forwardpath transfer function becomes G ( s) = s s + 3408.33 s + 5.8 10
2 ( 5.8 10 8 6 ) 266 Unitstep Response Unitstep response attributes: Maximum overshoot = 0 Rise time = 0.0208 sec Settling time = 0.0283 sec 1038 (a) Disturburnce toOutput Transfer Function Y (s) TL ( s)
For T ( s )
L
r =0 == = 1/
s s (1 + 0.01 s)(1 + 0.1 s ) + 20 K 2 (1 + 0.1s ) G (s)
c =1 1038 (b) 0 . 01 Thus 10 K Performance of Uncompensated System. K = 10, G ( s ) = 1
lim y ( t ) sY ( s )
t s 0 c = lim = 1 K 10 G ( s) = 200 s (1 + 0.01s ) (1 + 0.1s ) 5.19 dB. The Bode diagram of G ( j ) is shown below. The system is unstable. The attributes of the frequency response are: PM = 9.65 deg GM = (c) Singlestage Phaselead Controller Design
To realize a phase margin of 30 degrees, a = 14 and T = 0.00348. G (s)
c = 1 + aTs 1 + Ts = 1 + 0 . 0487 s 1 + 0 . 00348 s 267 The Bode diagram of the phaselead compensated system is shown below. The performance attributes are: PM = 30 deg GM = 10.66 dB M = 1. 95 BW = 131.6 rad/sec.
r (d) Twostage Phaselead Controller Design
Starting with the forwardpath transfer function G ( s) = s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348 s ) 200 ( 1+ 0.0487 s ) The problem becomes that of designing a singlestage phaselead controller. For a phase margin or 55 degrees, a = 7.385 and T = 0.00263. The transfer function of the secondstage controller is G ( s) c1 = 1 + aTs + Ts = 1 1 + 0 . 01845 + 0 . 00263 s s 1 Thus G ( s) = s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348s )( 1 + 0.00263s ) 200 ( 1+ 0.0487 s ) (1 + 0.01845 s ) The Bode diagram is shown on the following page. The following frequencyresponse attributes are obtained: PM = 55 deg GM = 12.94 dB M
r = 1.11 BW = 256.57 rad/sec Bode Plot [parts (b), (c), and (d)] 268 1039 (a) Twostage Phaselead Controller Design.
The uncompensated system is unstable. PM = 43.25 deg and GM = 18.66 dB. With a singlestage phaselead controller, the maximum phase margin that can be realized affectively is 12 degrees. Setting the desired PM at 11 deg, we have the parameters of the singlestage phaselead controller as a = 128.2 and T = 0 . 00472 . The transfer function of the singlestage controller
1 is G c1 ( s) = 1 + aT 1 s + T1 s = 1 + 0 . 6057 s s 1 1 + 0 . 00472 Starting with the singlestagecontroller compensated system, the second stage of the phaselead controller is designed to realize a phase margin of 60 degrees. The parameters of the secondstage controller are: b = 16.1 and T = 0 . 0066 . Thus,
2 G c2 (s) = 1 + bT s
2 1+T s
2 = 1 + 0 .106 s 1 + 0 . 0066 s 1 G (s)
c = G c 1 ( s )G c 2 ( s ) = 1 + 0 . 6057 s + 0 .106 s 1 + 0 . 00472 s 1 + 0 . 0066 s Forwardpath Transfer Function: 269 G ( s) = G c1 ( s ) G c2 ( s ) G p ( s) = s ( s + 2)( s + 5 ) ( s + 211.86 ) ( s + 151.5 ) = 1. 08
BW = 65.11 rad/sec 1,236,598.6 ( s + 1.651 )( s + 9.39 ) Attributes of the frequency response of the compensated system are: GM = 19.1 dB PM = 60 deg M
r The unitstep response is plotted below. The timeresponse attributes are: Maximum overshoot = 10.2% t
s = 0 .1212 sec t r = 0 . 037 sec (b) Singlestage Phaselag Controller Design.
With a singlestage phaselag controller, for a phase margin of 60 degrees, a = 0.0108 and T = 1483.8. The controller transfer function is 1 + 16 . 08 s G (s) = c 1 + 1483 . 8 s The forwardpath transfer function is G (s) = G c ( s )G p ( s ) = s s + + + + 6 .5 s s 0 . 0662 s 2 5 0 . 000674 BW = 1.07 rad/sec From the Bode plot, the following frequencyresponse attributes are obtained: PM = 60 deg GM = 20.27 dB M
r = 1. 09 The unitstep response has a long rise time and settling time. The attributes are: Maximum overshoot = 12.5% t
s = 12 . 6 sec t r = 2 .126 sec (c) Leadlag Controller Design.
For the leadlag controller, we first design the phaselag portion for a 40degree phase margin. The result is a = 0.0238 and T = 350 . The transfer function of the controller is
1 1 + 350 s The phaselead portion is designed to yield a total phase margin of 60 degrees. The result is b = 4.8 and T = 0 .2245 . The transfer function of the phaselead controller is
2 G c1 ( s) = 1 + 8 . 333 s 1 + 0 .2245 s The forwardpath transfer function of the leadlag compensated system is G c2 (s) = 1 + 1.076 s G ( s) =
Frequencyresponse attributes: Unitstep response attributes: s ( s + 2 ) ( s + 5 ) ( s + 0.00286 ) ( s + 4.454 ) PM = 60 deg GM = 13.07 dB M = 1. 05 BW = 3.83 rad/sec r
Maximum overshoot = 5.9% t
s 68.63 ( s + 0.12 )( s + 0.929 ) = 1. 512 sec t r = 0 . 7882 sec 270 Unitstep Responses. 1040 (a) The uncompensated system has the following frequencydomain attributes: PM = 3.87 deg The Bode plot of G GM = 1 dB M
r = 7 .73 BW = 4.35 rad/sec p ( j ) shows that the phase curve drops off sharply, so that the phaselead controller would not be very effective. Consider a singlestage phaselag controller. The phase margin of 60 degrees is realized if the gain crossover is moved from 2.8 rad/sec to 0.8 rad/sec. The attenuation of the phaselag controller at high frequencies is approximately 15 dB. Choosing an attenuation of 17.5 dB,
20 log 10 we calculate the value of a from a = 17 . 5 dB Thus a = 0.1334 The upper corner frequency of the phaselag controller is chosen to be at 1/aT = 0.064 rad/sec. Thus, 1/T = 0.00854 or T = 117.13. The transfer function of the phaselag controller is 1 + 15 .63 s G (s) = c 1 + 117 .13 The forwardpath transfer function is G ( s) = G c ( s ) G p ( s) = From the Bode plot of G ( j ) , the following frequencydomain attributes are obtained: PM = 60 deg GM = 18.2 dB M
r s (1 + 0.1 s ) (1 + 0.5 s ) (1 + 117.13 s ) (1 + 0.05 s ) = 1. 08 5 ( 1 + 15.63 s ) (1  0.05 s ) BW = 1.13 rad/sec The unitstep response attributes are: maximum overshoot = 10.7% t
s = 10 .1 sec t r = 2 .186 sec Bode Plots 271 1040 (b) Using the exact expression of the time delay, the same design holds. The time and frequency domain attributes are not much affected. 272 1041 (a) Uncompensated System.
Forwardpath Transfer Function: G ( s) =
The Bode plot of G ( j ) is shown below. 10 (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s )
GM = The performance attributes are: PM = 10.64 deg The uncompensated system is unstable. 2.26 dB (b) PI Controller Design.
Forwardpath Transfer Function: G ( s) = Ramperror Constant: K v = lim sG
s 0 10 K p s + K I s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s ) ( s ) = 10 K I = 0 .1 Thu s K I = 0 . 01 0.1 (1 + 100K P s ) ( ) G ( s) = s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s )
P The following frequencydomain attributes are obtained for various values of K KP 0.01 0.02 0.05 0.10 0.12 0.15 0.16 0.17 0.20 . Mr 2.54 2.15 1.52 1.17 1.13 1.14 1.16 1.18 1.29 PM (deg)
24.5 28.24 38.84 50.63 52.87 53.28 52.83 51.75 49.08 GM (dB)
5.92 7.43 11.76 12.80 12.23 11.22 10.88 10.38 9.58 BW (rad/sec)
0.13 0.13 0.14 0.17 0.18 0.21 0.22 0.23 0.25 The phase margin is maximum at 53.28 degrees when K P = 0 .15 . The forwardpath transfer function of the compensated system is G ( s) = s (1 + s ) (1 + 10 s )( 1 + 5 s ) ( 1 + 2 s ) 0.1 (1 + 15 s ) The attributes of the frequency response are: PM = 53.28 deg GM = 11.22 dB M
r = 1.14 BW = 0.21 rad/sec The attributes of the unitstep response are: Maximum overshoot = 14.1% t
r = 10 . 68 sec t s = 48 sec Bode Plots 273 1041 (c) Timedomain Design of PI Controller. 274 By setting K I = 0 . 01 and varying K P we found that the value of K P that minimizes the maximum overshoot of the unitstep response is also 0.15. Thus, the unitstep response obtained in part (b) is still applicable for this case. 1042 Closedloop System Transfer Function. Y (s) R (s ) = 1 s + ( 4 + k 3 ) s + ( 3 + k 2 + k3 ) s + k1
3 2 1 2 For zero steadystate error to a step input, k =1. For the complex roots to be located at 1 +j and 1 and solve for zero remainder.  j, we divide the characteristic polynomial by s s + ( 2 + k2 )
3 +2s+ 2 s + 2 s + 2 s + ( 4 + k3 ) s + ( 3 + k2 + k3 ) s + 1
2 2 s +
3 2s
3 2 + 2s
2 2 3 ( 2 + k ) s + (1 + k + k ) s + 1 ( 2 + k ) s + ( 4 + 2k ) s + 4 + 2k
2 3 3 3 ( 3+k
For zero remainder, 2  k 3 ) s  3  2 k3
Thus Thus k
3  3  2k3 = 0 0.5. = 1. 5
k
2 3 + k 2  k 3 = 0 = 1. 5 The third root is at steadystate error. Not all the roots can be arbitrarily assigned, due to the requirement on the 1043 (a) Openloop Transfer Function. G ( s) = X 1 (s ) E (s ) = k3 s s + (4 + k 2 ) s + 3 + k 1 + k 2 2 1 2 3 Since the system is type 1, the steadystate error due to a step input is zero for all values of k , k , and k that correspond to a stable system. The characteristic equation of the closedloop system is
3 2 s + ( 4 + k2 ) s + ( 3 + k1 +k 2 )s + k 3 = 0
and For the roots to be at 1 + j, 1  j, 10, the equation should be: 3 2 s + 12 s + 22 s + 20 = 0
2 Equating like coefficients of the last two equations, we have 4 + k = 12 3 Thus Thus Thus k + k 1 + k 2 = 22
k
3 2 1 =8 = 11 k = 20 k3 = 20 (b) Openloop Transfer Function. Y (s ) Gc ( s) 20 = = 2 E ( s ) ( s + 1) ( s + 3 ) s s + 12 s + 22 ( ) Thus Gc ( s) = 20 ( s + 1 ) ( s + 3 ) s s + 12 s + 22
2 ( ) 1044 (a) 275 0 25.92 A = 0 2.36 1 0 0 0 0 0 0 0 0 0 1 0 0 0.0732 B = 0 0.0976 The feedback gains, from k 1 to k 4 : 2.4071E+03
The A 4.3631E+02  B K 8.4852E+01 1.0182E+02 matrix of the closedloop system 1.5028E+02 0.0000E+00 0.0000E+00 2/3258E+02 3.1938E+01 1.0000E+00 6.2112E+00 0.0000E+00 4.2584E+01 0.0000E+00 0.0000E+00 7.4534E+00 0.0000E+00 1.0000E+00 8.2816E+00 9.9379E+00 The B vector 7.3200E02 0.0000E+00 0.0000E+00 9.7600E02 Time Responses: 1044 (b) The feedback gains, from k 1 to k 2 : 9.9238E+03
The A 1.6872E+03  B K 1.3576E+03 8.1458E+02 matrix of the closedloop system 276 7.0051E+02 0.0000E+00 0.0000E+00 9.6621E+02 1.2350E+02 1.0000E+00 0.0000E+00 1.6467E+02 9.9379E+01 0.0000E+00 0.0000E+00 5.9627E+01 0.0000E+00 1.0000E+00 1.3251E+02 7.9503E+01 The B vector 7.3200E02 0.0000E+00 0.0000E+00 9.7600E02 Time Responses: 1045 The solutions are obtained by using the poleplacement design option of the linsys program in the ACSP software package. (a) The feedback gains, from k 1 to k 2 : 6.4840E+01 5.6067E+00 2.0341E+01 2.2708E+00 The A  B K matrix of the closedloop system 3.0950E+02 4.6190E+02 0.0000E+00 0.0000E+00 1.0000E+00 0.0000E+00 0.0000E+00 1.1463E+02 1.4874E+01 0.0000E+00 0.0000E+00 1.0000E+00 3.6724E+01 1.7043E+02 1.477eE+01 3.6774E+01 The B vector 6.5500E+00 0.0000E+00 0.0000E+00 277 6.5500E+00 (b) Time Responses: x ( 0 ) =
0 .1 0 0 0
' With the initial states x ( 0 ) = 0 .1 0 0 0 , the initial position of ' from its stable equilibrium position. The steel ball is initially pulled toward the magnet, so x 1 or y1 is preturbed downward x 3 = y 2 is x 3 or y 2
is preturbed negative at first. Finally, the feedback control pulls both bodies back to the equilibrium position. With the initial states x ( 0 ) = 0 0 0 .1 0 , the initial position of ' downward from its stable equilibrium. For t > 0, the ball is going to be attracted up by the magnet toward the equilibrium position. The magnet will initially be attracted toward the fixed iron plate, and then settles to the stable equilibrium position. Since the steel ball has a small mass, it will move more actively. 1046 (a) Block Diagram of System. 278 u = k1 x1 + k 2 (  x1 + w1 ) dt
State Equations of Closedloop System: dx1 dt 2  k1 1 x1 0 + = dx2  k2 0 x2 k2 dt Characteristic Equation: 1 w1 0 w2 sI  A =
For s = s + 2 + k1 k2
s2 1 s = s + ( 2 + k1 ) s + k2 = 0
2 10, 10,
X (s) sI A = = + 20 s + 200 = 0
2 Thus k1 = 18 and k2 = 200 = X 1 (s) 200 W 1 ( s ) s 1+ 2 s
1 + s 1W 2 ( s )
1 + 18 = s + 200 s 2 = 200 W 1 ( s ) s
2 + sW 2 ( s ) + 20 s + 200 W1 ( s ) = 1 s W2 (s) W2 s W2 = const ant X ( s) = s s + 20 s + 200
2 ( 200 + W2s ) lim x( t) = lim sX ( s) = 1
t s 0 1046 (b) With PI C ontroller:
Block Diagram of System: 279 Set K P = 2 and KI = 200 . X (s ) = (K P s + K I ) W1 ( s ) + sW2 ( s ) s + 20 s + 200
2 = ( 2s + 200 ) W ( s ) + sW ( s )
1 2 s + 20 s + 200
2 Time Responses: 280 Chapter 11 THE VIRTUAL LAB Part 1) Solution to Lab questions within Chapter 11
1151 Open Loop Speed 1. Open loop speed response using SIMLab: a. +5 V input: The form of response is like the one that we expected; a second order system response with overshoot and oscilla tion. Considering an amplifier gain of 2 and K b = 0.1 , the desired set point should be set to 2.5 and as seen in the figure, the final value is approximately 50 rad/sec which is armature voltage divided by K b . To find the above response the systems parameters are extracted from 1131 of the text and B is calculated from 113 by having m as: m = Ra J m , Ra B + k b k m B= Ra J m  k b k m m = 0.000792kg m 2 / sec Ra m b. +15 V input: 282 c. 10 V input: 2. Study of the effect of viscous friction: 283 The above figure is plotted for three different friction coefficients (0, 0.001, 0.005) for 5 V armature input. As seen in figure, two important effects are observed as the viscous coefficient is increased. First, the final steady state velocity is decreased and second the response has less oscillation. Both of these effects could be predicted from Eq. (11.1) by increasing damping ratio . 284 3. Additional load inertia effect: As the overall inertia of the system is increased by 0.005 / 5.2 2 and becomes 1.8493 10 3 kg.m2 , the mechanical time constant is substantially increased and we can assume the first order model for the motor (ignoring the electrical subsystem) and as a result of this the response is more like an exponential form. The above results are plotted for 5 V armature input. 285 4. Reduce the speed by 50% by increasing viscous friction: As seen in above figure, if we set B=0.0075 N.s/m the output speed drop by half comparing with the case that B=0 N.s/m. The above results are plotted for 5 V armature input. 286 5. Study of the effect of disturbance: Repeating experiment 3 for B=0.001 N.s/m and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of open loop system is like the effect of higher viscous friction and caused to decrease the steady state value of speed. 287 6. Using speed response to estimate motor and load inertia: Using first order model we are able to identify system parameters based on unit step response of the system. In above plot we repeated the experiments 3 with B=0.001 and set point voltage equal to 1 V. The final value of the speed can be read from the curve and it is 8.8, using the definition of system time constant and the cursor we read 63.2% of speed final value 5.57 occurs at 0.22 sec, which is the system time constant. Considering Eq. (113), and using the given value for the rest of parameters, the inertia of the motor and load can be calculated as: J= m ( Ra B + K m K b ) 0.22(1.35 0.001 + 0.01) = = 1.8496 10  3 kg.m2 Ra 1.35 We also can use the open loop speed response to estimate B by letting the speed to coast down when it gets to the steady state situation and then measuring the required time to get to zero speed. Based on this time and energy conservation principle and knowing the rest of parameters we are able to calculate B. However, this method of identification gives us limited information about the system parameters and we need to measure some parameters directly from motor such as Ra , K m , K b and so on. So far, no current or voltage saturation limit is considered for all simulations using SIMLab software. 288 7. Open loop speed response using Virtual Lab: a. +5 V: 289 b. +15 V: c. 10 V: 290 Comparing these results with the part 1, the final values are approximately the same but the shape of responses is closed to the first order system behavior. Then the system time constant is obviously different and it can be identified from open loop response. The effect of nonlinearities such as saturation can be seen in +15 V input with appearing a straight line at the beginning of the response and also the effects of noise and friction on the response can be observed in above curves by reducing input voltage for example, the following response is plotted for a 0.1 V step input: 291 8. Identifying the system based on open loop response: Open loop response of the motor to a unit step input voltage is plotted in above figure. Using the definition of time constant and final value of the system, a first order model can be found as: G( s ) = 9 , 0.23s + 1 where the time constant (0.23) is found at 5.68 rad/sec (63.2% of the final value). 292 1152 Open Loop Sine Input 9. Sine input to SIMLab and Virtual Lab (1 V. amplitude, and 0.5, 5, and 50 rad/sec frequencies) a. 0.5 rad/sec (SIMLab): 293 b. 5 rad/sec (SIMLab): c. 50 rad/sec (SIMLab): 294 d. 0.5 rad/sec (Virtual Lab): e. 5 rad/sec (Virtual Lab): 295 f. 50 rad/sec (Virtual Lab): 10. Sine input to SIMLab and Virtual Lab (20 V. amplitude) a. 0.5 rad/sec (SIMLab): 296 b. 5 rad/sec (SIMLab): c. 50 rad/sec (SIMLab): 297 d. 0.5 rad/sec (Virtual Lab): e. 5 rad/sec (Virtual Lab): 298 f. 5 rad/sec (Virtual Lab): In both experiments 9 and 10, no saturation considered for voltage and current in SIMLab software. If we use the calculation of phase and magnitude in both SIMLab and Virtual Lab we will find that as input frequency increases the magnitude of the output decreases and phase lag increases. Because of existing saturations this phenomenon is more sever in the Virtual Lab experiment (10.f). In this experiments we observe that M = 0.288 and = 93.82 o for = 50. 299 1153 Speed Control 11. Apply step inputs (SIMLab)
In this section no saturation is considered either for current or for voltage. a. +5 V: 300 b. +15 V: c. 10 V: 301 12. Additional load inertia effect: a. +5 V: b. +15 V: 302 c. 10 V: 13. Study of the effect of viscous friction: 303 As seen in above figure, two different values for B are selected, zero and 0.0075. We could change the final speed by 50% in open loop system. The same values selected for closed loop speed control but as seen in the figure the final value of speeds stayed the same for both cases. It means that closed loop system is robust against changing in system's parameters. For this case, the gain of proportional controller and speed set point are 10 and 5 rad/sec, respectively. 14. Study of the effect of disturbance: Repeating part 5 in section 1151 for B=0.001 and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of closed loop system is not substantial like the one on the open loop system in part 5, and again i is shown the robustness of closed loop system against disturbance. Also, to study the effects of t conversion factor see below figure, which is plotted for two different C.F. and the set point is 5 V. 304 By decreasing the C.F. from 1 to 0.2, the final va lue of the speed increases by a factor of 5. 15. Apply step inputs (Virtual Lab) a. +5 V: 305 b. +15 V: c. 10 V: 306 As seen the responses of Virtual Lab software, they are clearly different from the same results of SIMLab software. The nonlinearities such as friction and saturation cause these differences. For example, the chattering phenomenon and flatness of the response at the beginning can be considered as some results of nonlinear elements in Virtual Lab software. 1154 Position Control 16. 160 o step input (SIMLab) 307 17. 0.1 N.m step disturbance 18. Examine the effect of integral control 308 In above figure, an integral gain of 1 is considered for all curves. Comparing this plot with the previous one without integral gain, results in less steady state error for the case of controller with integral part. 19. Additional load inertia effect (J=0.0019, B=0.004): 309 20. Set B=0: 21. Study the effect of saturation 310 The above figure is obtained in the same conditions of part 20 but in this case we considered 10 V. and 4 A. as the saturation values for voltage and current, respectively. As seen in the figure, for higher proportional gains the effect of saturations appears by reducing the frequency and damping property of the system. 22. Comments on Eq. 1113 After neglecting of electrical time constant, the second order closed loop transfer function of position control obtained in Eq. 1113. In experiments 19 through 21 we observe an under damp response of a second order system. According to the equation, as the proportional gain increases, the damped frequency must be increased and this fact is verified in experiments 19 through 21. Experiments16 through 18 exhibits an over damped second order system responses. 23. In following, we repeat parts 16 and 18 using Virtual Lab: Study the effect of integral gain of 5: 311 312 Ch. 11 Problem Solutions Part 2) Solution to Problems in Chapter 11 111. In order to find the current of the motor, the motor constant has to be separated from the electrical component of the motor. The response of the motor when 5V of step input is applied is: a) The steady state speed: 41.67rad/sec b) It takes 0.0678 second to reach 63% of the steady state speed (26.25rad/sec). This is the time constant of the motor. c) The maximum current: 2.228A 313 11.2 The steady state speed at 5V step input is 50rad/sec. a) It takes 0.0797 seconds to reach 63% of the steady state speed (31.5rad/sec). b) The maximum current: 2.226A c) 100rad/sec 314 113 a) b) c) d) 50rad/sec 0.0795 seconds 2.5A. The current When Jm is increased by a factor of 2, it takes 0.159 seconds to reach 63% of its steady state speed, which is exactly twice the original time period . This means that the time constant has been doubled. 315 114 Part 1: Repeat problem 111 with TL = 0.1Nm a) It changes from 41.67 rad/sec to 25 rad/sec. b) First, the speed of 63% of the steady state has to be calculated. 41.67  (41.67  25) 0.63 = 31.17 rad/sec. The motor achieves this speed 0.0629 seconds after the load torque is applied c) 2.228A. It does not change Part 2: Repeat problem 112 with TL = 0.1Nm a) It changes from 50 rad/sec to 30 rad/sec. b) The speed of 63% of the steady state becomes 50  (50  30) 0.63 = 37.4 rad/sec. The motor achieves this speed 0.0756 seconds after the load torque is applied c) 2.226A. It does not change. 316 Part 3: Repeat problem 113 with TL = 0.1Nm a) It changes from 50 rad/sec to 30 rad/sec. b) 50  (50  30) 0.63 = 37.4 rad/s The motor achieves this speed 0.0795 seconds after the load torque is applied. This is the same as problem 113. c) 2.5A. It does not change d) As TL increases in magnitude, the steady state velocity decreases and steady state current increases; however, the time constant does not change in all three cases. 317 115 The steady state speed is 4.716 rad/sec when the amplifier input voltage is 5V: 116 a) 6.25 rad/sec. b) 63% of the steady state speed: 6.25 0.63 = 3.938 rad/sec It takes 0.0249 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 1 Ampere. 318 117 a) 9.434 rad/sec. b) 63% of the steady state speed: 9.434 0.63 = 5.943 rad/sec It takes 0.00375 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 10 Amperes. d) When there is no saturation, higher Kp value reduces the steady state error and decreases the rise time. If there is saturation, the rise time does not decrease as much as it without saturation. Also, if there is saturation and Kp value is too high, chattering phenomenon may appear. 118 a) The steady state becomes zero. The torque generated by the motor is 0.1 Nm. b) 6.25  (6.25  0) 0.63 = 2.31 rad/sec. It takes 0.0249 seconds to reach 63% of its new steady state speed. It is the same time period to reach 63% of its steady state speed without the load torque (compare with the answer for the Problem 116 b). 119 The SIMLab model becomes The sensor gain and the speed input are reduced by a factor of 5. In order to get the same result as Proble m 116, the Kp value has to increase by a factor of 5. Therefore, Kp = 0.5. The following graphs illustrate the speed and current when the input is 2 rad/sec and Kp = 0.5. 319 1110 320 a) 1 radian. b) 1.203 radians. c) 0.2215 seconds. 1111 a) The steady state position is very close to 1 radian. b) 1.377 radians. c) 0.148 seconds. It has less steady state error and a faster rise time than Problem 1110, but has larger overshoot. 321 1112 Different proportional gains and the ir corresponding responses are shown on the following graph. As the proportional gain gets higher, the motor has a faster response time and lower steady state error, but if it the gain is too high, the motor overshoot increases. If the system requires that there be no overshoot, Kp = 0.2 is the best value. If the system allows for overshoot, the best proportional gain is dependant on how much overshoot the system can have. For instance, if the system allows for a 30% overshoot, Kp = 1 is the best value. 322 1113 Let Kp = 1 is the best value. As the derivative gain increases, overshoot decreases, but rise time increases. 323 1114 324 1115 There could be many possible answers for this problem. One possible answer would be Kp = 100 Ki= 10 Kd= 1.4 The Percent Overshoot in this case is 3.8%. 325 1116 0.1 Hz 0.2 Hz 326 0.5 Hz 1 Hz 327 2Hz 5Hz 328 10Hz 50Hz 329 As frequency increases, the phase shift of the input and output also increase. Also, the amplitude of the output starts to decrease when the frequency increases above 0.5Hz. 1117 As proportional gain increases, the steady state error decreases. 330 1118 Considering fast response time and low overshoot, Kp =1 is considered to be the best value. 331 1119 It was found that the best Kp = 1 As Kd value increases, the overshoot decreases and the rise time increases. 332 Appendix H GENERAL NYQUIST CRITERION H1 (a)
L( s ) = 5( s s(s  2) + 1 )( s  1 ) P =1 P =1 When = 0 : L ( j 0 ) = 90 o o L( j0 ) = =0 When = : L ( j ) = 180 L ( j) L ( j ) = 5( j  2)  j 1 + ( 2 ) =  5( + 2 j ) 1+ ( 2 ) When L ( j ) = 0 , = . The Nyquist plot of L ( j ) does not intersect the real axis except at the origin.
o o o 11 = ( Z  0.5 P  P )180 = ( Z  1.5 ) 180 =  90 Thus, Z=1. The closedloop system is unstable. The characteristic equation has 1 root in the righthalf splane. Nyquist Plot of
L ( j ): (b) L ( j ) =
s( s 50 + 5)( s  1 ) P =1
o P =1 When = 0 : L ( j 0 ) = 90 L( j0)
o = =0 When = : L ( j ) = 270 L ( j) 333 L ( j ) = 50
2 4  j 5 + ( 2 ) = 50 4 + j 5 + 2 16 + 4 2 (5+ )
2 ( 2 ) 2 For Im L ( j ) = 0 , = . Thus, the Nyquist plot of L(s) intersects the real axis only at the origin where = . 11 = ( Z  0.5 P  P )180 = ( Z  1.5 ) 180 = 90 o o o Thus, Z = 2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. The Nyquist plot of
L ( j ): (c) L ( s ) = s s + 3s + 1
3 ( 3( s + 2) ) P =1 P=2 When = 0 : L ( j 0 ) = 90 o o L( j0 ) = =0
4 2 When = : L ( j ) = 270 3( j + 2)
4 L ( j) L ( j ) = (  3 2 ) + j
or = 3( j + 2)  3 (4 ( 4  3 2 ) )  j 2 2 + Setting Im L ( j ) = 0,  3  2 = 0
4 2 = 3 . 56
2
o = 1.89 rad/sec.
o o L ( j 1.89 ) =3 11 = ( Z  0.5 P  P )180 = ( Z  2.5 ) 180 =  90 Thus, Z = 2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. 334 Nyquist Plot of L ( j ): (d) L ( s ) = 100 s ( s + 1) s + 2
2 ( ) P =3 o o P=0
L( j0 ) When = 0 : L ( j 0 ) = 90 = When = : L ( j ) = 360 L ( j) = 0 The phase of L ( j ) is discontinuous at = 1.414 rad/sec. 11 = 35.27 + 270  215.27
o o ( o ) = 90 o 11 = ( Z  1.5) 180 = 90
o o Thus, P 11 = 360 180 o o = 2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. Nyquist Plot of
L ( j ): 335 (e) L(s ) = s s + 2 s + 2 s + 10
3 2 ( s  5s + 2
2 )
L( j0 ) P =1 P=2 When = 0 : L ( j 0 ) = 90 o o = =0
2 4 2 2 When = : L ( j ) = 180
2 L ( j) ( 2  )  j5 (  2 )  j (10  2 ) ( 2  )  j5 L ( j ) = =  2 ) + j (10  2 ) ( (  2 ) + (10  2 )
4 2 2 4 2 2 2 2 2 Setting L( j) = 0,
4 ( 2  )(10  2 ) + 5( 2 2 4  2 2 )= 0
/ sec or  3 .43 + 2 .86 = 0 2 Thus, = 1.43 o r 2.0 1. = 1.2 rad / sec or L ( j1.2 ) =  0 . 7 L ( j1.42 ) = 0 . 34
2
o o 1.42 r ad 11 = ( Z  0.5 P  P )180 = ( Z  2.5 )180 = 90 o Thus, Z = 2 . The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. Nyquist Plot of
L ( j ): (f) L(s ) =  s 1
2 2 ( ) ( s + 2 ) = 0.1s  0.2 s + 0.1s + 0.2 s ( s + s + 1) s ( s + s + 1)
3 2 2 P =1 P=0 When = 0 : L ( j 0 ) = 90 o o L( j0 ) = = 0 .1 When = : L ( j ) = 180 L( j) 336 ( 0.2 + 0.2 ) + 0.1 j ( 1 + ) = (1+ ) ( 0.2 + j 0.1 )   j ( 1 ) L ( j ) =  + j (1  ) + ( 1 )
2 2 2 2 2 2 2 4 2 2 2 Setting Im L ( j ) = 0 , = 2 .
2
o Thus,
o = 1.414 rad/sec
o L ( j 1.414 ) =  0 .3 11 = ( Z  0.5 P  P )180 = ( Z  0.5 ) 180 =  90 The closedloop system is stable. Nyquist Plot of
L ( j ): Thus, Z = 0 H2 (a) L(s ) = K ( s  2) s s 1
2 ( ) P =1 P =1
o o For stability, Z = 0. 11 =  ( 0.5 P + P ) 180 = 270 For K > 0, 11 = 90 . For K < 0, 11 = +90
o o The system is unstable.
o 270 The system is unstable. Thus the system is unstable for all K. 337 Nyquist plot (b) L( s ) =
s(s K + 10 )( s  2) P =1 P =1 For stability, Z = 0. 11 =  ( 0.5 P + P ) 180 = 1.5 180 =  270
o o o For K > 0, 11 = 90 o .
o The system is unstable. 270
o For K < 0, 11 = 90 Nyquist Plot of
L ( j ): . The system is unstable for all values of K . 338 (c) L ( s ) = s s + 3s + 1
3 ( K ( s + 1) )
o o P =1 P = 2 For stability, Z = 0. 11 =  ( 0.5 P + P 1 ) 180 =  450
o o For K > 0, 11 = 90 For K < 0, 11 = +90 . The system is unstable. when K > 1. 11 = 270
o when K < 1. The Nyquist plot of L ( j ) crosses the real axis at K, and the phase crossover frequency is 1.817 rad/sec. The system is unstable for all values of K. Nyquist Plot of
L ( j ): (d) L ( s ) = K s  5s + 2
2 s s + 2 s + 2 s + 10
3 2 ( ( ) ) P =1 P = 2 For stability, Z = 0. 11 =  ( 0.5 P + P ) 180 = 2.5 180 =  450
o o o The Nyquist plot of L ( j ) intersects the real axis at the following points: = : L ( j 0 ) = 0 = 1.42 rad / sec: L ( j 1.42 ) =  0 . 34 = 1.2 rad / sec: L ( j 1.4 ) = 0 . 7 K K 339 0 1.43 < < K K < 1.43 < 2 . 94 11 = 90 11 = 270 11 = 90 11 = 90
o o o o 2 . 94 K <K <0 Since 11 does not equal to  450 Nyquist Plot of
L ( j ): o for any K, the system is unstable for all values of K . (e) L(s ) = K s 1
2 2 ( )( s + 2) s ( s + s + 1)
L ( j ) P =1 P=0 For stability, Z = 0. The Nyquist plot of 11 =  ( 0.5 P + P ) 180 =  90
o o intersects the real axis at: 340 = :
=
2 rad / sec: L ( j ) L( j = K 2 ) = 3K K>0
K 11 = 90 11 = 90
1 3 o o o Unstable Unstable Unstable Stable < 1 1 < K <  
1 3 11 = 270 11 = 90 <K <0 o The system is stable for  1 3 <K <0. (f) L(s ) = ) s ( s + 1) ( s + 4 )
K s  5s+1
2 2 ( P =3 P=0 For stability, Z = 0. 11 =  ( 0.5 P + P ) 180 = 270
o o The Nyquist plot of L ( j ) plot intersects the real axis at = :
< = 0 . 92
0 K K K L ( j ) rad / sec: =0
L ( j 0 . 92 ) = 1. 3 K
o o < 0 . 77 11 = 90  180 11 = 90
o o =  270 o Stable Unstable Unstable > 0 . 77 <0 11 = 270 The system is stable for 0 < K < 0.77. 341 Nyquist Plot: H3 (a) = ( Z  0.5 P  P )180 = ( Z  1.5 ) 180 = 90
o o 11 o Thus, Z = 2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. (b) 11 = ( Z  0.5 P  P )180 = ( Z  1 ) 180 =  180 o o o Thus, Z = 0 . The closedloop system is stable. (c) = ( Z  0.5 P  P )180 = ( Z  2 )180 = 360
o o 11 o Thus, Z = 0 . The closedloop system is stable. (d) = ( Z  0.5 P  P )180 = Z 180 = 360
o o 11 o Thus, Z = 2 The closedloop system is unstable. The characteristic equation has two roots in the righthalf splane. (e) = ( Z  0.5 P  P )180 = ( Z  1 ) 180 =  162  18
o o o 11 o Thus, Z = 0 . The closedloop system is stable. (f) 11 = ( Z  0.5 P  P )180 = ( Z  0.5 ) 180 =  90 o o o Thus, Z = 0 The closedloop system is stable. H4 (a) The stability criterion for 1 / L ( s ) is the same as that for
o L ( s ).
o Thus, 11 = ( Z  0.5 P  P )180 = ( Z  1.5 ) 180 From Fig. HP4 (a), 11 = 270
o . Thus , Z = 0. The closedloop system is stable. 342 (b) The stability criterion for 1 / L ( s ) is the same as that for 11 = ( Z  0.5 P  P )180 = ( Z  1 ) 180 o L ( s ).
o Thus, From Fig. HP4 (b), 11 = 180 H5
o . Thus, Z = 0. The closedloop system is stable. Thus, 11 = 270
o (a) = ( Z  0.5 P  P )180 = ( Z  1.5 ) 180
11 o For stability, Z = 0. 0 <
K K K < 0 .5 11 = 90 o o Unstable Stable Unstable
o > 0 .5 <0 11 = 270 11 = 90 The system is stable for K > 0.5. (b) = ( Z  0.5 P  P )180 = ( Z  1 ) 180
o 11 o For stability, Z = 0. Thus, 11 = 180 o 0 K K < K < 0 .5 11 = 90 o o Unstable Stable Unstable
o > 0 .5 <0 11 = 270 11 = 90 The system is stable for K > 0.5. 343 (c) = ( Z  0.5 P  P )180 = ( Z  0.5 ) 180
o 11 o For stability, Z = 0. Thus, 11 = 90
o o o o 0 < K < 0 .5
K 11 = 90 11 = 270 11 = 90 11 = 90
o Stable Unstable Stable Unstable 0 .5 K K < >1 <1 <0 The system is stable for K > 1. (d) = ( Z  0.5 P  P )180 = ( Z  2 ) 180
o 11 o For stability, Z = 0. Thus, 11 = 360
o o o 0 K K < K < 0 .5 11 = 0 11 = 0 Unstable Stable Unstable > 0 .5 <0 11 = 360
o The system is stable for K > 0.5. H6 (a) + 4 Ks + ( K + 5) s + 10 = 0 Ks ( 4 s + 1 ) L (s) = P =0 eq 3 s + 5 s + 10
s
3 2 P =2 =0 =0 When = 0 : L eq ( j 0 ) = 90 When = : Leq ( j ) =
4 2 o o L eq ( j0 )
eq L eq ( j ) =  90
2 L ( j )
2 K ( j  4 ) 10 + j (5  )
2 2 = K ( j  4 ) 10  j 5  100 + Thus,
o 2 ( (5 )
2 2 ) 2 Setting Im L eq ( j ) =0  5  2 .5 = 0 = 5 .46 = 2 .34 rad/sec.
o L eq ( j 2 . 34 ) = 2 . 18 K For stability, 11 =  ( 0.5 P + P ) 180 = 360 Thus for the system to be stable, the  1 point should be encircled by the Nyquist plot once. The system is stable for K > 1/2.18 = 0.458. 344 Routh Tabulation:
s s s
3 1 4K 4K
2 K +5
10 K K
2 2 >0 1 + 20
4K K  10 + 5 K  2 .5 > 0 + 5 .45)(
K s o 10 (K  0 .458 ) >0 For system stability, K > 0.458. (b) s 3 + K ( s 3 + 2s 2 +1 ) = 0
Leq ( s) = K s + 2s + 1
3 ( ) s 3 P = 3
o P =0
L
eq When = 0: When = : Leq ( j ) =
1 L eq ( j 0 ) =  270 L eq ( j ) = 0
o ( j0 ) = L eq ( j ) =K Setting =K shows that its real part is
Im L ( j ) K 1  2 ( 2 )  j
3 3  j K 3 + j ( 1  2 2 ) = 3 L eq ( j 0 . 707 ) eq =0  2 3 2 =0 = , = 0 . 707 rad/sec The Nyquist plot is a straight line, since the equation of always equal to K for all values of . For stability, =  ( 0.5 P + P ) 180 = 270
o 11 o L eq ( j ) K>0 K < 0 11 = 90 11 = 90
o o Unstable Unstable The system is unstable for all values of K. Routh Tabulation:
s s s
3 1+ K 2K 0 K K K 2 >0 < 1 >0 1 1  K
2K s o K K The system is unstable for all values of K . 345 (c) s ( s + 1) ( s2 + 4) + K ( s 2 + 1) = 0
Leq ( s) = K s +1
2 s ( s + 1) s + 4
2 ( ( ) ) P = 3 P =0 For stability, 11 =  ( 0.5 P + P ) 180 = 270
o o 0 K < K < 11 =  270 11 = 270
o o Stable Unstable <0 (d) Leq ( s) =
Leq ( j ) = s s + 2 s + 20
2 ( 10 K ) P =1 2 P =0 10 K  2 + j 20  2 ( 2 ) = 10 K 2  j 20  4 + 4 2 ( 20  )
2 ( 2 ) 2 Setting
L
eq Im L eq ( j ) K 4 =0 = 4.47 ( j 4.47 ) = For stability, 11 =  ( 0.5 P + P ) 180 =  90
o o K 0 >4 <
K 11 = 270 <4 11 = 90 o o Unstable Stable The system is stable for 0 < K < 4. Routh Tabulation
s s s
3 1 2 40 20 10 K K K 2 1  10
2 <4 >0 For stability, 0 < K < 4 s o 10 K K 346 (e) s ( s + 2s + s +1) + K ( s + s + 1) = 0
3 2 2 Leq ( s) = s s + 3s + 3
3 ( K ( s + 2) ) P =1 = K P=2 L eq ( j 0 ) =  90 o L eq ( j ) = 0  270 o Leq ( j ) = Setting ( K ( j + 2)
4  3
( j ) 2 ) + j 3 ( 2 4  3 2 ) + j ( 4  3 2  6 ) ( 4  3 2 ) 2 + 4 2 Im L eq =0 4  3 2  6 = 0 The real positive solution is = 2 . 09 rad/sec
L eq ( j 2 . 09 ) = 0 . 333 K For stability, 11 =  ( 0.5 P + P ) 180 = 450
o o K K 0 >0 < 3 >
K 11 = 90 11 = +90 > 3 o o o Unstable Unstable Unstable 11 = 270 The system is unstable for all values of K . Routh Tabulation s s s 4 1 3 3 + K 2 K 3 3 2 3 K 3  K 2 2K K < 3 s 1 ( 3 + K )  2 K ( 3 + K )
s
o 1 2K K >0 The conditions contradict. The system is unstable for all values of K. 347 Appendix I DISCRETEDATA CONTROL SYSTEMS I1 (a)
F (z) = (b) (ze )
3 ze3 2 F (z ) = 2 j ze 2 j ze  2 2 j ( z  e 2 j ) ( z  e 2 j ) 2 1 (c)
z z F (z ) = z  e  (2 j )  z  e  (2+ j ) 2j 1 (d)
F (z ) = ze (z + e ) (z  e )
2 2 2 3 I2 (a)
F (z ) =
 j 2T ze j 2 T ze  2 2 2 j ( z  e j 2T ) ( z  e  j 2T ) T (b)
F (z) =1  z 1 +z 2 z 3 +z 4 L Add Eqs. (1) and (2). We have (1 + z ) F ( z ) = 1
1 (1) z 1 F (z ) = z 1 z 2 +z 3 z 4 +L (2) Or, F (z) =z (z + 1) I3 (a)
Z
2 5 T 2 10T T e z T e z 1 ( 1) 2 2 z = = + 2 aT 2 ( s + 5) 3 2 ! a z  e a = 5 2 ( z  e 5T ) ( z  e 5T )3 (b)
F (s) = = 1 s (s z z
3 + 1) 
(z =
Tz 1 s  1 s
2 +
2 1 s
3  1 s +1 
z z F (z) 1  1) =  2 + T z(z + 1)
3 2( z  1)
10 e
2 T (c)
F (s) =
s(s 10 10 25 s + 5) 2  10 z 25 ( s + 5)  
(s + 5)
5T 5 T 2 F (z ) = 10 z 25( z  1) 25 z  e ( 5 T ) (z e ) 2 Tze 2 (d)
F (s ) = s s +2
2 ( 5 ) = 2.5 s  2.5 s s +2
2 F (z) = 2 .5 z z 1  2 .5 z ( z z
2  cos T )  2 z cos T + 1 348 I4 (a)
F (z) = 10 z (z  1 )( z  0 .2 ) = 12 . 5 z z 1  12 . 5 z z  0 .2 f (k ) = 12 . 5 1  ( 0 .2 ) k k = 0 , 1, 2, L (b)
F (z ) = ( z  1) ( z z
2 + z +1 ) = ( 0.1667 + j 0.2887 ) z ( 0.1667  j 0.2887) z  3( z  1 ) ( z + 0.5 + j0.866) ( z + 0.5  j0.866 )
z   (0.1667  j 0.2887)e
j 2 k 3 f ( k ) = 0.333  (0.1667 + j 0.2887) e = 0.333  0.333cos  j 2 k 3 2 k 3  0.576sin 2 k 3 = 0.333 + 0.666cos 2 k + 240 0 k = 0,1, 2 , L 3 (c)
F (z) = z (z  1 )( z + 0 . 85) = 0 . 541 z z 1  0 . 541 z z + 0 .85 f (k ) = 0 .541 1  ( 0 . 85) k k = 0 , 1, 2 , L (d)
F (z) = 10 (z  1 )( z  0 . 5) = 20 z 1  20 z  0 .5 f (0) =0 f (k ) = 20 1  ( 0 .5) k 1 k = 0 , 1, 2 , L I5 (a)
lim f ( k ) = lim 1  z
k z 1 ( 1 ) F ( z ) = lim z
z 1 0.368
2  1.364 z + 0.732
z
1 =1 Expand F(z) into a power series of k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 f(k) 0.000000 0.000000 0.368000 0.869952 1.285238 1.484260 1.451736 1.261688 1.026271 0.844277 0.768362 0.798033 0.894075 1.003357 1.082114 . k 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 f(k) 1.109545 1.089310 1.041630 0.991406 0.957803 0.948732 0.960956 0.984269 1.007121 1.021225 1.023736 1.016836 1.005586 0.995292 0.989487 349 (b)
F (z) = 10 z (z  1 )( z + 1 ) The function 1  z ( 1 ) F ( z) has a pole at z = 1 , so the finalvalue theorem cannot be applied. The response f(k) oscillates between 0 and 10 as shown below. k 1 2 3 4 5 6 7 8 9 10 f(k) 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 k 11 12 13 14 15 16 17 18 19 20
z (z z f(k) 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 0.000000 10.000000 (c)
k lim f (k ) = lim z 1  1 z
1 F (z) = lim z 1  0 . 5) = 2 F(z) is expanded into a power series of k 1 2 3 4 5 6 7 8 9 10 f(k) 1.000000 1.500000 1.750000 1.875000 1.937500 1.968750 1.984375 1.992188 1.996094 1.998047
z (z 1 by long division. k 11 12 13 14 15 16 17 18 19 20 f(k) 1.999023 1.999512 1.999756 1.999878 1.999939 1.999969 1.999985 1.999992 1.999996 1.999998 (d)
F (z) =  1 )( z  1. 5) F (z) has a pole outside the unit circle.
F ( z ) . F ( z ) is f (k ) is unbounded. The finalvalue theorem cannot be applied to
z
1 expanded into a power series of . f(k) 0.000000 1.000000 2.500000 4.750000 8.125000 13.187500 k 11 12 13 14 15 16 f(k) 113.33078 170.995117 257.492676 387.239014 581.858521 873.787781 k 1 2 3 4 5 6 350 7 8 9 10 20.781250 32.171875 49.257813 74.886719
x (0) 17 18 19 20 =
x (1 ) 1311.681641 1968.522461 2953.783691 4431.675781 Taking the ztransform on both
z I6 (a) x( k + 2 )  x ( k + 1) + 0 .1 x ( k ) = u s ( k ) z  zx ( 1)  zX z
2 =0. sides,
z X (z)
2 2 x(0) (z) + zx ( 0 ) + 0 .1 X ( z ) = 10 z z 1 + 1.455 z z  0.1127  z X ( z) = transform,
x( k ) ( z  1) ( z  z + 0.1 ) = 1 11.455z z  0.8873 Taking the inverse z = 10 + 1.455 ( 0 .1127 ) k  11 .455 = 1, ( 0 . 8873) k k = 0 , 1, 2 , L (b) x( k + 2)  x (k ) = 0 x (0) x (1 ) =0
2 Taking the ztransform on both sides, we have z X ( z )  z x(0)  zx(1)  X ( z ) = 0
2 2 X ( z) = z
2 z 1 x ( k ) = cos k 2 k = 0,1,2, L I7 (a)
P (1 ) P( 2) = (1+ r) P (0)  u = ( 1 + r ) P ( 1)  u M where P (1 ) P(0 ) = = amoun t owed P0 after the f irst p eriod. = amoun t borr owed i nitial ly P (k + 1 ) = (1 + r ) P ( k )  u u = amoun t paid each period inclu ding p rincip al and inter est. (b) By direct substitution,
P (2) P ( 3) = ( 1 + r ) P (1 )  u = (1 + r ) = (1 + r ) P ( 2 )  u = ( 1 + r ) = (1 + r )
N 2 P (0) P (0) (1+ r)u  u  (1 + r )
2 3 u  (1 + r ) u  u P (N ) P (0) u (1+ r) M N 1 + (1 + r ) N 2 + L + (1 + r ) + 1 = 0 Solving for u from the last equation, we have
u = (1+ r)
N 1 N P 0 (1 + r )
k =0 = (1 (1 +r ) + r) N N P r
0 k 1 + 1 ) = (1 + r ) P ( k )  u , (c) Taking the ztransform on both sides of the difference equation,
we have P( k k =0 P( k + 1) z k = (1 + r ) P( k ) z
k =0 k  uz z 1 351 Or, zP ( z )  zP 0 = (1 + r ) P ( z )  uz z 1 Thus, P (z) = zP z 0 (1+ r)  uz (z  1) z  (1 + r ) Taking the inverse ztransform and setting k = N, we have
P (N ) = P 0 (1 + r ) N + u r  u (1 r + r) N =0 Solving for u from the last equation, we have
u = (1 (1 + r) + r) N N P0 r 1 (d) For
u P 0 = $15
(1 (1 ,000 , r
N N = 0 . 01 , =
( 1. 01 ) and N
48 = 48 1 month s, = +r ) + r) P0 r ( 15000 )( 0 . 01 )
48 1 = ( 1. 612 )( 15000 )( 0 . 01 ) 1. 612 ( 1. 01 ) 1 = $395 .15 I8 (a)
G( z) = 5.556 z 1  5.556 z  0.1 (b) G( z ) = (c) G( z ) = (d) G( z ) = 2z z  1 (z  0.5) 2
1 20 z z  0.5 4z z 1   20z z  0.8 4z  + 80 z z 1 2z z  0.5 2z 2 ( z  0.5 ) 2  I9 k lim f (k ) = lim z 1  1 z F (z) = lim z (z z 1  0 . 5) =
z 2
1 F(z) is expanded into a power series of k 1 2 3 4 5 6 7 8 9 10 f(k) 1.000000 1.500000 1.750000 1.875000 1.937500 1.968750 1.984375 1.992188 1.996094 1.998047 by long division. k 11 12 13 14 15 16 17 18 19 20 f(k) 1.999023 1.999512 1.999756 1.999878 1.999939 1.999969 1.999985 1.999992 1.999996 1.999998 (d) 352 F (z) = z (z  1 )( z  1. 5) F (z) has a pole outside the unit circle. f (k ) is unbounded. The finalvalue theorem cannot be applied to k 1 2 3 4 5 6 7 8 9 10
F ( z ) . F ( z ) is expanded into a power series of z 1 . f(k) 0.000000 1.000000 2.500000 4.750000 8.125000 13.187500 20.781250 32.171875 49.257813 74.886719
x (0) k 11 12 13 14 15 16 17 18 19 20 =
x (1 ) f(k) 113.33078 170.995117 257.492676 387.239014 581.858521 873.787781 1311.681641 1968.522461 2953.783691 4431.675781 218 (a)
sides, x( k + 2 )  x ( k + 1) + 0 .1 x ( k ) = u s ( k ) z  zx ( 1)  zX
z
2 = 0 . Taking the ztransform on both
z z X (z) X (z ) 2 2 x(0) (z) + zx ( 0 ) + 0 .1 X ( z ) = +
1.455 z z =  z 1 z  z + 0 .1 k = 10 z z 1  0 .1127  z 1 11 .455 z z  0 . 8873 Taking the inverse z transform,
x( k ) = 10 + 1.455 ( 0 .1127 )  11 .455 = 1, ( 0 . 8873) k k = 0 , 1, 2 , L (b)
have x( k + 2)  x (k ) = 0 x (0) x (1 ) =0 Taking the ztransform on both sides, we
z z
2 2 z X (z) 2 z 2 x ( 0 )  zx ( 1 )  X ( z ) =0 X (z) = 1 x (k ) = cos k 2 k = 0 , 1, 2 , L 219 (a)
P (1 ) P( 2) = (1+ r) P (0)  u = ( 1 + r ) P ( 1)  u M where P (1 ) P(0 ) = = amoun t owed P0 after the f irst p eriod. = amoun t borr owed i nitial ly P (k + 1 ) = (1 + r ) P ( k )  u
P (2) P ( 3) u = amoun t paid each period inclu ding p rincip al and inter est. (b) By direct substitution,
= ( 1 + r ) P (1 )  u = (1 + r ) = (1 + r ) P ( 2 )  u = ( 1 + r ) = (1 + r )
N
2 P (0) P (0) (1+ r)u  u  (1 + r )
2 3 u  (1 + r ) u  u P (N ) P (0) u (1+ r) M N 1 + (1 + r ) N 2 + L + (1 + r ) + 1 = 0 353 Solving for u from the last equation, we have
u = (1+ r)
N 1 N P 0 (1 + r )
k =0 = (1 (1 +r ) + r) N N P r
0 k 1 + 1 ) = (1 + r ) P ( k )  u , (c) Taking the ztransform on both sides of the difference equation,
we have P( k k =0 P( k + 1) z k = (1 + r ) P( k ) z
k =0 k  uz z 1
uz (z Or,
zP ( z )  zP
0 = (1 + r ) P ( z )  uz z 1 Thus, P (z)
= zP z 0 (1+ r)   1) z  (1 + r ) Taking the inverse ztransform and setting k = N, we have
P (N ) = P 0 (1 + r ) N + u r  u (1 r + r) N =0 Solving for u from the last equation, we have
u = (1 (1 + r) + r) N N P0 r 1 (d) For
u P 0 = $15
(1 (1 ,000 , r
N N = 0 . 01 , =
( 1. 01 ) and N
48 = 48 1 month s, = +r ) + r) P0 r ( 15000 )( 0 . 01 )
48 1 = ( 1. 612 )( 15000 )( 0 . 01 ) 1. 612 ( 1. 01 ) 1 = $395 .15 I9 I10 354 I11 355 I12 (a) Discrete state equations: 0 1 2 3 1 A= B= 0 1 ( sI  A ) = e e
t s 1 2 s + 3
 2t ( sI  A ) 1 = 1 s + 3 1 ( s +1)( s + 2) 2 s e + 2 e e
T ( t) = L 2 e  t  e 2 t ( sI  A ) 1 =  t 2 e + 2 e2 t T = 1 sec. 2 e  T  e 2 T (T ) = t 2 t T 2 T  e + 2e 2 e + 2 e 0.2325 0.6 (T ) = 0.465 0.0972 (T ) = e 2 T T 2 T ( t) = L 1 t 2 t 1 1 e  e ( sI  A)1 B = L1 = t ( s + 1)( s + 2) s  e + 2 e 2 t 0.2325 T = 1 sec. (T ) =  0.0972 e  T  e 2 T  e T + 2 e 2 T (b) 2 e NT  e 2 NT ( NT ) =  NT 2 NT  2e + e
e e
NT  NT 2 NT + 2e e
2 NT N 1 k=0 2 e N  e 2 N = N 2 N  2e + e e e N N e 2 N 2 N + 2e x ( NT ) = ( NT )x (0) + [ ( N  k  1) T ] (T )u (kT ) I13 (a) Discrete state equations: 1 1 s s2 B= = ( sI  A ) = ( s I  A ) 1 = 1 s 0 s 0 s 0 1 s 1 t 1 T 1 0.001 ( t) = (T ) = 0 1 0 1 = 0 1 1 s 2 t T 0.001 1 1 1 ( t) = L ( sI  A) B = L = (T ) = 1 1 = 1 1 s 0 s
1 1 s 1 ( NT ) = A= 0 0 0 1 (b) 1 0 N 1 k=0 NT 1 0.001N = 0 1 1 x ( NT ) = ( NT )x (0) + [ ( N  k  1) T ] (T )u (kT ) 356 [( N  k  1) T ] = 1 ( N  k  1) T 1 0.001( N  k  1) 0 = 0 1 1 0.001( N  k ) [( N  k  1) T ] (T ) = 1 I14 (a) Transfer function:
X( z ) U (z ) = [ zI  (T ) ] (T ) =
1 z  0.6 0.465 1 z + 0.0972 0.2325 0.2325 0.2325 = z + 0.0972  0.0972 ( z ) 0.465 z  0.6  0.0972 0.2325 1 X( z ) U (z ) = 0.2325 z 0.0972( z + 0.6673) ( z ) 1 ( z ) = z  0 . 5028
2 ( z ) = z  0.5028 z + 0.04978
2 (b) Characteristic equation:
z + 0 . 04978 = 0 I15 (a) Transfer function:
X( z ) U (z ) = [ zI  (T ) ] (T ) =
1 1 0.001z z  1 0.001 0.001 0 1 = (z ) z  1 (z ) z 1 1
2 ( z ) = ( z  1) (b) Characteristic equation:
( z ) = ( z 1) = 0
2 I16 State diagram: ( z ) = z 1 z 1 6z 2 z 2 + 5z 2  10 z 3  5z 3 = 2 z 2  15 z 3 =0 Characteristic equation: 2z + 15 = 0 357 I17 State diagram: State equations: x ( k + 1) = 2 x ( k ) + x ( k ) 1 2 3
x (k
2 Output equation:
y (k ) + 1) = x 3 ( k ) + 1) = 0 . 1 x 1 ( k )  0 .2 x 2 ( k )  0 .1 x 3 ( k ) + r ( k )
Y (z) R( z ) z 1 + 0 .1 z
1 2 = x1( k ) x (k
3 Transfer function: = +z
3 2 + 0 .2 z + 0 .2 z 3 = z z
3 2 +2 + 0 .1 z + 0 .2 z + 0 .2 I18
Openloop transfer function: G( z ) = Closedloop transfer function:
Y (z) R( z ) Y ( z) E( z) = 1 z ( 1 ) Z s ( s + 1) = z  0.368 1 0.632 = G (z) 1+ G ( z) = 0 . 632 z  0 .264 + 0 . 632 Discretedata state equation:
x [( k + 1 ) T ] = 0 .264 x ( kT ) r ( kT ) I19 (a) = z + 1. 5 z  1 = 0 This is a secondorder system, n = 2. F(1) = 1.5 > 0 F(1) = 1.5 < 0 Thus, for n = 2 = even, F(1) < 0. The system is unstable. The characteristic equation roots are at z = 0.5 and z = 2. F(z) has one root outside the unit circle.
F (z)
2 Let Or z = w w +1 1 . The characteristic equation becomes ( w + 1)
F (z)
3 2 2 + 1.5 w  1  ( w  1) = 0 or 1.5 w + 4 w  1.5 = 0
2 2 2 ( ) 2 w + 1 + 1.5 w + 1  1 = 0 w1 w1 The last coefficient of the last equation is negative. Thus, the system is unstable. (b) = z + z + 3 z + 0 .2 = 0 This is a thirdorder system, n = 3. 358 Let z = w w
3 +1 1 . The characteristic equation becomes
2 w + 1 + w + 1 + 3 w + 1 + 0.2 = 0 w1 w1 w1 Or + 1 ) + ( w + 1 ) ( w  1 ) + 3 ( w + 1 )( w  1 ) + 0 .2( w  1 ) = 0 or 5 .2 w + 0 .4 w  0 .4 w + 2 . 8 = 0 Since there is a negative sign in the characteristic equation in w, the equation has at least one root in the righthalf wplane.
(w
3 2 2 3 3 2 Routh Tabulation:
w w w w
3 5 .2 0 .4  0 .4
2 .8 2 1  36 .8
2 .8 0 Since there are two sign changes in the first column of the Routh tabulation, F(z) has two roots outside the unit circle. The three roots are at: 0.0681, 0.466 + j1.649 and 0.466  j1.649. (c) = z  1.2 z  2 z + 3 = 0 w +1 Let z = . The characteristic equation becomes w 1
F (z)
3 2 3 2 w + 1  1.2 w + 1  2 w + 1 + 3 = 0 w1 w 1 w1 Or Routh Tabulation:
w w w w
3 0 .8 w 3  5 .2 w + 15 .2 w  2 .8 = 0
2 0 .8 15 .2 2  5.2
14. 77  2 .8 1 0  2 .8 There are three sign changes in the first column of the Routh tabulation. Thus, F(z) has three roots outside the unit circle. The three roots are at 1.491, 1.3455 + j0.4492, and 1.3455  j0.4492. (d) = z  z  2 z + 0 .5 = 0 w +1 Let z = . The characteristic equation becomes w 1
F (z)
3 2 w + 1  w + 1  2 w + 1 + 0.5 = 0 w1 w1 w1 3 2 Or  1.5 w + 2 . 5 w + 7 . 5 w  0 . 5 = 0
3 2 359 Routh Tabulation: 3 w  1. 5
w w w
2 7 .5 2 .5 7 .2  0 .5 1 0  0 .5 Since there are two sign changes in the first column of the Routh tabulation, F(z) has three zeros outside the unit circle. The three roots are at z = 1.91, 1.1397, and 0.2297. I20 Taking the ztransform of the state equation, we have
zX ( z ) = ( 0 . 368  0 . 632
X (z) R (z ) K ) X ( z ) + KR ( z ) K Or, =
z z The characteristic equation is Stability Condition:
0 . 368
3 2  0 .368 + 0 . 632  0 .368 + 0 . 632
K K K =0 1
< K The root is z = 0.368  0.632K < 2 .165  0 . 632 <1 or I21 z + z + 1. 5 Kz  ( K + 0 . 5) = 0 w +1 Let z = . The characteristic equation becomes w 1 w + 1 + w + 1 + 1.5 K w + 1  ( K + 0.5) = 0 w 1 w 1 w1 3 2 Or ( 1. 5 + 0 . 5 K Routh Tabulation:
w w w
3 )w 3 + ( 5. 5 + 1. 5 K ) w + ( 0 . 5  4. 5 K ) w + ( 2 . 5 K + 0 . 5) = 0
2 1.5 5 .5 1 + 0 .5 K + 1. 5 K
K 0 .5  4. 5 K + 0 .5
(K 2 2 .5 K
2 1  14
5 .5 4K + 0 .5 + 1. 5 K  0.07)( K + 3.57) < 0 w 0 2 .5 K 2.5 K + 0 .5 > 0 Stability Condition: 0 .2 < K < 0 . 07 I22 (a) Forwardpath transfer function: G ho Gp ( z ) = 1  z ( 1 )Z s = 1  z 1 K Z 0.667  0.444 + 0.444 = K (0.00476 z + 0.004527) ) s2 ( 2 s s + 1.5 z  1.8607 z + 0.8607 ( s + 1.5) K
2 Characteristic equation:
z
2 2 For T = 0.1 sec. z  1.8607 z + 0 . 8607 + 0 .00476 Kz + 0 . 004527 + ( 0 . 00476 K  1.8607 ) z + 0 .8607 + 0 . 004527 K K =0 =0 360 Let z = w w +1 1 . The characteristic equation becomes
0 . 009287 w
2 (b) + ( 0 .2786  0 .009054 K ) w + 3 .7214  0 . 000233 K = 0 For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditons for stability are: 0.2786  0.009054K > 0 or K < 30.77 3.7214  0.000233K > 0 or K < 15971.67 Thus, for stability, 0 < K < 30.77 T = 0.5 sec. Forwardpath transfer function: K ( 0 . 09883 z + 0 . 07705) G G (z ) = ho p 2 z  1.4724 z + 0 .47237 Characteristic equation:
z
2 + ( 0 . 09883 K  1.4724 )z + ( 0 . 07705 K + 0 .47237 ) =0 Let z = w w +1 1 . The characteristic equation becomes
0 .17588 Kw
2 + ( 1. 05526  0 .1541 K )w + 2 .94477  0 . 02178 K =0 For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are:
1. 05526 2 . 9447  0 . 1541  0 . 02178 < 6 . 8479 K K > 0 or > 0 or K K < 6 . 8479 < 135 .2 Stability condition: 0 < K (c) T = 1 sec. Forwardpath transfer function:
G G (z ) ho p = K ( 0 .3214 z z
2 + 0 .19652
z )  1.2231 + 0 .22313 + ( 0 .22313 + 0 .19652 =0 Characteristic equation:
z
2 + ( 0 . 3214 K  1.2231 )z K ) Let z = w w +1 1 . The characteristic equation becomes
2 0 . 5179 Kw + ( 1. 55374  0 . 393 K )w + 2 .4462  0 .12488 K =0 For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are :
1. 55374 2 .4462  0 . 393 K > 0  0 .12488 K > 0 < 3 . 9535 or or K K < 3 . 9535 < 19 . 588 Stability condition: 0 < K I23 (a) (b) Roots: 1.397 0.3136 + j0.5167 Unstable System. Roots: 0.3425 0.6712 + j1.0046 Unstable System. 0.3136 j0.5167 0.06712 j1.0046 361 (c) (d) Roots: 0.4302 Unstable System. Roots: 0.5 0.8115 Stable System. 0.7849 + j1.307 0.7849 j1.307 0.0992 j0.7708 0.0992 + j0.7708 I24 (a) Forwardpath Transfer Function: T = 0.1 sec. 5 = 1  z 1 Z 2.5  1.25 + 1.25 = 0.02341z + 0.021904 1 G ( z ) = (1  z ) Z 2 ) s 2 s s + 2.5 z 2  1.8187z + 0.8187 ( s ( s + 2) Closedloop Transfer Function:
Y (z) R( z ) = G (z) 1+ G ( z) = 0 . 0234 z z
2 + 0 . 021904
z  1. 7953 + 0 .8406 (b) UnitStep Response y(kT) (c) Forwardpath Transfer Function: T = 0.05 sec. 5 = 0.0060468 z + 0.0058484 1 G ( z ) = (1  z ) Z 2 s ( s + 2) z 2  1.9048z + 0.9048
Closedloop Transfer Function:
Y (z) R( z ) = 0 . 006046 z
2 z + 0 . 005848
z 4  1. 8988 + 0 . 91069 362 Unitstep Response y(kT) I25 (a)
Y ( z) = 1  z ( 1 )Z s 1
3 U ( z) 
K T
t U ( z ) = 10 E( z)  Kt 1  z
U (z) Th us U (z) ( 1 )Z s  1) 1
2 U ( z) U (z) = 10 E (z)
2 z 1 = 10( z z 1+ K tT E (z) Y ( z) = 1  z G( z) = Y ( z) ( 1 )T z ( z + 1) 10( z 1)
3 2( z  1)
2 z  1 + K tT E ( z) = 5T ( z + 1) E( z) Error Constants:
K p ( z  1) ( z  1 + K T )
t = lim
K z 1 G (z) 1 T
2 =
lim ( z
z 1 K v = 1 T lim ( z
z 1 1 )G ( z ) = 10 T K T
t 2 2 = 10 K
t a =  1) 2 G (z) =0 Closedloop Transfer Function: Y ( z) R (z ) = z + K tT + 5T  2 z + 5T  K tT + 1
2 2 2 (b) Forwardpath Transfer Function:
G( z) = 5T ( z + 1)
2 ( z  1) ( z  1 + K T )
t ( 5T ( z + 1)
2 ) ( ) (c) Characteristic Equation: T = 0.1 sec.
F (z ) = z + Kt T + 5T  2 z + 5T  K tT + 1 = z + ( 0.1Kt  1.95) z + ( 1.05  0.1Kt ) = 0
2 2 2 2 ( ) ( ) 363 For stability, from Eq. (662), where Thus,
a
0 F (1 ) a
1 > 0, F ( 1 ) > 0, a0 < a2
or K = 1. 05  0 .1 K t
F (1 ) a0 = 0 .1 K t  1. 95
F (  1) = 0 .1 > 0
1. 05 = 4  0 .2 K t > 0
or 0.5 < K t t < 20 =  0 .1 K t < a 1 = 1 < 20 . 5 Stability Condition:
0 .5 < K t < 20 =
0 . 05 ( z z
2 (d) Unitstep Response: K t =5
Y (z) R( z) + 1)  1.45 z + 0 . 55 (e) Unitstep Response K t =1 =
0 . 05 ( z z
2 Y (z) R( z ) +1)  1. 85 z + 0 . 95 364 I26 (a) T = 0.1 sec.
Gp (s ) = s + 37.06 s + 141.2 E (z ) ( s + 37.06 z + 141.2 ) m ( z) 1  30 + 14.2 + 29.7 + 0.418 = (1  z ) Z 2 E( z) s s + 4.31 s + 32.7 s
1000( z + 20) m ( z)
2 = 1 z ( 1 )Z s 100( s + 20) 2 = 1 z ( 1 ) 30 z z 1
2 2 + 14.2 z ( z 1)
z
2 + 29.7 z z  0.65 = + z  0.038 0.418 z
2 = 3 . 368 z z
3 + 1. 7117  0 . 3064
z 3 . 368 z (z + 1. 7117 z z  0 . 3064  0 . 038
)  1. 6876 z + 0 . 71215  0 . 024576  1 )( z  0 . 65)( (b) Closedloop Transfer Function: T = 0.1 sec.
m ( z )
R (z ) = 3. 368 z z
3 3 2 + 1. 7117
z z
2 2 z  0 . 3074
z z + 1. 6805 + 2 .424  0 . 331 Characteristic Equation:
z + 1. 6805 + 2 .424  0 . 331 = 0  0 . 903 
j 1. 3545 Roots of Characteristic Equation:
z = 0 .125 ,  0 . 903 + j1. 3545 , z The complex roots are outside the unit circle = 1, so the system is unstable. 365 (c) T = 0.01 sec.
m ( z )
E (z) 0 . 04735 z z
3 2 Forwardpath Transfer Function Closedloop Transfer Function = = + 0 . 005974
z
2 2 3z z  0 . 03663  0 . 69032  0 . 03663  0 . 72695
z  2 . 6785
z
3 + 2 . 3689
2 m ( z )
R (z ) 0 . 047354 z + 0 . 005875
z 3z  2 . 6312 + 2 . 3748 Characteristic Equation:  2 . 6312 Characteristic Equation Roots:
z
3 z 2 + 2 . 3748 +
j 0 .27 , z  0 . 72695 = 0
0 . 921 z = 0 . 789 , 0 . 921  j 0 .27 T = 0.001 sec. Forwardpath Transfer Function m ( z ) 1.43 10 =
E (z) 6 z 3 + 4. 98 10
3 4
2 z 2  2 . 384 10
z 7 z  4. 89 10 4 z
6  2 . 9635 z + 2 .9271
z
2  0 .9636
7 Closedloop Transfer Function m ( z ) 1.43 10 =
R (z ) z 3 + 4. 98 10
z
3 4
2  2 . 384 10
z z  4. 89 10 4  2 . 963 z + 2 . 9271  0 . 9641 Characteristic Equation:
z
3  2 .963 z 2 + 2 . 9271 z  0 . 9641 = 0 Characteristic Equation Roots:
z = 0 . 99213 , 0 . 98543 + j 0 . 02625 , 0 . 98543  j 0 . 02625 (d) Error Constants:
K p = lim z 1 G ho G p ( z ) = lim 3 . 368 z (z 2 + 1.7117 z z  0 . 3064  0 . 038
2 z 1  1)( z  0 .65)( =  lim =
) z Kv = 1 T lim ( z
z 1  1) 2 G ho G p ( z ) 3. 368 z (z + 1. 7117
z  0 . 3064
) z 1  0 . 65)(
2  0 . 038 = 14. 177 T Ka = 1 T
2 lim ( z  1) Gho Gp ( z) = 2 z 1 1 T
2 ( 3.368 z lim
z 1 + 1.7117 z  0.3064 ( z  1)( z  0.038) ) ( z  1) = 0 Steadystate Errors: Step Input: Ramp Input:
e ss = = 1 1+ K 1 K v p =0
T 14. 177 e ss = 366 Parabolic Input: e ss = 1 K a = I27 (a) Forwardpath Transfer Function: (no zeroor der hold) T = 0.5 sec.
G (z) =
z 0 .1836 zK
2  1. 0821 z + 0 . 0821 =
(z 0 .1836 zK  1 )( z  0 . 0821 ) T = 0.1 sec.
G (z) = 0 . 0787 zK z
2  1. 6065 z + 0 . 6065 = 0 . 0787 zK (z  1)( z  0 .6065) 367 (b) Openloop Transfer Function: (with zeroorder hold) T = 0.5 sec.
G (z) = K ( 0 . 06328 z z
2 + 0 . 02851 + 0 . 08021 )  1. 0821 = 0 . 06328 K ( z (z + 0 .4505)
) z  1 )( z  0 . 0821 T = 0.1 sec.
G (z) = K ( 0 . 00426 z z
2 + 0 .003608
z )  1. 6065 + 0 . 6065 = 0 . 00426 K ( z (z + 0 .8468 )  1 )( z  0 .6065) I28 Forwardpath Transfer Function: G( z) = 0.0001546 K z + 3.7154 z + 0.8622
2 z z  2.7236 z
3 ( ( 2 ) + 2.4644 z  0.7408 ) 368 I29 (a) P (z ) = z 3 1 Q (z) = z 2 + 1. 5 z  1 = ( z  0 . 5)( z + 2) The system is unstable for all values of K . 369 I30 (a) Bode Plot: The system is stable. (b) Apply wtransformation,
z = 2 2 + wT  wT = G ho G ( z )
G (w )
2 + wT 2  wT Then G ho G(w) z= = 10( 1  0 .0025
w (1 + w ) w ) 2 The Bode diagram of G ho is plotted as shown below. The gain and phase margins are determined as follows: GM = 32 dB PM = 17.7 deg. 370 Bode plot of G ho G (w ) : I31 2 16.67 N 1  e Ts = 0.000295 ( z + 3.39 z + 0.714) G hoG ( z ) = Z s s ( s + 1)( s + 12.5) ( z  1) ( z  0.9486 )( z  0.5354 ) The Bode plot of G ho G(z) is plotted as follows. The gain margin is 17.62 dB, or 7.6. Thus selecting an integral value for N, the maximum number for N for a stable system is 7. Bode Plot of G ho G ( z ) 371 I32 (a) G (s)
c =2+ 200 s Backwardrectangular Integration Rule: Gc ( z ) = 2 + 200T z 1 = 2 z  2 + 200T z 1 = 2 + ( 200T  2 ) z 1 z
1 1 Forwardrectangular Integration Rule: Gc ( z ) = 2 + 200Tz z 1 = ( 2 + 200T ) z  2 ( 2 + 200T )  2 z
z 1 = 1z
1 1 Trapazoidal Integration Rule: Gc ( z ) = 2 + 200T ( z + 1) 2 ( z  1) = ( 4 + 200T ) z + 200T 2 ( z  1) 2 = ( 4 + 200T ) + ( 200T  2 ) z
2 1z 1 ( 1 ) (b) G (s)
c = 10 + 0 .1 s The controller transfer function does not have any integration term. The differentiator is realized by backward difference rule. G c ( z ) = 10 +
5 s 0.1 ( z  1) Tz = (10T + 0.1) z  0.1
z = (10T + 0.1)  0.1z 1 (c) G (s)
c = 1 + 0 .2 s + Backwardrectangular Integration Rule: Gc ( z ) = 1 + 0.2 ( z  1) Tz + 5T = ( z  1) ( T + 0.2 )  0.2 z
T 1 + 5Tz 1 1 1 z Forwardrectangular Integration Rule: Gc ( z ) = 1 + 0.2 ( z  1) Tz + 5Tz z 1 = ( T + 0.2 )  0.2 z
T 1 + 5T 1 z
1 Trapezoidal Integration Rule: 0.2 ( z  1) Tz 5T ( z + 1) 2 ( z  1) Gc ( z ) = 1 + + = ( T + 0.2 )  0.2z
T 1 + 5T 1 + z ( ) 2 (1  z )
1 1 372 I33 (a) G (s)
c = 10 s + 12 T = 0.1 sec Gc (z ) = 1  z ( 1 ) Z s ( s + 12) = 0.8333( 1  z ) Z s  s + 12 10
1 1 1 = 0.8333 1  z 10 ( s + 1.5 ) s + 10 ( 1 ) z 1  z  e z z 1.2 = 0.5825 z  0.301 (b) G c ( s ) = T = 1 sec Gc (z ) = 1  z = 1 z
s s ( 1 ) Z s ( s + 10) = ( 1  z ) Z 1 10 ( s + 1.5 ) 8.5 1.5 s + s + 10 8.5 ( 1 ) z 1 + z  e 1.5 z T = 0.1 sec 1 = 10 ( z  0.9052) z  0.368 (c) G (s)
c = + 1. 55 Gc (z ) = 1  z ( 1 ) Z s + 1.55 = ( 1  z ) z  e 1
1 z
0.155 = z 1 z  0.8564 (d) G (z)
c = 1 + 0 .4 s 1 + 0 . 01 s Gc (z ) = 1  z ( 1 )Z 1 0.025 z 0.975z z  0.975 = 40 ( 1  z ) Z z  1 + z  e10 = 40 z  0.0000454 s (1 + 0.01 s ) 1 + 0.4 s
b I34 (a) Not physically realizable, since according to the form of Eq. (1118), (b) Physically realizable. (c) Physically realizable. (d) Physically realizable. (e) Not physically realizable, since the leading term is 0.1z. (f) Physically realizable.
(a)
G (s)
c 0 0 but a 0 = 0. I35 = 1 + 10 s K P =1 K D = 10 Thus Gc ( z ) = ( T + 10) z  10
Tz 373 G ho Gp ( z ) = 1  z ( 1 ) Z s4 = 2T ( z + 1) 2 G ( z ) = G c ( z ) Gho G p ( z ) = ( z  1) 2T ( z + 1) [(T + 10 ) z  10 ] z ( z  1)
3 2 2 By trial and error, when T = 0.01 sec, the maximum overshoot of When T = 0.01 sec, G( z) = 0.02 ( z + 1) (10.01 z  10 ) z ( z  1)
2 y ( kT ) is less than 1 percent. Y ( z) R (z ) = 0.02 ( z + 1)( 10.01 z  10 ) z  1.7998 z + 1.0002 z  0.2
3 2 When the input is a unitstep function, the output response y ( kT ) is computed and tabulated in the following for 40 sampling periods. The maximum overshoot is 0.68%, and the final value is 1. Sampling Periods k y ( kT )  I36 (a) G ho Gp ( z ) = ( 1  z 1 ) Z 1 s3 2 = 2T ( z + 1) ( z  1 )2 G ( z ) = G c ( z ) Gho G p ( z ) = K PTz + KD ( z  1 ) 2T Tz 2T ( z + 1) ( z  1)
2 2 2 = ( K PT + KD ) z2 + K P Tz  K D z ( z  1) Characteristic Equation: z + 2 K PT + K DT  1 z + 2K PT + 1 z  2 K DT = 0
3 2 2 2 2 ( ) ( ) For two roots to be at z = 0.5 and 0.5, the characteristic equation should have z  z + 0 .25 as a 2 factor. Dividing the characteristic equation by z  z + 0 .25 and solving for zero remainder, we get 374 4K T
P 2 + 2 K D T  0 .25 = 0 =
0 . 0139 T
2 2 a nd  0 . 5 K P T  2 .5 K D T + 0 .25 = 0
2 Solving for K P and K D from these two equations, we have
P K K D = 0 . 0972 T f or T The third root is at z = 1  2 K P T  2 K D T = 0 . 7778 = 0 . 01 sec The forwardpath transfer function is G( z) = 0.2222 ( z + 1) ( z  0.8749) z ( z  1)
0 .2222 z z
3 2 2 Y (z) R( z ) = + 0 . 0278
z
2 z  0 .1944
z  1. 7778 + 1. 0278  0 .1944 Unitstep Response: (b) (b) KP = 1, T = 0.01 sec
G (z) = G c ( z ) G ho G p ( z ) = 2T = 0 . 02 + +   + +  T K
D z 2 Tz K D z z 1 2 0 . 01 KD z z z 2 0 . 01 z  KD 1 2 The unitstep response of the system is computed for various values of tabulated below to show the values of the maximum overshoot. KD Max overshoot (%) 1.0 14 5.0 0.9 6.0 0.67 7.0 0.5 8.0 0.38 9.0 0.31 K D . The results are 9.1 0.31 9.3 0.32 9.5 0.37 10.0 0.68 375 I37 (a) Phaselead Controller Design:
G ( z ) = G ho Gp ( z ) = 1  z ( 1 )Z s 4
3 = 0.02 ( z + 1) z ( z  1) 2 T = 0.1 sec Closedloop Transfer Function: Y (z ) R (z ) = Gho Gp ( z) 1 + G ho Gp ( z ) = 0.02 ( z + 1) z  1.98 z + 1.02
2 The system is unstable. 4 (1  0.05 w ) w
2 2 With the wtransformation, z = T T +w
2 = 20 20 +w w G ( w) = w From the Bode plot of G ( j w ) the phase margin is found to be 5.73 degrees. For a phase margin of 60 degrees, the phaselead controller is
G (w )
c = 1 + aTw 1 + Tw = 1 1 + 1.4286 + 0 . 0197 w w The Bode plot is show below. The frequencydomain characteristics are: PM = 60 deg GM = 10.76 dB
M
r = 1.114 The transfer function of the controller in the zdomain is Gc ( z ) = 21.21 ( z  0.9222 ) ( z + 0.4344) (b) Phaselag Controller Design:
Since the phase curve of the Bode plot of
G ( j
w ) is always below 180 degrees, we cannot design a phaselag controller for this system in the usual manner. 376 Bode Plots for Part (a): G ho G p ( z ) =  + 1 z
1
Z 4500 K s 361 .2 s 2 + =   K 0 . 002008 z 1 z z 0 . 001775 0 . 697 377 I38 (a) Forwardpath Transfer Function:
Kv = 1 T lim
z 1 [ ( z  1) G ho G p ( z ) = lim
z 1 K ( 2.008 z + 1.775) z  0.697 = 1000 Thus K v = 80 .1 (b) Unitstep Response: Maximum overshoot = 60 percent. (c) Deadbeatresponse Controller Design: (K = 80.1)
G ho Gp ( z ) =
1 1 0.16034 z + 0.14217 ( z  1) ( z  0.697 )
1 G ho G p ( z 1 ) = Q(z P(z ) ) = 0 . 16034 z + 0 .14217 + 0 . 697
z z 2 1  1. 697 z 1 2 Q (1 ) = 0 . 3025 Digital Controller: Gc ( z ) = P (z ) Q (1)  Q (z )
1 1 = = 1  1.697 z + 0.697 z
1 1 2 2 0.3025  0.16034 z  0.14217z 3.3057 ( z  1 ) ( z  0.697 ) z  0.53 z  0.47
2 Forwardpath transfer function: G ( z ) = G c ( z ) Gho G p ( z ) = 3.3057 ( z  1 ) ( z  0.697 ) z  0.53 z  0.47
2 Closedloop system transfer function: Unitstep response:
Y (z) M (z) = 0 . 53 z z + 0 .47
2 = 0 . 53 z 1 +z 2 +z 3 +L 378 Deadbeat Response: I39 G p (s) = 2500 s(s + 25)
1 T = 0.05 sec G hoG ( z ) = 1  z ( )Z 1 1 2.146 z + 1.4215 = ( z  1 ) ( z  0.2865 ) s ( s + 25 ) 2500
2 1 2 G ho ( ) = 2.146 z + 1.4215 z G( z ) = P ( z ) 1  1.2865 z + 0.2865z
1 Q z 1 2 Q (1 ) = 3 . 5675 Deadbeat Response Controller Transfer Function: Gc z ( )=Q
1 ( ) = 1  2.865z + 0.2865z (1)  Q ( z ) 3.5675  2.146 z  1.4215 z
P z
1 1 2 1 1 2 G (z)
c = (z  1 )( z  0 .285)
2 3 . 5675 z  2 .146 z  1.4215 Forwardpath Transfer Function:
G (z) = G c ( z ) G ho G p ( z ) = 2 .146 z 3 . 5675 z
2 + 1.4215
z  2 . 146  1.4215 Closedloop System Transfer Function:
M (z) = 0 . 6015 z
1 + 0 . 3985
2 z Unitstep response: Y (z) = 1 + 0 . 6015 z +z 2 +z 3 +L 379 I40 The characteristic equation is z + (  1.7788 + 0.1152 k1 + 22.12 k 2 ) z + 0.7788 + 4.8032 k1  22.12 k 2 = 0
2 For the characteristic equation roots to be at 0.5 and 0.5, the equation should be
z
2  z + 0 .25 = 0 Equating like coefficients in the last two equations, we have  1.7788 + 0 .1152
0 . 7788 k k
1 + 22 .12 k 2 =  1 + 4. 8032 1  22 . 12 k 2 = 0 .25
k
1 Solving for the value of k and k
1 2 from the last two equations, we have = 0 . 058 and k 2 = 0 . 035 . 380 ...
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