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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 ...
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