of University Wales Swansea SCHOOL OF ENGINEERING EG 243: Control Systems
Root Locus
1 Plot and calibrate the root locus diagrams of the following systems as a function of the open-loop gain K:
a)
2
K ; s (s + 1)
b)
K (s + 2 ) ; s (s + 1)
c)
K (s + 2 ) s(s + 1)(s + 5)
Use the root locus technique to determine the range of values of K for which the system with open-loop transfer function Go(s) is closed-loop stable.
Go (s ) =
K (s - 1)(s + 2)(s + 8)
3
Plot the root locus of Go (s ) =
s (s + 9 )
2
K (s + 1)
as a function of K.
4
A position control system has plant transfer function G (s ) =
4 . s (s + 4 )
A cascade proportional plus derivative compensator with transfer function D(s) = KP + TDs is added to the system so that the closed-loop poles of the compensated system have a natural frequency of n = 10 rad s-1 and damping ratio = 0.5. Determine the proportional gain, KP and derivative time, TD required to achieve this response. Calculate the steady-state velocity error of the compensated system. Hint: use the root locus angle criterion to find the location of the compensator zero when the closed-loop poles are positioned as specified.
1
1
Plot and calibrate the root locus diagrams of the following systems as a function of the open-loop gain K:
a)
K ; s (s + 1)
b)
K (s + 2 ) ; s (s + 1)
c)
K (s + 2 ) s(s + 1)(s + 5)
See L17-20. Root locus sketched using 5 rules. L19, S7, S10, S12, S14. a) See L17, S14-17, which is a similar problem.
Go (s ) =
K ; s(s + 1)
Gc (s ) =
K s2 + s + K
=
K
(s + 0.5)2 + (K - 0.25)
At K = 0.25 there are concurrent poles at = -0.5, which is the break-away point
Rule 1,2: there are 2 branches; n=2; poles at s = 0 and 1. Symmetry Rule 3: the root locus is on the real axis for 1 < s < 0
Rule 4 there are no finite zeros; m=0 2 infinite zeros The branch start and end points are: 1 at [0, - 0.5 + j]; 2 at [-1, -0.5 - j] Rule 5
using L19, S14/15 the asymptotes are at 90 and 270 with real-axis
intercept at 0 =
finite poles - finite zeros = (- 1 + 0) - (0) = -0.5
n-m 2-0
1
numKGH=1; denKGH=conv([1,0],[1,1]) rlocus(numKGH,denKGH)
denKGH =
1
0
2
b)
Go (s ) =
K (s + 2 ) ; s (s + 1)
Gc (s ) =
s 2 + (1 + K )s + 2 K
K (s + 2 )
Rule 1,2: there are 2 branches; n=2; poles at s = 0 and 1. Symmetry Rule 3: the root locus is on the real axis for 1 < s < 0
Rule 4 there is 1 finite zero; m = 1 1 infinite zero The branch start and end points are: 1 at [0, ]; 2 at [-1, -2] Rule 5
at 0 =
finite poles - finite zeros = (- 1 + 0) - (- 2) = +1
n-m 2 -1
using L19, S14/15 the asymptote is at 180 with real-axis intercept
It is possible to calibrate the sketch by finding the break-away [K maximum on real-axis section] and break-in [K minimum on real-axis section] points. For points on the root locus, 1 + K G(s) H(s) = 0 K G(s) H(s) = -1
- x2 + x K (x + 2) ; u sin g x for typing ease = -1; K = K G (s )H (s ) = x( x + 1) x+2 dK - x 2 + x *1 - ( x + 2 )[- (2 x + 1)] - x 2 - 4 x - 2 = = dx ( x + 2)2 ( x + 2)2 dK = 0, giving max and min values, when x 2 + 4 x + 2 = 0; dx x = -2 2 = -3.414 or - 0.586 Checked by transition method x = -2 - 2 : x = -2 + 2 :
(
)
[(
)]
(6 + 4 2 ) + (- 2 - 2 ) = 4 + 3 2 = 5.828 (- 2 - 2 ) + 2 2 (6 - 4 2 ) + (- 2 + 2 ) = - 4 - 3 2 = 0.172 K =- (- 2 + 2 ) + 2 2
K =-
Previous solution was x = -2 3
3
b) numKGH=[1,2]; denKGH=conv([1,0],[1,1]);
rlocus(numKGH,denKGH)
c) numKGH=[1,2]; denKGH=conv([1,1,0],[1 5]) denKGH = 1 6 5
0
rlocus(numKGH,denKGH)
4
c)
Go (s ) = Gc (s ) =
s 3 + 6 s 2 + (5 + K )s + 2 K
K (s + 2 ) ; s(s + 1)(s + 5) K (s + 2 )
Gc (s ) =
K (s + 2 ) s (s + 1)(s + 5) + K (s + 2)
Rule 1,2: there are 3 branches; n=3; poles at s = 0, 1and -5. Rule 3: the root locus is on the axis for 1 < s < 0 and 5 < s <-2
Rule 4 there is 1 finite zero; m = 1 2 infinite zeros The branch start and end points are: 1 at [0, -j] 2 at [-1, +j] 3 at [-5, -2] Rule 5
intercept at 0 = using L19, S14/15 the asymptotes are at 90 and 270 with real-axis
finite poles - finite zeros = (- 5 - 1 + 0) - (- 2) = -2 ; the zero!
n-m 3 -1
m
The breakaway point can be found using the transition method.
n 1 1 U sin g = ; x + zi j =1x + p j i =1
1 1 1 1 = + + x + 2 x x +1 x + 5
Use Matlab to find roots of the resulting cubic. First construct each term. a=conv([1 1 0],[1 2]) a= 1 3 2 0 b=conv([1 1 0],[1 5]) b= 1 6 5 0 c=conv([1 2 0],[1 5]) c= 1 7 10 0 d=conv([1 3 2],[1 5]) d= 1 8 17 10 sum=a-b+c+d sum = 2 12 24 10
roots(sum) ans = -2.7211 + 1.2490i; Sketch on previous page
-2.7211 - 1.2490i;
-0.5578 break-away
5
2
Use the root locus technique to determine the range of values of K for which the system with open-loop transfer function Go(s) closed-loop is stable.
Go (s ) =
K (s - 1)(s + 2)(s + 8)
Rule 1,2: there are 3 branches; n=3; poles at s = +1, -2 and -8. Rule 3: the root locus is on the axis for s < -8 and 2 < s <+1 3 infinite zeros 3 at [-8, -]
Rule 4 there is no finite zero; m = 0 The branch start and end points are: 1 at [1, +] 2 at [-2, -] Rule 5
using L19, S14/15 the asymptotes are at 180 and 60 with realaxis intercept at 0 =
finite poles - finite zeros = (+ 1 - 2 - 8) = -3
n-m 3-0
The break-away point between poles at +1 and 2 is determined by transition.
n 1 1 U sin g = ; x + zi j =1x + p j i =1 m
0=
1 1 1 + + x -1 x + 2 x + 8
0 = ( x + 2 )( x + 8) + ( x - 1)( x + 8) + ( x - 1)( x + 2 ) 0 = x 2 + 10 x + 16 + x 2 + 7 x - 8 + x 2 + x - 2 0 = 3 x 2 + 18 x + 6 x=
(
) (
) (
)
- 6 36 - 8 = -3 7 = -3 2.65 2
Break-away is at 0.35. [-5.65 is outside the range]. The j crossings are determined from the Routh Array. See L12, P8. Nise, p337 Closed-loop characteristic equation [CLCE] is:
CLCE = 1 + Go (s )H (s ) = 1 +
i.e.
K =0 (s - 1)(s + 2)(s + 8)
s3 + 9s2 + 6s2 16 + K = 0
6
s 3: s 2: s 1: s 0:
a3 a2 b1 c1
a1 a0 b2 c2
1 9 70 - K
6 -16+K 0 0
a1 a2 - a3 a0 6 * 9 - (- 16 + K ) = 70 - K ; b2 = 0 a2 9 K = 70 makes a row of zeros at s1. Then the s2 row is a factor of the original CLCE polynomial and the roots provide the j crossings. b1 =
i.e. 9s2 +(-16 + 70) = 9s2 + 54 = 0;s2 = -6;s = j6 = j2.45
The root locus also crosses the imaginary axis s = 0 when K = 16 Stability range is 16 < K < 70
numKGH=1; a=conv([1 -1],[1 2]) denKGH=conv([a],[1 8]) >> rlocus(numKGH,denKGH) a = 1 1 -2 denKGH = 1 9
6 -16
7
3
Plot the root locus of Go (s ) =
s 2 (s + 9 )
K (s + 1)
as a function of K.
Rule 1: Rule 3: Rule 4 Rule 5
intercept at 0 =
there are 3 branches; n=3; 2 poles at s = 0, and -9. the root locus is on the axis for 9 < s <-1 there is one finite zero; m=1 2 infinite zeros
using L19, S14/15 the asymptotes are at 90 with real-axis
n-m 3 -1 2 The branch start and end points are: 1 at [+0, -4 - j] 2 at [-0, -4 + j] 3 at [-9, -1]
The break-away point is determined by transition:
n 1 1 U sin g = ; x + zi j =1 x + p j i =1 m
finite poles - finite zeros = (- 0 - 0 - 9) - (- 1) = - 8 = -4
1 1 1 1 = + + x +1 x x x + 9
x x - = 2; x +1 x + 9 x 2 + 6 x + 9 = 0;
x( x + 9 ) - x( x + 1) = 2( x + 1)( x + 9 )
( x + 3)2 = 0;
x = -3 twice
There is also break-away at s = 0 from 2 poles at origin. Characteristic equation = 1 + Go(s) = 1 + i.e. s +9s + K(s + 1) = 0; 27 + 81 + K(-3 + 1) = 0;
3 2
s (s + 9 ) Put s = -3 to find K. K = 27
2
K (s + 1)
=0
When K = 27 the CLCE s3 +9s2 + 27s + 27 = (s + 3)3 = 0 Thus there are 3 closed-loop poles at s = -3, one on each branch.
8
There are no crossings of the imaginary axis so the system is stable for all values of gain, K.
9
4
A position control system has plant transfer function G (s ) =
4 . s (s + 4 )
A cascade proportional plus derivative compensator with transfer function D(s) = KP + TDs is added to the system so that the closed-loop poles of the compensated system have a natural frequency of n = 10 rad s-1 and damping ratio = 0.5. Determine the proportional gain, KP and derivative time, TD required to achieve this response. Calculate the steady-state velocity error of the compensated system. Hint: use the root locus angle criterion to find the location of the compensator zero when the closed-loop poles are positioned as specified.
Kp 4TD s + 4 K p + TD s TD OLTF Go (s ) = G (s )D(s ) = = s (s + 4 ) s (s + 4 ) There are 2 open-loop poles [p1 and p2] and 1 zero [z]; place the zero.
(
)
n = 10 rad s-1 and = 0.5 d = n = 5 d = n(1- 2) = 100.75 = 53 = cos-1 = 0.5; = 60
n
d
d For closed-loop poles to be at 5 53 the root locus angle criterion must be satisfied. See L18, S1-3. A
closed-loop pole O j53
z
O
2
-5
X X
1 origin O
Follow L18, Ex9.2. Angle criterion is: -p1 -p2 + z = (2r + 1)*180 Take r = 0 Ap1 = n = 10 p1 = 180- 1 = 180 = 120; -1 p2 = 180 - 2 = 180 -tan (53) = 180 - 83.4 = 96.6; Ap2 = 76 z = - 180+ 120 + 96.6 = + 36.6 Az = (6.432 + 75) = 10.78
10
Oz = d + d tanz = 5 + 53 tan 36.6 = 5 + 6.43 = 11.43 = From L18, E7-8:
Kp TD
(s + p j )
n
K = 4TD =
j =1 m i =1
(s + zi )
=
pole lengths = Ap1 * Ap2 = 10 * 76 = 8.09 Az 10.78 zero lenghts
Kp = Oz*TD = 11.43*2.025 = 21.35
TD = K = 2.025 s;
Go (s ) =
8.09(s + 11.43) s (s + 4 )
K v = lim s 0 sGo (s ) = 1 1 = = 0.043 K v 23.12
1 4 0
8.09 *11.43 = 23.12 4
Steady-state velocity error =
numKGH=[1 11.43]; denKGH=conv([1 0],[1 4]) rlocus(numKGH,denKGH)
[4.3 %]
denKGH =
This solution has branches between [0, -11.43] and [-4, -] This solution differs from previous work.
11