CONTROL SYSTEMS ENGINEERING D227

S.A.E. SOLUTIONS

TUTORIAL 8 STABILITY AND THE S PLANE SELF ASSESSMENT EXERCISE No.1 1. Sketch the approximate root locus diagram for a closed loop system with unity feed back when

the forward loop transfer function is: 3) 2)(s (s s

kG ++= Determine the poles and the approximate point where the locus cuts the imaginary axis. Comment on the stability. (0, -2, -3 and j 3 ) The characteristic equation is G + 1 = 0

so the correct form is 013) 2)(s s(s

k =+++ s(s+2)(s+3) + k = 0 The poles are hence at 0, -2 and -3 The mean value is (0 -2 -3)/3 = -5/3 Drawing the three asymptotes with two mirrored at 60o reveals that they cut the imaginary axis at 2.9. One pole moves into the upper quadrant so the system could become unstable with high enough gain. 2. Plot the root locus with respect to k for a system with a forward transfer function

153s

kG 2 ++= s and unity feed back. The two loci run from poles at -0.232 and -1.434 and converge at -0.833 running off at 90o. The system is always stable.

3. Plot the root locus with respect to k for a system with a forward transfer function

1)5ss(3s

kG 2 ++= and unity feed back. Comment on the difference to the result for Q2 There is one extra pole at 0,0 giving an intercept at 0.55 and angles of 60o and 300o. The extra s term produces the possibility of instability at high gains.

The Complete Plot

The Graphical Construction

4. A system with unity feed back has a forward transfer function of 4)2s4)(s (s

kG(s) 2 +++= Determine the largest value of k for which the system is stable. Poles are at -4, (-1 + 3) and (-1 +3) Intercept is (-4 -1 -1)/3 = -2 Angles for 3 poles are 60o to the real axis The roots are given by

R1( )k ( )8 k

13 2

R2( )k .12

( )8 k

13 2 ...1

2i 3 ( )8 k

13

R3( )k .12

( )8 k

13 2 ...1

2i 3 ( )8 k

13

k = 50 is the limit for stability.

5.

( )( )

( )( ) 160s808s

s180808sss8080

s180G

d2

d

2d

d

i

ocl

++++=

+++++==

The characteristic equation is ( ) 0160s808s d2 =+++

The roots are 2ddd 100209440-4- ++

The full plot is shown.

SELF ASSESSMENT EXERCISE No.2 1. Using the Routh Hurwitz criterion, verify that k = 60 where the root locus cuts the imaginary axis for the characteristic equation s3 + 6s2 + 11s + 6 + k = 0 The characteristic equation simplifies to is s3 + 6s2 + 11s + 6 + k = 0 a = 1 b = 6 c = 11 d = 6 + K R = c - ad/b = 11 - 1 x (6 + K)/6 At the limit of stability R = 0 11 = 1 x (6 + K)/6 = 1 +K/6 10 = K/6 K = 60 2. Determine if the closed loop systems described by the following transfer functions are stable. a. G(s) = 1/(2s3 + 3s2 +0.8s + 1) b. G(s) = 1/(s3 + 12s2 +2s + 1000) a. The C.E. is s3 + 12s2 + 2s + 1000 = 0 a = 1 b = 12 c = 2 d = 1000 R = c - ad/b = 2 - 1 x 1000/12 R = 2 83.3 = -81.3 so the system is unstable b. The C.E. is 2s3 + 3s2 +0.8s + 1 = 0 a = 2 b = 3 c = 0.8 d = 1 R = c - ad/b = 0.8 - 2 x 1/3 R = 0.8 0.666 = 0.113 so the system is stable 3. A system with velocity feed back has a transfer function of

5K)s2K200(110ss30KG(s) 23 ++++=

Determine the value of which makes the system stable when K = 40. The C.E. is s3 + 10s2 + 200(1 + 2K)s + 5K a = 1 b = 10 c =200(1+ 2K) =200(1+ 80a) d = 5 x 40 = 200 R = 0 = c - ad/b = 200(1+ 80) - 1 x 200/10 0 = 200 + 16000 - 20 0 = 180 + 16000 = -180/16000 = -0.01125 4. A system with velocity feed back has a transfer function of

20K)sK500(110s2s10KG(s) 23 ++++=

Determine the value of which makes the system stable when K= 4. The C.E. is 2s3 + 10s2 + 500(1 + K)s + 20K a = 2 b = 10 c =500(1+ K) = 500(1+ 4a) d = 20 x 4 = 80 R = 0 = c - ad/b = 500(1+ 4) - 2 x 80/10 0 = 500 + 2000 - 16 0 = 484 + 2000 = -484/2000 = -0.242