Homework Set # 12- Solutions
Spring 2013 Control Systems I
Name: ________________________________________________________
Aux-70 For the control system with the loop gain G(s)H(s) given below a) Draw the Root Locus
diagram. b) Calculate the value of K that such that the system becomes marginally stable. c)
Calculate the j intercept of the root locus.
K 5(s  1)(s  4)
G(s) H (s) 
( s  2) 2
Aux-70 Solution
a) The root locus is shown below.
5
b) Use the Routh Hurwitz criterion to calculate the value of K where the root locus crosses the jω
axus . First derive the Characteristic Equation
C. E.  1  G(s) H (s)  1 
K 5(s  1)(s  4)
 (s  2) 2  K 5(s  1)(s  4)
( s  2) 2
 (5K  1)s 2  (4  25K )s  20 K  4  0
The Routh array
s2
5K  1 20 K  4
1
s
4 - 25K
s 0 20 K  4
For the s1 row to be zero, K must be: 4  25K  0
c) Set up and solve the auxiliary equation for j
A(s)  (5K  1)s 2  20 K  4  0 for K = 0.16
or
K  4 / 25  0.16
1.8s 2  7.2  0
s   j 7.2 / 1.8  2
b) K = 0.16
c) s   j 7.2 / 1.8  2
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Aux-71 (a) Sketch the root locus for the control system shown. (b) Is there a value of K where
the system is stable or marginally stable.
R(s)
+_
C(s)
Aux_71 Solution First determine the loop gain:
K ( s  2)
G( s ) H ( s ) 
s( s  1)( s  5)
From the loop gain we can see that there are poles at s = 0, -1, and +5. There also is a zero at -2.
Since there are two poles and one zero in the loop gain function there must be two zeros at
infinity that is at ±90.
Next determine the value of K where the system becomes stable or marginally stable. The
C . E.  s( s  1)( s  5)  K ( s  2)  s 3  4s 2  ( K  5) s  2 K  0
Clearly the s2 term is negative and no value of K will change that sign, therefore the system is
unstable.
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Aux_72 a) Sketch the root locus for the control system shown. b) If the system becomes
marginally stable for some value of K, determine that value of K and the poles on the j axis
where the marginal instability exists.
R(s)
+_
Solution Aux-72 a) First determine the loop gain:
From the loop gain we can see that there are poles at
s = -1, and +5. There also is a zero at -2.
Since there are two poles and one zero in the loop gain
function there must be a zero at infinity that is at -180.
C(s)
G( s)H ( s) 
K ( s  2)
( s  1)( s  5 )
b) Next determine the value of the poles on the j where it becomes marginally unstable
The Characteristic equation is:
C . E.  ( s  1)( s  5)  K ( s  2)  s 2  4s  5  Ks  2 K  s 2  ( K  4) s  2 K  5  0
Using Routh Hurwitz criterion.
s2
1
2K  5
s1 K  4
s0 2K  5
We can see that if K = 4 then the s1 row is all zeros. That
being the case we can write and solve the auxiliary
equation for the j axis roots.
We can see that if K = 4 then the s1 row is all zeros. That being the case we can write and solve
the auxiliary equation for the j axis roots.
K=4
A( s )  s 2  2 K  5  s 2  3
s = j3
sj 3
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Aux_73
gain;
DorfE7.8
For the following unity feedback control system with the following forward loop
K ( s  1)
s 2 ( s  9)
a) How may (if any) zeros are at infinity?
b) If there are zeros at infinigy, what are the angles the asymptotes make with the real axis?
c) Where do the asymptotes intersect the real axis?
d) If the root locus intersects or breaks away from the real axis, where is the breadaway point?
e) Sketch the root locus and indicate all the pole/zero, intersection and breakaway points.
Solution Aux-76
G c ( s )G ( s ) 
Root Locus
8
6
4
Imaginary Axis
2
0
-2
-4
-6
-8
-10
-8
-6
-4
-2
0
2
Real Axis
Answers (Circle Correct Answer)
a) Zeros at infinity;
1
b) Asymptote angles: ±1800
2
±900
3
±60 and ±1800
4
±45 and ±1350
c) Asumptote intersection point on the real axix: s = -1, -2, -3, -4, -5, -6, -7
d) Root Locus real axis breakaway point: s = -1, -2, -3, -4, -5, -6, -7
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Aux-74 Given the control system shown below. a) Accurately sketch the root locus of the
system. Label all the poles and zeros and show arrows in the direction the roots are moving as K
varies. b) Calculate σa, the asymptote breakaway point on the real axis. c) Sketch the
asymptotes. d) Calculate the root locus breakaway point from the real axis.
R
+
C
Solution Aux-74
a)
Root Locus
25
20
15
Imaginary Axis
10
5
0
-5
-10
-15
-20
-25
-35
-30
-25
-20
-15
-10
-5
0
5
10
15
Real Axis
b) Asymptote breakaway point is at: σa = -4.33 (s = -4.33)
c) The asymptotes are shown on the root locus plot above.
d) The root locus breakaway point is at s = -1.492
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