Homework Set #11 Solutions
Spring 2013 Control Systems 1
Name: ________________________________________________
Aux-63
Dorf E6.24
Consider the system represented in state variable form
x  Ax  Bu
where
y  Cx  Du
0 1 0
0
A= 0 0 1 B= 0 C= 1 0 0 D= 0
-k -k -k
1
a) Determine the system transfer function G(s) using the above matricies.
b) Using the Routh - Hurwitz process for what values of k is the system stablr?
Solution Aux_63
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Aux_64 - Sketch the Root Locus and determine the breakaway points for
K
K
G( s)H ( s) 
 3
s ( s  1)( s  2) s  3s 2  2s
T ( s) 
C( s )
K ( s  1)
 2
R( s ) s  2s  K ( s  1)
Solution Aux_64 - Since there are three poles and no zeros in the loop gain function,
there must be three zeros at infinity. These zeros must lie on asymptotes at  45 and 180. We have poles at s = 0 and s = -1, therefore the root locus lies on the real axis
between 0 and –1. There is no root locus between –1 and –3 because there are an
even number of pole/zero entities to the right of that region. The pole at –3 terminates
at infinity at -180. Therefore there must be a breakaway point between 0 and –1.
To calculate the breakaway points, if there are any, solve the expression
(numerator) x (derivative of the denominator) – (denominator) x derivative of the
numerator)
K
K
G( s ) H ( s ) 
 3
s( s  1)( s  2) s  3s 2  2s
Breakaway  1(3s 2  6s  2)  ( s3  3s 2  2s )( 0)  3s 2  6s  2  0
Solving for the roots s for the above equation
s1 = -0.42265
s2 = -1.5773
Since s2 = -1.5773 falls between –1 and –3 where no root locus exists this root cannot be the
breakaway point. The root s1 = -0.42265 falls between 0 and –1 where the root locus exists,
therefore this must be the
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breakaway point.
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Aux_65 – (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)
Aux_65 Solution
G( s ) H ( s ) 
+_
K ( s  2)
s( s  1)( s  5)
C(s)
K ( s  2)
s( s  1)( s  5)
(a) First determine the loop gain:
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 a zero at infinity
that is at -180.
(b) Next determine the value of K where the system becomes stable or marginally stable from the
Characteristic Equation
C . E.  s( s  1)( s  5)  H ( 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_66 – (a) For the control system shown below plot the root locus of the system characteristic
equation as the parameter “a” varies. (b) If the plot intersects the real axis what is the intercept
point?
R(s) +
C(s)
5
s( s  a)
_
Aux_66 Solution –
C.E.  1 
(a) First write the characteristic equation:
Rewrite the C.E in the form of 1  a
5
 0 or s 2  as  5  0
s( s  a)
Num
Den
Group the terms that are not multiplied by “a” and divide the equation by these terms.
(s 2  5)  as  0 and
The roots of the loop gain are:
1 a
s
s2  5
0
s   j 5 The pole of the C.E. is s  0
Root Locus
2.5
2
1.5
Imaginary Axis
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
-2.5
-2
-1.5
-1
-0.5
0
Real Axis
(b) The intercept point is s = -
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5
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Aux_67– (a) For the loop gain function below sketch the root locus diagram. (b) Approximately
calculate the values of the three breakaway points.
GH 
K ( s  1.5)( s  1.6)
s( s  1)( s  2)( s  3)
Aux_67 Solution
(a)
Root Locus
4
3
2
Imaginary Axis
1
0
-1
-2
-3
-4
-3
-2.5
-2
-1.5
-1
-0.5
0
Real Axis
(b) The breakaway points are approximately s = -0.636, -1.55, and -2.35
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Aux_68 Work part (a) in problem E7.1 in the text book.
Solution Aux_68
For the Characteristic Equation
C.E. = 1 +
s(s + 4)
s2  4 s

s2  2 s  2 s2  2 s  2
the root locus is:
Root Locus
1.5
1
Imaginary Axis
0.5
0
-0.5
-1
-1.5
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
Real Axis
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Aux_69 An example of using the Routh-Hurwitz criterion for simple design problems,
consider that the characteristic equation of a closed-loop control system is
s3  3Ks 2  ( K  2)s  4  0
It is desired to find the range of K so that the system is stable.
Aux-69 Solution
s3
s
2
1
(K + 2)
3K
4
3K(K + 2) - 4
3K 2  6 K  4
0 or
3K
3K
0
s
4
2
From the s row, the condition of stability is K  0, and from the s1 row, the condition of stability
is
3K 2  6 K  4  0
or
K  2.528
or
K > 0.528
When the conditions of K > 0 and K > 0.528 are compared, it is apparent that the latter
requirement is more stringent. Thus, for the closed-loop system to be stable, K must satisfy
K > 0.528
The requirement for K < -2.528 is disregarded since K cannot be negative.
s1
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