EE C128 / ME C134 Spring 2014 HW5 - Solutions
UC Berkeley
Homework 5 - Solutions
Note: Each part of each problem is worth 3 points and the homework is worth a total of 24 points.
1. State Space Analysis
Given the system represented in state space as follows:


 
−1 −7 6
−5


 
ẋ = −8 4
8  x + −7 r
4
7 −8
5
h
i
y = −9 −9 −8 x
convert the system to one where the new state vector, z, is given by


−1 −7 6


z = −8 4
8 x
4
7 −8
Solution: Let


 
−1 −7 6
−5
h
i


 
T = A = −8 4
8  ; B = −7 and C = −9 −9 −8
4
7 −8
5
Then, we can rewrite the equations as
ẋ = Ax + Br
(1)
y = Cx
(2)
z = Tx
(3)
Using Equation (3), we can solve for x and subsitute in Equations (1) and (2) as follows,
x = T −1 z
T −1 ż = AT −1 z + Br
Therefore
ż = T AT −1 z + T Br
(4)
y = CT −1 z
(5)
Using Equations (4) and (5), we can now solve for the new system with the new state vector z as
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EE C128 / ME C134 Spring 2014 HW5 - Solutions
shown below.
UC Berkeley


−1 −7 6


T AT −1 = −8 4
8
4
7 −8


84


T B =  52 
−109
h
i
CT −1 = −13.8 −3.65 −13
Therefore,

−1 −7

ż = −8 4
4
7
h
y = −13.8



6
84



8  z +  52  r
−8
−109
i
−3.65 −13 z
2. Diagonalizing A State Space System
Diagonalize the following system:


 
−5 −5 4
−1


 
ẋ =  2
0 −2 x +  2  r
0 −2 −1
−2
h
i
y = −1 1 2 x
Solution: Eigenvalues are −1, −2 and −3, since,
det(λI − A) = (λ + 3)(λ + 2)(λ + 1)
Solving for Eigenvectors, Ax = λx, we get,
 
 
 
1
1
1
 
 
 
u1 = 0 ; u2 = 1 ; u3 = −2
1
2
−2
Thus,
h
P = u1
P −1
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

1
1
1
i


u2 u3 = 0 1 −2
1 2 −2


−2 −4 3


= 2
3 −2
1
1 −1
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EE C128 / ME C134 Spring 2014 HW5 - Solutions
UC Berkeley
Now, ż = P −1 AP z + P −1 Bu and y = CP z,




−1 0
0
−12




ż =  0 −2 0  z +  8  u
0
0 −3
−3
h
i
y= 1 4 7 z
3. Routh-Hurwitz Stability Criterion
How many roots of the following polynomial are in the right half-plane, in the left half-plane, and
on the jω -axis.
P (s) = s5 + 3s4 + 5s3 + 4s2 + s + 3
Solution: The Routh-Hurwitz table is given as follows
Since there are 2 sign changes, there are 2 RHP poles, 3 LHP poles and no poles on the jω-axis..
4. Stability Of Closed Loop Unity Feedback System
Determine whether the unity feedback system shown below is stable if
G(s) =
240
(s + 1)(s + 2)(s + 3)(s + 4)
Solution: The closed loop transfer function T (s) is given by
T (s) =
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s4
+
10s3
240
+ 35s2 + 50s + 264
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EE C128 / ME C134 Spring 2014 HW5 - Solutions
UC Berkeley
The Routh-Hurwitz table is given as follows
Since there are 2 sign changes, there are two closed loop poles in the RHP and the closed loop system
is unstable.
5. Routh-Hurwitz Stability Criterion
Consider the following Routh table. Notice that the s5 row was originally all zeros. Tell how many
roots of the original polynomial were in the right half-plane, in the left half-plane, and on the jω-axis.
Solution:
Even Polynomial (6th order) : 1 sign change, therefore
RHP: 1 pole; LHP: 1 pole; jω-axis: 4 poles
Rest (1st order): 0 sign changes, therefore
RHP: 0 poles; LHP: 1 pole jω-axis: 0 poles
Total (7th order):
RHP: 1 pole; LHP: 2 poles; jω-axis: 4 poles
6. Routh-Hurwitz Stability Criterion
Consider the figure given in Problem 4. Find the range of K for closed-loop stability for the following
cases when the transfer function G(s) is given by
G(s) =
K(s − a)
s(s − b)
(a) a < 0, b < 0
(b) a < 0, b > 0
(c) a > 0, b < 0
(d) a > 0, b > 0
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EE C128 / ME C134 Spring 2014 HW5 - Solutions
UC Berkeley
Solution: The characteristic equation for all cases is s2 + (K − b)s − Ka = 0. Using Routh-Hurwitz
analysis, we find the following relations for closed loop stability:
K −b>0
(6)
−Ka > 0
(7)
(a) a < 0, b < 0
Using Relations (6) and (7), we find that K > 0 for the system to be stable.
(b) a < 0, b > 0
Using Relations (6) and (7), we find that K > b for the system to be stable.
(c) a > 0, b < 0
Using Relation (6), we find that K > b, and from Relation (7) we find that K < 0. Therefore,
b < K < 0.
(d) a > 0, b > 0
Since both Relations (6) and (7) cannot be satisfied simultaneously, there is no such K, when
the system will be stable.
7. Stability Of Closed Loop Unity Feedback System
Consider the figure given in Problem 4. Given G(s) as below, find the following
G(s) =
K(s + 4)
s(s + 1.2)(s + 2)
(a) The range of K that keeps the system stable.
(b) The value of K that makes the system oscillate.
(c) The frequency of oscillation when K is set to the value that makes the system oscillate.
Solution: The closed loop transfer function is given by:
T (s) =
5s3
+
16s2
5K(s + 4)
+ (12 + 5K)s + 20K
Using Routh-Hurwitz analysis, we find the following relations for closed loop stability:
192 − 20K > 0
(8)
20K > 0
(9)
(a) For closed loop stability, both Equations (8) and (9) need to be satisfied simultaneously. Thus
0 < K < 9.6.
(b) For oscillations, at least one pair of poles must lie on the imaginary axis. K = 9.6 makes one
pair of poles purely imaginary.
(c) Let K = 9.6. Now, the characteristic equation becomes 5s3 +16s2 +60s+192 = 0. Factorization
√
√
yields (32s2 + 384)(s2 + 12) = 0. Roots of (s2 + 12) = 0 are ±2 3. Therefore, ω = 2 3 rad/s.
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EE C128 / ME C134 Spring 2014 HW5 - Solutions
UC Berkeley
8. Stability of Closed Loop Feedback System
Find the range of K to keep the system shown below stable
Solution: The closed loop transfer function is given by:
T (s) =
K(s2 + 2s + 1)
s3 + 2s2 + (K + 1)s − K
The Routh-Hurwitz table is given below:
Thus, for stability, K < 0 and K > −2/3. Therefore,
2
− <K<0
3
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