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:

x =

−1 −7 6

−8 4 8

4 7 −8

x +

−5

−7

5

ry =

[−9 −9 −8

]x

convert the system to one where the new state vector, z, is given by

z =

−1 −7 6

−8 4 8

4 7 −8

x

Solution: Let

T = A =

−1 −7 6

−8 4 8

4 7 −8

; B =

−5

−7

5

and C =[−9 −9 −8

]

Then, we can rewrite the equations as

x = 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−1z

T−1z = AT−1z +Br

Therefore

z = TAT−1z + TBr (4)

y = CT−1z (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 UC Berkeley

shown below.

TAT−1 =

−1 −7 6

−8 4 8

4 7 −8

TB =

84

52

−109

CT−1 =

[−13.8 −3.65 −13

]Therefore,

z =

−1 −7 6

−8 4 8

4 7 −8

z +

84

52

−109

ry =

[−13.8 −3.65 −13

]z

2. Diagonalizing A State Space System

Diagonalize the following system:

x =

−5 −5 4

2 0 −2

0 −2 −1

x +

−1

2

−2

ry =

[−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,

u1 =

1

0

1

;u2 =

1

1

2

;u3 =

1

−2

−2

Thus,

P =[u1 u2 u3

]=

1 1 1

0 1 −2

1 2 −2

P−1 =

−2 −4 3

2 3 −2

1 1 −1

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EE C128 / ME C134 Spring 2014 HW5 - Solutions UC Berkeley

Now, z = P−1APz + P−1Bu and y = CPz,

z =

−1 0 0

0 −2 0

0 0 −3

z +

−12

8

−3

uy =

[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) =240

s4 + 10s3 + 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) =5K(s+ 4)

5s3 + 16s2 + (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

3< K < 0

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