Homework 4 - Solutions - University of California, Berkeley

EE C128 / ME C134 Spring 2014 HW4 - Solutions

UC Berkeley

Homework 4 - Solutions

Note: Each part of each problem is worth 3 points and the homework is worth a total of 42 points.

1. State Space Representation To Transfer Function Find the transfer function and poles of the system represented in state space below.

8 -4 1 -4

x = -3 2 0 x + -3 u(t)

5 7 -9

4

0

y = 2 8 -3 x; x(0) = 0 0

Solution: G(s) = C(sI - A)-1B

(s - 2)(s + 9) -(4s + 29)

(s - 2)

(sI

-

A)-1

=

s3

-

s2

1 - 91s

+

67

-3s - 27

5s - 31

(s2 + s - 77)

-3

7s - 76 (s2 - 10s + 4)

(s - 2)(s + 9) -(4s + 29)

(s - 2)

-4

C(sI-A)-1B

=

s3

-

s2

1 - 91s

+

67

2

8

-3

-3s - 27

5s - 31

(s2 + s - 77)

-3 -3

7s - 76 (s2 - 10s + 4) 4

-44s2 + 291s + 1814 G(s) = s3 - s2 - 91s + 67

2. Transfer Function Analysis For Mechanical Systems For the system shown below, do the following: (a) Find the transfer function G(s) = X(s)/F (s).

(b) Find , n, % OS, Ts, Tp and Tr.

Solution: (a) Writing the equation of motion yields, (5s2+5s+28)X(s) = F (s). Solving the transfer function,

X (s) F (s)

=

5s2

+

1 5s

+

28

=

s2

+

1 5

s

+

28 5

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EE C128 / ME C134 Spring 2014 HW4 - Solutions

UC Berkeley

(b) Clearly, n2 = 28/5 rad/s and 2n = 1. Therefore, n = 2.37, = 0.211.

4

Ts

=

n

=

8.01

sec

Tp = n

= 1.36 sec 1 -2

%OS = e-/ 1-2 ? 100 = 50.7%

Tr

=

1 n

(1.76

3

- 0.4172

+ 1.093

+ 1)

=

0.514

sec

3. Transfer Function From Unit Step Response For each of the unit step responses shown below, find the transfer function of the system.

Solution:

K

(a) This is a first-order system of the form: G(s) =

. Using the graph, we can estimate the

s+a

1

K

time constant as T = 0.0244 sec. But, a = = 40.984, and DC gain is 2. Thus = 2. Hence,

T

a

K = 81.967.

Thus,

81.967 G(s) =

s + 40.984

K (b) This is a second-order system of the form: G(s) = s2 + 2ns + n2 . We can estimate the

percent overshoot and the settling time from the graph.

%OS

=

(13.82-11.03) 11.03

? 100

=

25.3%

Ts = 2.62sec

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EE C128 / ME C134 Spring 2014 HW4 - Solutions

UC Berkeley

We can now calculate and n from the given information.

- ln(%OS/100)

=

= 0.4

2 + ln2(%OS/100)

4 n = Ts = 3.82

K DC Gain = 11.03. Therefore, n2 = 11.03. Hence, K = 160.95. Substituting all values, we get

160.95 G(s) = s2 + 3.056s + 14.59

K (c) This is a second-order system of the form: G(s) = s2 + 2ns + n2 . We can estimate the

percent overshoot and the peak time from the graph.

%OS

=

(1.4-1.0) 1.0

?

100

=

40%

Tp = 4

We can now calculate and n from the given information.

- ln(%OS/100)

=

= 0.28

2 + ln2(%OS/100)

n = Tp

= 0.818 1 - 2

K DC Gain = 1.0. Therefore, n2 = 1.0. Hence, K = 0.669. Substituting all values, we get

0.669 G(s) = s2 + 0.458s + 0.669

4. State Space Analysis A system is represented by the state and output equations that follow. Without solving the state equation, find the characteristic equation and the poles of the system.

023

0

x = 0 6 5 x + 1 u(t)

142

1

y= 1 2 0 x

Solution:

1 0 0 0 2 3 s -2 -3

(sI - A) = s 0 1 0 - 0 6 5 = 0 s - 6 -5

001

142

-1 -4 s - 2

Characteristic Equation: det (sI - A) = s3 - 8s2 - 11s + 8

Factoring yields poles: 9.111, 0.5338 and - 1.6448

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EE C128 / ME C134 Spring 2014 HW4 - Solutions

UC Berkeley

5. Block Diagram To Transfer Function Reduce the system shown below to a single transfer function, T (s) = C(s)/R(s).

Solution: Push G2(s) to the left past the summing junction.

Collapse the summing junctions and add the parallel transfer functions.

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EE C128 / ME C134 Spring 2014 HW4 - Solutions Push G1(s)G2(s) + G3(s) to the right past the summing junction.

UC Berkeley

Collapse the summing junctions and add feedback paths.

Applying the feedback formula,

T (s) =

G3(s) + G1(s)G2(s)

1 + H(s)[G3(s) + G1(s)G2(s)] + G2(s)G4(s)

6. Block Diagram To Transfer Function For the system shown below, find the poles of the closed-loop transfer function, T (s) = C(s)/R(s).

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