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.

x =

8 −4 1

−3 2 0

5 7 −9

x +

−4

−3

4

u(t)

y =[2 8 −3

]x; x(0) =

0

0

0

Solution: G(s) = C(sI−A)−1B

(sI−A)−1 =1

s3 − s2 − 91s+ 67

(s− 2)(s+ 9) −(4s+ 29) (s− 2)

−3s− 27 (s2 + s− 77) −3

5s− 31 7s− 76 (s2 − 10s+ 4)

C(sI−A)−1B =1

s3 − s2 − 91s+ 67

[2 8 −3

](s− 2)(s+ 9) −(4s+ 29) (s− 2)

−3s− 27 (s2 + s− 77) −3

5s− 31 7s− 76 (s2 − 10s+ 4)

−4

−3

4

G(s) =

−44s2 + 291s+ 1814

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 ζ, ωn, % 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)=

1

5s2 + 5s+ 28=

15

s2 + s+ 285

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EE C128 / ME C134 Spring 2014 HW4 - Solutions UC Berkeley

(b) Clearly, ω2n = 28/5 rad/s and 2ζωn = 1. Therefore, ωn = 2.37, ζ = 0.211.

Ts =4

ζωn= 8.01 sec

Tp =π

ωn√

1− ζ2= 1.36 sec

%OS = e−ζπ/√

1−ζ2 × 100 = 50.7%

Tr = 1ωn

(1.76ζ3 − 0.417ζ2 + 1.093ζ + 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:

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

s+ a. Using the graph, we can estimate the

time constant as T = 0.0244 sec. But, a =1

T= 40.984, and DC gain is 2. Thus

K

a= 2. Hence,

K = 81.967.

Thus,

G(s) =81.967

s+ 40.984

(b) This is a second-order system of the form: G(s) =K

s2 + 2ζωns+ ω2n

. 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 ζ and ωn from the given information.

ζ =− ln(%OS/100)√π2 + ln2(%OS/100)

= 0.4

ωn =4

ζTs= 3.82

DC Gain = 11.03. Therefore,K

ω2n

= 11.03. Hence, K = 160.95. Substituting all values, we get

G(s) =160.95

s2 + 3.056s+ 14.59

(c) This is a second-order system of the form: G(s) =K

s2 + 2ζωns+ ω2n

. 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 ζ and ωn from the given information.

ζ =− ln(%OS/100)√π2 + ln2(%OS/100)

= 0.28

ωn =π

Tp√

1− ζ2= 0.818

DC Gain = 1.0. Therefore,K

ω2n

= 1.0. Hence, K = 0.669. Substituting all values, we get

G(s) =0.669

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.

x =

0 2 3

0 6 5

1 4 2

x +

0

1

1

u(t)

y =[1 2 0

]x

Solution:

(sI−A) = s

1 0 0

0 1 0

0 0 1

−0 2 3

0 6 5

1 4 2

=

s −2 −3

0 s− 6 −5

−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 UC Berkeley

Push G1(s)G2(s) +G3(s) to the right past the summing junction.

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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EE C128 / ME C134 Spring 2014 HW4 - Solutions UC Berkeley

Solution: Push 2s to the left past the pickoff point and combine the parallel combination of

2 and 1/s.

Push (2s+ 1)/s to the right past the summing junction and combine summing junctions.

Hence,

T (s) =2(2s+ 1)

1 + 2(2s+ 1)Heq(s)where Heq = 1 +

s

2s+ 1+

5

2s.

T (s) =4s2 + 2s

6s2 + 13s+ 5

7. Finding Constants To Meet Design Specifications

For the system shown below, find the values of K1 and K2 to yield a peak time of 1.5 seconds and

a settling time of 3.2 seconds for the closed loop system’s step response.

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EE C128 / ME C134 Spring 2014 HW4 - Solutions UC Berkeley

Solution: The closed-loop transfer function is given by:

T (s) =10K1

s2 + (10K2 + 2)s+ 10K1

We know, ζωn =4

Ts= 1.25. And ωn

√1− ζ2 =

π

Tp= 2.09. Therefore, poles are at −ζωn ±

ωn√

1− ζ2 = −1.25± j2.09. Hence, ωn =√

1.252 + 2.092 =√

10K1 and (10K2 + 2)/2 = 1.25.

Therefore, K1 = 0.593 and K2 = 0.05.

8. Signal-Flow Graphs

Draw a signal-flow graph for the following equation.

x =

0 1 0

0 0 1

−2 −4 −6

x +

0

0

1

r(t)y =

[1 1 0

]x

Solution:

x1 = x2

x2 = x3

x3 = −2x1 − 4x2 − 6x3 + r

y = x1 + x2

9. Signal-Flow Graphs

Using Mason’s rule, find the transfer function, T (s) = C(s)/R(s), for the system represented by the

following figure.

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EE C128 / ME C134 Spring 2014 HW4 - Solutions UC Berkeley

Solution:

Closed-loop gains: G2G4G6G7H3;G2G5G6G7H3;G3G4G6G7H3;G6H1;G7H2

Forward-path gains: T1 = G1G2G4G6G7;T2 = G1G2G5G6G7;T3 = G1G3G4G6G7;T4 = G1G3G5G6G7

Nontouching loops 2 at a time: G6H1G7H2

∆ = 1− [H3G6G7(G2G4 +G2G5 +G3G5) +G6H1 +G7H2] + [G6H1G7H2]

∆1 = ∆2 = ∆3 = ∆4 = 1

T (s) =T1∆1 + T2∆2 + T3∆3 + T4∆4

∆

T (s) =G1G2G4G6G7 +G1G2G5G6G7 +G1G3G4G6G7 +G1G3G5G6G7

1−H3G6G7(G2G4 +G2G5 +G3G4 +G3G5)−G6H1 −G7H2 +G6H1G7H2

10. State Space Representation and Signal-Flow Graphs

Represent the system shown below in state space form and draw its signal-flow graph.

G(s) =s+ 3

s2 + 2s+ 7

Solution: Writing the state equations,

x1 = x2

x2 = −7x1 − 2x2 + r

y = 3x1 + x2

x =

[0 1

−7 −2

]x +

[0

1

]r

y =[3 1

]x

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EE C128 / ME C134 Spring 2014 HW4 - Solutions UC Berkeley

11. Canonical Forms

Represent the system given in Problem 10 in controller canonical form and observer canonical form.

Solution:

Controllable canonical form:

x =

[−2 −7

1 0

]x +

[1

0

]r

y =[1 3

]x

Observable canonical form:

x =

[−2 1

−7 0

]x +

[1

3

]r

y =[1 0

]x

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