Post on 11-Apr-2018
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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