Homework 4 - Solutions - University of California, Berkeley
[Pages:9]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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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,
2(2s + 1)
s
5
T (s)
=
1 + 2(2s + 1)Heq(s)
where
Heq
=1+
2s + 1
+
. 2s
4s2 + 2s T (s) = 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)
=
s2
+
10K1 (10K2 + 2)s
+
10K1
4 We know, n = Ts = 1.25.
And n
1 - 2 = = 2.09.
Tp
Therefore, poles are at -n ?
n 1 - 2 = -1.25 ? j2.09. Hence, n = 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.
010
0
x = 0 0 1 x + 0 r(t)
-2 -4 -6
1
Solution:
y= 1 1 0 x
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) = T11 + T22 + T33 + T44
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.
s+3 G(s) = s2 + 2s + 7 Solution: Writing the state equations,
x1 = x2 x2 = -7x1 - 2x2 + r y = 3x1 + x2
01
0
x =
x+ r
-7 -2
1
y= 3 1 x
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