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11/16/
ECE 382 Midterm 3 Solution
Problem 1 (30 points) In the above system, the plant transfer function is G(s) = (^) (s+2)^12.
(a) (5 pts) For the uncompensated system with C(s) = 1, find the dominant closed-loop pole(s), the settling time ts (5% criterion), and the steady-state error for tracking unit step input. Solution: The system has characteristic equation 1 + G(s) = 0, which is equivalent to (s + 2)^2 + 1 = 0. So the dominant closed-loop poles are − 2 ± j. The settling time ts = 32 = 1.5 s. The system is of type 0, with Kp = G(0) = 0.25. Thus, ess = 1/(1 + 0.25) = 0.8.
(b) (5 pts) Given the two design objectives: (i) damping ratio ξ =
√ 3 2 ; (ii) settling time (5% criterion) is ts = 1 second, find the desired dominant closed-loop poles locations sd. Solution: sd = −σd ± jωd = −ξωn ± j
1 − ξ^2 ωn. Since ts = (^) ξω^3 n = (^) σ^3 d = 1, we must have σd = 3. On the
other hand, ωd/σd =
1 − ξ^2 /ξ = 1/
3, which implies ωd = σd/
- Thus, the desired closed-loop poles are sd = − 3 ± j
(c) (12 pts) Design a lead compensator of the form
C 1 (s) = K 1
s + 3 s + p
by finding K 1 and p so that the compensated system has dominant closed-loop poles at sd computed in (b). Solution: We first compute the angle of defficiency:
φ = 180◦^ − [− 26 (sd + 2)] = 420◦^ = 60◦.
We need to choose −p so that it and −3 span φ = 60◦^ angle with respect to sd = −3 + j
3 (only one of the two desired closed-loop poles is used for convenience). Since −3 is directly below sd, this is indeed very easy:
p = z +
3 ∗ tan(φ) = 3 + 3 = 6.
To find K 1 , we used the magnitude condition:
1 + K 1
sd + 3 sd + 6
(sd + 2)^2
= 0 ⇒ K 1 = 8.
To sum up, the lead compensator is C 1 (s) = 8 s s+3+.
(d) (8 pts) Design a compensator C(s) to satisfy not only the two design objectives in part (b), but also: (iii) the steady-state error for tracking unit step input is ess < 0 .1. Solution: We can add a lag compensator to the lead compensator designed in part (c). For the system compensated by C 1 (s), the open-loop transfer function is
C 1 (s)G(s) =
8(s + 3) (s + 6)(s + 2)^2
It is easy to see that the static error constant is Kp = lims→ 0 C 1 (s)G(s) = 1. Thus, ess = (^) 1+1^1 = 0.5, which
does not satisfy requirement (iii). Indeed, we need to have ess,new = (^) 1+K^1 p,new < 0 .1, i.e., Kp,new > 9. For example, we can set the goal for Kp,new = 10. Then the lag compensator
C 2 (s) = s + 0. 1 s + 0. 01
can be used. To sum up, the compensator that achieves all three design objectives is a lag-lead compensator given by
C(s) = C 1 (s)C 2 (s) = 8
(s + 0.1)(s + 3) (s + 0.01)(s + 6)
Problem 2 (20 points) In the above system, the plant has an unknown stable transfer function G(s). Suppose the plant is first isolated from other parts of the system, and subject to the input signal of 5 sin(20t). By measuring the plant output, it is found that the plant has the steady-state response of 7.5 cos(20t).
(a) (10 pts) Find the plant’s frequency response G(jω) at ω = 20. Solution: The steady-state output is 7.5 cos(20t) = 5 ∗ 1 .5 sin(20t + 90◦). Therefore,
G(j20) = 1. 56 90 ◦^ = j 1. 5.
(b) (10 pts) Suppose the closed-loop system is stable under the compensator C(s) = (^) s+20^10. Determine the steady-state output yss(t) of the closed-loop system under the input u(t) = 2 sin(20t). Solution: The closed-loop system has transfer function
C(s)G(s) 1 + C(s)G(s)
which is assumed to be stable. Thus, the frequency response of the closed-loop system at ω = 20 is
C(j20)G(j20) 1 + C(j20)G(j20)
10 j20+20 ·^ j^1.^5 1 + (^) j20+20^10 · j 1. 5
As a result, under the input u(t) = 2 sin(20t), the steady-state output is
yss(t) = 2 ∗ 0 .3721 sin(20t + 29. 74 ◦) = 0.7442 sin(20t + 29. 74 ◦)
At ω 2 = 0.1, 20 log |G(jω 2 )| ≃ −60 db, and 6 G(jω 2 ) ≃ − 90 ◦.
At ω 3 = 10, 20 log |G(jω 3 )| ≃ −60 db, and 6 G(jω 3 ) ≃ 90 ◦.
At ω 4 = 1000, 20 log |G(jω 3 )| ≃ −60 db, and 6 G(jω 3 ) ≃ − 90 ◦.
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Magnitude (dB)
10 −4^10 −3^10 −2^10 −1^100 101 102
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0
45
90
Phase (deg)
Bode Diagram
Frequency (rad/sec)
