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10/19/
ECE 382 Midterm 2
Problem 1 (20 points) In the above system, C(s) = 10 and G(s) = (^) s (^2) +6^1 s. Find the maximum overshoot and the settling time (5% criterion) of the closed-loop system. Solution: The closed-loop transfer function is
Y (s) U (s)
C(s)G(s) 1 + C(s)G(s)
s^2 + 6s + 10
which is a standard second order system. We have ω^2 n = 10, hence ωn =
10; and 2ζωn = 6, hence ζ = 3/
- Therefore, the maximum overshoot is
Mp = e
− √ 1 ζ−ζ 2 π = e−^3 π^ = 8−^5 = 0.008%.
The settling time (5% criterion) is
ts =
ζωn
Problem 2 (20 points) Suppose in the system above, C(s) = K and G(s) = (^) s (^3) +s (^21) +2s− 4.
(a) (2 pts) Determind the system type. Solution: The system is of type 0.
Find the steady-state errors of the system for tracking the unit step input for the following values of K: Solution: (9 pts) We first find the range of K for stability. The characteristic equation is equivalent to s^3 + s^2 + 2s + (K − 4) = 0. Construct the Routh array:
s^3 : 1 2 s^2 : 1 K − 4 s^1 : 6 − K s^0 : K − 4
Thus, for the first column to have the same sign, the range of K for stability is 4 < K < 6.
(b) (3 pts) K = 3: Solution: Since K is outside the range (4, 6), the closed-loop system is unstable. Thus, the steady-state tracking error is infinite.
(c) (3 pts) K = 5: Solution: In this case, the closed-loop system is stable. Thus, the position error constant is
Kp = lim s→ 0 C(s)G(s) = −K/4 = − 1. 25.
and for tracking unit step input, ess = (^) 1+^1 Kp = −4.
(d) (3 pts) K = 7: Solution: Since K is outside the range (4, 6), the closed-loop system is unstable. Thus, the steady-state tracking error is infinite.
Problem 3 (20 points) In the above system, assume G(s) = (^) s (^3) +s (^2) +2^1 s− 0. 5 , and C(s) is a PI compensator:
C(s) = 1 +
K
s
, K ≥ 0.
(a) (15 pts) Find the range of K ≥ 0 so that the closed-loop system is stable. Solution: The system has characteristic equation:
1 + C(s)G(s) = 1 +
s + K s
s^3 + s^2 + 2s − 0. 5
⇒ s^4 + s^3 + 2s^2 + 0. 5 s + K = 0
Construct the Routh array:
s^4 : 1 2 K s^3 : 1 0. s^2 : 1.5 K s^1 : 0.^751 .− 5 K s^0 : K
For the closed-loop system to be stable, the first column cannot change sign. Thus, the range of K for stability is: 0 < K < 7 .5.
(b) (5 pts) Write the steady-state error ess of the system for tracking the unit ramp input in terms of K. By changing K, what is the smallest absolute value of such tracking error |ess| one can possibly achieve? Solution: Since the open-loop transfer function C(s)G(s) = (^) s(s (^3) +ss 2 + (^) +2Ks− 0 .5) has exactly one pole at the origin, the system is of type 1. The velocity error constant is
Kv = lim s→ 0 sC(s)G(s) = − 2 K.
Thus, for tracking unit ramp input, ess = (^) K^1 v = − (^21) K. This expression of ess only holds if the closed-loop
system is stable, i.e., if 0 < K < 7 .5. Thus, the smallest possible |ess| is 151 , which is achieved by choosing K arbitrarily close to 7.5.
Problem 4 (20x points) In the plot above, find the transfer function G(s) in the right system so that the two systems have the same root loci. Namely, they have the same set of closed-loop poles for all K ≥ 0.
Solution: See above. −3.5^ −3^ −2.5^ −2^ −1.5^ −1^ −0.5^0 0.5^1 1.
−
−
−
0
1
2
3
Root Locus
Real Axis
Imaginary Axis
