ELEN90055 Control Systems Worksheet 4 – Solutions to Starred Problems Semester 2
1. The transfer function of the second order system is usually written as ωn2 ωn2
H(s) = s2 +2ζωns+ωn2 = (s+ζωs)2 +ωn2(1−ζ2) So the characteristic equation is
(s+ζωn)2 +ωn2(1−ζ2) = 0 and for 0 ≤ ζ < 1, the root of the characteristic equation are
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s=−ζωn±jωn 1−ζ2.
× 𝑗𝜔𝑑 𝜔𝑛 𝜃
ts = 4.6 ≤9.2 ζωn
Figure 1: σ 1
If the overshoot is less than or equal to 0.17 then ζ ≥ 0.5 and θ = sin−1 ζ = 30◦).
Figure 2: ζ
1.8 tr ≈ ω
≤ 0.6 ⇒ 𝑗3
From Franklin, Powel & Emami-Naeini“Feedback control of dynamic systems (slide 7, L7-21.pdf).
See also Goodwin, Graebe and Salgado eq. (4.8.15) in sec. 4.8 for the exact formula relating overshoot to damping coefficient.
Figure 3: ωn
So the region in the complex s-plane for which the poles could be placed are shown below
Figure 4: 2
• If the settling time is 9.2sec, then ζ ωn = 0.5 and therefore the poles should be exactly on the line ζ ωn = 0.5.
• To make the rise time small, we need to make ωn large as tr = 1.8 . ωn
• The overshoot condition is satisfied if the poles are located in the highlighted area in Figure 2.
The location of the poles that satisfy the above conditions are shown in Figure 5.
Figure 5: Problem 1(b).
tr=1.8≤2 ⇒ ωn≥0.9. ωn
Mp ≤10% ⇒ ζ ≥0.6 ⇒ θ ≥36.8◦
The region of the complex plane where all the dominant poles of the closed-loop transfer
function should be located is highlighted in Figure 6
0.9𝑗 36.8°
Figure 6: Problem 2.
3. Rise time is a measure of how fast the system reacts to a change in its input, and settling
time is a measure of how fast the system can settle down to steady state with a desired miss 3
ratio (that is, how fast the system’s transient decays). So depending on the application, both rise time and settling time can be used as measures of how fast the response is. In other words, depending on how we define a faster response, the controller can be designed such that we have a lower rise time or a lower settling time.
In this specific problem, we have two responses in which the rise times are close to each other (0.5sec and 1sec) while the settling times are very different (4sec and 10sec). Because of this big difference in the settling times, we say that the response with tr = 1sec and ts = 4sec is faster. See the following step responses to compare the responses of these two systems.
Step Response
0 2 4 6 8 10 12
Time (seconds)
Figure 7: Problem 3a.
Step Response
0 2 4 6 8 10 12
Time (seconds)
Figure 8: Problem 3b.
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