Question

A unity feedback system has an open loop transfer function G(s) =25/s(s+8). What is the percentage of it’s peak overshoot?

(a) 0.13%

(b) 1.13%

(c) 1.26%

(d) 1.52%

This question was posed to me in an international level competition.

Question is from Integral Square Error & its Minimization in portion Error Analysis of Control Systems

Answer 1

The correct answer is:

(b) 1.13%

Explanation:

To find the percentage peak overshoot of a unity feedback system, we first need to analyze the system's damping ratio (ζ), which is related to the overshoot.

Given the open-loop transfer function:
G(s)=25s(s+8)G(s) = \frac{25}{s(s+8)}G(s)=s(s+8)25​

The characteristic equation of the system is derived from the closed-loop transfer function:

T(s)=G(s)1+G(s)=25s(s+8)+25T(s) = \frac{G(s)}{1 + G(s)} = \frac{25}{s(s+8) + 25}T(s)=1+G(s)G(s)​=s(s+8)+2525​

This can be written as a second-order system:

T(s)=Ks2+8s+KT(s) = \frac{K}{s^2 + 8s + K}T(s)=s2+8s+KK​

where K=25K = 25K=25. The standard form for a second-order system is:

T(s)=Ks2+2ζωns+ωn2T(s) = \frac{K}{s^2 + 2ζω_n s + ω_n^2}T(s)=s2+2ζωn​s+ωn2​K​

where ωnω_nωn​ is the natural frequency and ζζζ is the damping ratio. By comparing the coefficients:

2ζωn=8andωn2=252ζω_n = 8 \quad \text{and} \quad ω_n^2 = 252ζωn​=8andωn2​=25

Thus:

ωn=25=5andζ=82×5=0.8ω_n = \sqrt{25} = 5 \quad \text{and} \quad ζ = \frac{8}{2 \times 5} = 0.8ωn​=25​=5andζ=2×58​=0.8

Now, the percentage peak overshoot MpM_pMp​ is given by:

Mp=100e−ζπ1−ζ2M_p = 100 e^{-\frac{ζ \pi}{\sqrt{1 - ζ^2}}}Mp​=100e−1−ζ2​ζπ​

Substituting ζ=0.8ζ = 0.8ζ=0.8:

Mp=100e−0.8π1−0.82≈1.13%M_p = 100 e^{-\frac{0.8 \pi}{\sqrt{1 - 0.8^2}}} \approx 1.13\%Mp​=100e−1−0.82​0.8π​≈1.13%

Thus, the correct answer is 1.13%.