I need to answer of this Control System Question For a unity feedback system where the plant is defined asKG(s)s(s +3)(s +5)a. Sketch the Nyquist Counter path and Nyquist diagram. Clearly show the real and imag-inary axis intercept points and the low and high frequency asymptotes.b. Using the Nyquist criterion, obtain the range of K in which the system can be stable,unstable, and also find the value of gain K for marginal stability.c. Calculate the frequency of oscillation for marginal stability.

Answered: For a unity feedback system where the…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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I need to answer of this Control System Question

For a unity feedback system where the plant is defined as
K
G(s)
s(s +3)(s +5)
a. Sketch the Nyquist Counter path and Nyquist diagram. Clearly show the real and imag-
inary axis intercept points and the low and high frequency asymptotes.
b. Using the Nyquist criterion, obtain the range of K in which the system can be stable,
unstable, and also find the value of gain K for marginal stability.
c. Calculate the frequency of oscillation for marginal stability.

Expert Solution

Step 1

Given data :

$G (s) = \frac{k}{s (s + 3) (s + 5)}$

Step 2

(a). Condition for phase cross over frequency,

$∠ G (j ω) = - 180 °$

Given transfer function convert in frequency domain,

$- 90 ° - \tan^{- 1} (\frac{ω}{3}) - \tan^{- 1} (\frac{ω}{5}) = - 180 ° \tan^{- 1} (\frac{ω}{3}) + \tan^{- 1} (\frac{ω}{5}) = 90 ° \tan^{- 1} (\frac{\frac{ω}{3} + \frac{ω}{5}}{1 - \frac{ω}{3} \times \frac{ω}{5}}) = 90 ° (\frac{\frac{8 ω}{15}}{1 - \frac{ω^{2}}{15}}) = \frac{1}{0} 1 - \frac{ω^{2}}{15} = 0 ω = \sqrt{15} r a d / s e c$

So phase cross over frequency is $ω_{p c} = \sqrt{15} r a d / s e c$

Now calculation for magnitude,

$|G (j ω)| = \frac{k}{ω \sqrt{ω^{2} + 9} \sqrt{ω^{2} + 25}}$

Putting values $ω_{p c} = \sqrt{15} r a d / s e c$ in the above expression,

$|G (j ω)| = \frac{k}{\sqrt{15} \sqrt{15 + 9} \sqrt{15 + 25}} = \frac{k}{\sqrt{15 \times 24 \times 40}} = \frac{k}{120}$

Hence Nyquist counter path and diagram is shown below,

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