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Consider the system given below where K is a constant gain, Gp is the plant, and Ge is a compensator. The Bode Plots of a Gp

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Answer #1

From the Bode frequency response, it is observed that the initial roll off of the magnitude response is approximately 20 dB /decade.

One pole at origin is responsible for this decay of magnitude. Therefore there is one pole at origin.

It is observed that the high frequency roll off of the magnitude response is approximately 60 dB /decade. This means that there are a total 3 poles. Therefore the number of excess poles = 3 - 1 = 2.

From the initial magnitude response, it is observed that the initial magnitude at w = 0.1rad/sec is 0 dB approximately.

i.e. 20log(K) + 20log(0.1) = 0 dB = 1

=> 20 log(K) = -20 dB => K = 0.1.

From the bode response, the gain margin is computed as given below

Bode Diagram 20 -60 -80 100 90 6-135 225 270 10- 100 Frequency (rad/s) 10 10

GM = - gain in dB at phase crossover frequency = -(-40 dB) = 40 dB

With K = 1, the magnitude response shifts upwards.

Therefore the new gain margin = 40- 20dB = 20 dB.

The transfer function is approximately given below based on the given information

T(s) = 0.1 /(s*(s+3)^2)

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Consider the system given below where K is a constant gain, Gp is the plant, and Ge is a compensator. The Bode Plots of a Gp is given below. Problem 1: Bode Diagram 20 2 40 -60 80 -100 90 135 180 a 2...
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