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5. GH(s) is a minimum-phase system which has the Bode plot shown below. It is desired to increase the phase margin by 40 degr

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Answer #1

from the given bode plot, the phase margin is 10 degrees approximately.

therefore the final phase margin requirement = 10 + 40 = 50 degrees.

the ratio of pole to zero alpha is given by alpha = (1-sin(50))/(1+sin(50)) = 0.1325.

hould locate the frequency at which the uncompensated system has a logarithmic magnitude of -20log(1/sqrt(alpha))

Select this frequency as the new gain crossover frequency since the compensator provides a gain of  20log(1/sqrt(alpha)) at ωmax.

-20log(1/sqrt(alpha)) = -8.7778 dB

Therefore wmax = 10 rad/sec approximately.

we already know,

urmax =w@wew-TVa

Therefore 10 = 1/tau*sqrt(0.1325) => tau = 1/(10*sqrt(0.1325)) = 0.2747

Therefore C(s) =  (0.2747s+1)/(0.2747*0.1325*s+1) =   (0.2747s+1)/(0.0364s+1)

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please show steps 5. GH(s) is a minimum-phase system which has the Bode plot shown below. It is desired to increase the phase margin by 40 degrees and also increase the closed-loop system bandwidth....
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