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Control Systems

Stability Analysis from Bode Plot

The asymptoti...

Question

The asymptotic magnitude Bode plot of an open loop system G(s) with K > 0 and all poles and zeros on the left hand side of the s-plane is shown in the figure. It is completetly sysmmetric about $ω_{c}$ . The minimum absolute angle contribution by G(s) is given by

{78.6}^{o}

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90^{o}

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{101.4}^{o}

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180^{o}

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Solution

The correct option is D $180^{o}$
Transfer function of the system,

$T (s) = \frac{k (s + 1)}{s^{2} (\frac{s}{100} + 1)}$

$∠ T (j ω) = - 180^{o} + t a n^{- 1} ω - t a n^{- 1} \frac{ω}{100}$ .

Put the minimum value of $ω$ , then will get the minimum absolute phase being $180^{o}$

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The asymptotic magnitude Bode plot of an open loop system G(s) with K > 0 and all poles and zeros on the left hand side of the s-plane is shown in the figure. It is completetly sysmmetric about ωc. The minimum absolute angle contribution by G(s) is given by

The correct option is D 180oTransfer function of the system, T(s)=k(s+1)s2(s100+1) ∠T(jω)=−180o+tan−1ω−tan−1ω100. Put the minimum value of ω, then will get the minimum absolute phase being 180o

The asymptotic magnitude Bode plot of an open loop system G(s) with K > 0 and all poles and zeros on the left hand side of the s-plane is shown in the figure. It is completetly sysmmetric about $ω_{c}$ . The minimum absolute angle contribution by G(s) is given by

The correct option is D $180^{o}$
Transfer function of the system,

$T (s) = \frac{k (s + 1)}{s^{2} (\frac{s}{100} + 1)}$

$∠ T (j ω) = - 180^{o} + t a n^{- 1} ω - t a n^{- 1} \frac{ω}{100}$ .

Put the minimum value of $ω$ , then will get the minimum absolute phase being $180^{o}$