The figure below shows the bode magnitude and phase plots of a stable transfer function
$G (s) = \frac{n_{0}}{s^{3} + d_{2} s^{2} + d_{1} s + d_{0}}$

Consider the negative unity feedback configuration with gain $k$ in the feedforward path. The closed loop in stable for $k < k_{0}$ . The maximum value of $k_{0}$ is

0.1

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Solution

The correct option is A 0.1
$F o r G (s)$
$M_{d B} (ω_{p c}) = 20 d B$

When cascaded with $k$ ,
$G M_{d B} = - 20 d B - 20 l o g_{10} (k) > 0 d B$
$20 + 20 l o g_{10} (k) < 0$
$20 l o g_{10} (k) < - 20$
$k < 10^{- 1} = 0.10$
$S o, k_{0} = 0.10$

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The figure below shows the bode magnitude and phase plots of a stable transfer function G(s)=n0s3+d2s2+d1s+d0 Consider the negative unity feedback configuration with gain k in the feedforward path. The closed loop in stable for k<k0. The maximum value of k0 is 0.1

The correct option is A 0.1For G(s) MdB(ωpc)=20 dB When cascaded with k, GMdB=−20 dB−20log10(k)>0 dB 20+20log10(k)<0 20log10(k)<−20 k<10−1=0.10 So,k0=0.10

The correct option is A 0.1
$F o r G (s)$
$M_{d B} (ω_{p c}) = 20 d B$

When cascaded with $k$ ,
$G M_{d B} = - 20 d B - 20 l o g_{10} (k) > 0 d B$
$20 + 20 l o g_{10} (k) < 0$
$20 l o g_{10} (k) < - 20$
$k < 10^{- 1} = 0.10$
$S o, k_{0} = 0.10$