Question

The system \[ G(s)=\frac{K}{(s+1)^4} \] has a gain margin of 20 dB. The value of \(K\) is ________. (rounded off to two decimal places)

Hint:

Find the phase crossover frequency where the phase is \(-180^\circ\), then use the gain margin formula \(GM=-20\log_{10}|G(j\omega_{pc})|\).

Answer 1

Correct Answer: 0.4

Step 1: Recall what gain margin means.
Gain margin is measured at the phase crossover frequency \(\omega_{pc}\), the frequency at which the phase of \(G(j\omega)\) equals \(-180^\circ\). At that frequency,
\[ GM(\text{dB}) = -20\log_{10}\left|G(j\omega_{pc})\right| \]

Step 2: Write the phase of \(G(j\omega)\).
\[ G(j\omega)=\frac{K}{(1+j\omega)^4} \] The phase contributed by each factor \((1+j\omega)\) in the denominator is \(-\tan^{-1}\omega\), and there are four identical factors, so
\[ \angle G(j\omega) = -4\tan^{-1}\omega \]

Step 3: Find the phase crossover frequency.
Set the phase equal to \(-180^\circ\):
\[ -4\tan^{-1}\omega_{pc}=-180^\circ \] \[ \tan^{-1}\omega_{pc}=45^\circ \] \[ \omega_{pc}=\tan(45^\circ)=1\ \text{rad/s} \]

Step 4: Find the magnitude of \(G(j\omega)\) at this frequency.
\[ \left|G(j\omega_{pc})\right|=\frac{K}{\left|1+j\omega_{pc}\right|^4}=\frac{K}{\left(\sqrt{1^2+1^2}\right)^4}=\frac{K}{\left(\sqrt2\right)^4}=\frac{K}{4} \]

Step 5: Apply the gain margin condition.
\[ 20=-20\log_{10}\left(\frac{K}{4}\right) \] \[ \log_{10}\left(\frac{K}{4}\right)=-1 \] \[ \frac{K}{4}=10^{-1}=0.1 \] \[ K=0.4 \]

Final Answer:
The value of \(K\) is \[ \boxed{0.4} \]