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: Locate the frequency where the phase turns to $-180^\circ$.
Each first order factor $(1+j\omega)$ in the denominator subtracts $\tan^{-1}\omega$ from the phase. With four such factors stacked, the total phase is $-4\tan^{-1}\omega$. Gain margin is always read at the frequency where this phase first reaches $-180^\circ$, so
\[ 4\tan^{-1}\omega_{pc}=180^\circ \ \Rightarrow\ \tan^{-1}\omega_{pc}=45^\circ \ \Rightarrow\ \omega_{pc}=1 \]

Step 2: Write the magnitude in decibels at that frequency.
In dB form,
\[ |G(j\omega)|_{dB}=20\log_{10}K-4\times20\log_{10}\sqrt{1+\omega^2} \] At $\omega_{pc}=1$, $\sqrt{1+\omega_{pc}^2}=\sqrt2$, so
\[ |G(j\omega_{pc})|_{dB}=20\log_{10}K-80\log_{10}\sqrt2 = 20\log_{10}K-40\log_{10}2 \]

Step 3: Use the definition of gain margin in dB.
Gain margin in dB is the negative of the magnitude in dB at the phase crossover, so
\[ 20 = -\left(20\log_{10}K-40\log_{10}2\right) \] \[ 20 = -20\log_{10}K+40\log_{10}2 \]

Step 4: Solve for $K$.
Using $\log_{10}2\approx0.3010$,
\[ 40\log_{10}2\approx12.04 \] \[ 20\log_{10}K = 12.04-20=-7.96 \] \[ \log_{10}K=-0.398 \] \[ K=10^{-0.398}\approx0.4 \]

Step 5: Cross check with the direct ratio method.
Since $|1+j|^4=4$, the condition $20=-20\log_{10}(K/4)$ gives $K/4=0.1$, so $K=0.4$, matching the decibel calculation.
\[ \boxed{0.4} \]