GM and PM from Bode plot of 1/[s(0.1s+1)(0.2s+1)]

$$G(s)=\frac{1}{s(1+0.1s)(1+0.2s)}$$

Type-1 system, $K=1$, corner frequencies $\omega_1 = 1/0.2 = 5$ rad/s and $\omega_2 = 1/0.1 = 10$ rad/s.

Magnitude plot (asymptotic)

• At $\omega=0.1$: $20\log(1/0.1) = 20$ dB; at $\omega=1$: 0 dB.
• At $\omega=5$: $20\log(1/5) = -14$ dB.
• At $\omega=10$: $-14 - 40\log 2 = -26$ dB.
• At $\omega=100$: $-26 - 60 = -86$ dB.

Phase plot

$$\phi(\omega) = -90^\circ - \tan^{-1}0.1\omega - \tan^{-1}0.2\omega$$

 dB
  20 |\ -20dB/dec
   0 |-\---------------------- w
     |  \ wgc~1
 -14 |   \___ w=5  -40
 -26 |        \__ w=10
     |           \  -60
 deg
 -90 |--__
-180 |-------x---------- wpc=7.07
-270 |            ----___

Phase crossover frequency and gain margin

At $\omega_{pc}$, $\phi=-180^\circ$: $\tan^{-1}0.1\omega+\tan^{-1}0.2\omega = 90^\circ \Rightarrow 0.02\omega^2 = 1$.

$$\begin{aligned} \omega_{pc} &= \sqrt{50} = 7.07\ \text{rad/s}\\ |G(j\omega_{pc})| &= \frac{1}{7.07\sqrt{1+0.5}\sqrt{1+2}} = \frac{1}{15} = 0.0667\\ GM &= 20\log\frac{1}{0.0667} = 20\log 15 = 23.52\ \text{dB} \end{aligned}$$

Gain crossover frequency and phase margin

From the asymptotic plot, the magnitude is 0 dB at $\omega\approx1$ rad/s. Solving $|G(j\omega)|=1$ exactly gives $\omega_{gc}=0.977$ rad/s.

$$\begin{aligned} \phi(\omega_{gc}) &= -90^\circ - \tan^{-1}(0.0977) - \tan^{-1}(0.1954)\\ &= -90^\circ - 5.58^\circ - 11.05^\circ = -106.6^\circ\\ PM &= 180^\circ + \phi(\omega_{gc}) = 73.4^\circ \end{aligned}$$

(Using the asymptotic value $\omega_{gc}=1$: $PM = 180^\circ-107.0^\circ = 73^\circ$.)

Answer: GM = 23.5 dB at $\omega_{pc}=7.07$ rad/s; PM ≈ 73.4° at $\omega_{gc}\approx0.98$ rad/s. Both margins are positive, so the closed-loop system is stable.