Polar plot of 1/[s(1+s)(1+2s)]

Frequency response

$$G(j\omega)H(j\omega)=\frac{1}{j\omega(1+j\omega)(1+j2\omega)}$$ $$|GH|=\frac{1}{\omega\sqrt{1+\omega^2}\sqrt{1+4\omega^2}},\qquad \angle GH=-90^\circ-\tan^{-1}\omega-\tan^{-1}2\omega$$

Rationalised form:

$$GH=\frac{-3\omega^2 - j\omega(1-2\omega^2)}{9\omega^4+\omega^2(1-2\omega^2)^2}$$

Key points

• Start ($\omega\to0^+$): magnitude $\infty$ at $-90^\circ$. The real part tends to $-(T_1+T_2)=-(1+2)=-3$, so the plot comes from $-j\infty$ along the asymptote $\text{Re}=-3$.
• Negative real-axis crossing: imaginary part zero when $1-2\omega^2=0\Rightarrow\omega_{pc}=\dfrac{1}{\sqrt2}=0.707$ rad/s.

$$\begin{aligned} GH(j0.707)&=\frac{-3(0.5)}{9(0.25)+0}=-\frac{1.5}{2.25}=-0.667 \end{aligned}$$

• End ($\omega\to\infty$): magnitude 0 at $-270^\circ$; the plot reaches the origin tangent to the positive imaginary axis.

              Im
              |
     -0.667   |
 ----x--o-----+--------- Re
    -1  |\_/  |   (ends at origin
        | w=0.707   from 2nd quad.)
     ___|     |
    /   |     |
   |  Re=-3 asymptote
   |   w -> 0+
   v  -j inf

Comment on stability

• Open-loop poles: $0,-1,-0.5$, so $P=0$.
• The polar plot cuts the negative real axis at $-0.667$, which is to the right of $-1$; with its mirror image and the infinite clockwise arc for the pole at the origin, the point $-1+j0$ is not encircled: $N=0$, $Z=0$.

The closed-loop system is stable.

Relative stability:

$$GM=\frac{1}{0.667}=1.5\ (3.52\ \text{dB}),\qquad PM=180^\circ-168.6^\circ=11.4^\circ\ \text{at}\ \omega_{gc}=0.572\ \text{rad/s}$$

The margins are small, so the response will be quite oscillatory; the gain can be raised by a factor of 1.5 before instability (Routh: $2s^3+3s^2+s+K$ needs $K<1.5$).

Answer: the plot crosses the negative real axis at −0.667 (ω = 0.707 rad/s); the closed-loop system is stable with GM = 1.5 (3.5 dB) and PM ≈ 11°.