Nyquist stability of 1/[s²(1+2s)(1+s)]

Open-loop data

$$G(s)H(s)=\frac{1}{s^2(1+2s)(1+s)}$$

Poles: double pole at $s=0$ (type-2), and $-0.5$, $-1$. No poles in the RHP: $P=0$. The double pole at the origin is bypassed by a small semicircle in the Nyquist contour.

Section I: $\omega=0^+\to\infty$

$$|GH|=\frac{1}{\omega^2\sqrt{1+4\omega^2}\sqrt{1+\omega^2}},\qquad \angle GH=-180^\circ-\tan^{-1}2\omega-\tan^{-1}\omega$$

• The phase is always below $-180^\circ$, so the plot lies in the second quadrant at low frequency. For small $\omega$: $GH\approx-\dfrac{1}{\omega^2}+j\dfrac{3}{\omega}$ (real part $\to-\infty$, imaginary part $\to+\infty$).
• It cuts the positive imaginary axis where $\tan^{-1}2\omega+\tan^{-1}\omega=90^\circ\Rightarrow2\omega^2=1\Rightarrow\omega=0.707$, at $GH=+j0.943$.
• It then enters the first quadrant and reaches the origin at $-360^\circ$.
• It never cuts the negative real axis for finite $\omega>0$.

Section II: infinite semicircle

$s=Re^{j\theta}$, $R\to\infty$: $GH\to0$, maps to the origin.

Section III: $\omega=-\infty\to0^-$

Mirror image of section I (third and fourth quadrants).

Section IV: indentation at the origin

$s=\epsilon e^{j\theta}$, $\theta:-90^\circ\to+90^\circ$:

$$GH\approx\frac{1}{\epsilon^2}e^{-j2\theta}$$

Angle goes from $+180^\circ$ to $-180^\circ$: a circle of infinite radius, 360° clockwise (two half circles), through the positive real axis.

             Im
   w>0      | j0.943
  (from -inf|+j inf)
    \_______|__
            |  \_ to origin
 ----x------+------ Re
    -1      |  _/
     _______|_/
    /  w<0  |
  plus a full infinite
  clockwise circle at w=0

Nyquist criterion

The $-1$ point lies between the $\omega>0$ branch (above it) and the $\omega<0$ branch (below it), and the 360° infinite clockwise arc closes the curve around it. So $-1+j0$ is encircled twice clockwise: $N=2$.

$$Z=P+N=0+2=2$$

The closed-loop system has two poles in the RHP and is unstable.

Check by Routh: $1+GH=0\Rightarrow 2s^4+3s^3+s^2+1=0$. The $s$ term is missing, so the system cannot be stable; the roots are $-1.04\pm j0.50$ and $0.29\pm j0.54$ (two in the RHP). A type-2 system with only extra lags needs a zero (lead action) to be stabilised.

Answer: P = 0, N = 2, Z = 2 – the closed-loop system is unstable for any positive gain.