Nyquist stability of 60/[(s+1)(s+2)(s+5)]

Open-loop poles

$G(s)H(s)=\dfrac{60}{(s+1)(s+2)(s+5)}$ has poles at $-1,-2,-5$: none in the right half plane, so $P=0$. There is no pole on the $j\omega$ axis, so no indentation is needed.

Frequency response

Denominator: $(s+1)(s+2)(s+5)=s^3+8s^2+17s+10$. With $s=j\omega$:

$$G(j\omega)H(j\omega)=\frac{60}{(10-8\omega^2)+j(17\omega-\omega^3)}$$ $$|GH|=\frac{60}{\sqrt{1+\omega^2}\sqrt{4+\omega^2}\sqrt{25+\omega^2}},\quad \angle GH=-\tan^{-1}\omega-\tan^{-1}\frac{\omega}{2}-\tan^{-1}\frac{\omega}{5}$$

Axis crossings

• Imaginary axis (real part of denominator zero): $10-8\omega^2=0\Rightarrow\omega=1.118$ rad/s, $GH=-j3.41$.
• Negative real axis (imaginary part zero): $17\omega-\omega^3=0\Rightarrow\omega=\sqrt{17}=4.123$ rad/s.

$$GH(j4.123)=\frac{60}{10-8(17)}=\frac{60}{-126}=-0.476$$

Nyquist plot

               Im
         w<0   |  ___
            ___|_/   \
 ----o--x--+---+------o--- Re
    -1 -0.476  |     6 (w=0)
           \___|_    /
               | \__/ w>0
               | -j3.41

• $\omega=0\to\infty$: starts at $+6$, goes through $-j3.41$, cuts the negative real axis at $-0.476$, reaches the origin at $-270^\circ$.
• $\omega=-\infty\to0$: mirror image.
• Infinite semicircle of the $s$-plane maps to the origin.

Nyquist criterion

$Z=P+N$. The plot cuts the real axis at $-0.476$, to the right of $-1$, so the point $-1+j0$ is not encircled: $N=0$.

$$Z=0+0=0$$

No closed-loop poles in the RHP, so the closed-loop system is stable.

Relative stability: $GM=1/0.476=2.1$ (6.44 dB), PM ≈ 25° at $\omega_{gc}=2.84$ rad/s. The system becomes unstable if the gain 60 is raised above $60\times2.1=126$ (Routh: $8\times17>10+K \Rightarrow K<126$).

Answer: N = 0, P = 0, Z = 0 – the closed-loop system is stable (GM ≈ 6.4 dB).