Nyquist Diagram and Gain Margin of $1/(s(s+2)^2)$: 23-Chem-A6 Process Dynamics and Control, May 2018, Q8

Problem 8: Nyquist Diagram and Gain Margin of $1/(s(s+2)^2)$ (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $G(s)=\dfrac{1}{s(s+2)^2}$ — an integrator ($1/s$) in series with a repeated first-order pair at $s=-2$; unity gain assumed in the loop for the margin.

Find. (1) the qualitative Nyquist locus with its asymptotes and real-axis crossing; (2) the gain margin.

Problem 8(1): coming up from $\mathrm{Im}\to-\infty$ along the vertical asymptote $\mathrm{Re}=-\tfrac14$ (as $\omega\to0^+$), the locus crosses the negative real axis at $-\tfrac{1}{16}$ (at $\omega=2$), then curls into the origin at $\omega\to\infty$ (approaching along the $-270^\circ$ direction).

Approach. Substitute $s=j\omega$, separate the denominator into real and imaginary parts, find the phase-crossover frequency where the imaginary part vanishes (locus on the real axis), evaluate $|G|$ there, and take $\mathrm{GM}=1/|G(j\omega_{pc})|$.

(1) Frequency response. $G(j\omega)=\dfrac{1}{j\omega\,(j\omega+2)^2}$. Expand $(j\omega+2)^2=(4-\omega^2)+4j\omega$; multiplying by $j\omega$ gives the denominator $$D(j\omega)=-4\omega^2+j\,\omega(4-\omega^2).$$ So $G=1/D$.
(1) Asymptotes and axis crossing. As $\omega\to0^+$, $\mathrm{Re}\,G\to-\tfrac14$ and $\mathrm{Im}\,G\to-\infty$ (vertical asymptote at $\mathrm{Re}=-0.25$). As $\omega\to\infty$, $|G|\to0$ (phase $\to-270^\circ$). The locus meets the real axis where $\mathrm{Im}\,D=0$: $\omega(4-\omega^2)=0\Rightarrow \omega_{pc}=2$ (the finite crossover).
(2) Value at phase crossover. At $\omega=2$, $D=-4(2)^2+j\,2(4-4)=-16$, so $$G(j2)=\frac{1}{-16}=-0.0625\quad(\text{real, negative; phase }=-180^\circ).$$
(2) Gain margin. $$\boxed{\mathrm{GM}=\frac{1}{|G(j\omega_{pc})|}=\frac{1}{1/16}=16\ (=24.1\ \text{dB}).}$$ The open-loop gain can be increased $16\times$ before the crossing reaches $-1$ and the closed loop goes marginally stable (at which point $k_{cu}=16$, $\omega_u=2$).

Quantity	Value
Low-frequency asymptote	$\mathrm{Re}=-\tfrac14$, $\mathrm{Im}\to-\infty$
Phase-crossover frequency	$\omega_{pc}=2$ rad/time
Real-axis crossing	$G=-1/16=-0.0625$
Gain margin	$\mathrm{GM}=16$ ($24.1$ dB)
Ultimate gain / frequency	$k_{cu}=16,\ \omega_u=2$

Concept & Theory

Topic. Nyquist plot of a type-1 system; phase crossover; gain margin from the negative-real-axis intercept.

Key relations. Phase crossover: $\mathrm{Im}\,G(j\omega)=0$; $\mathrm{GM}=1/|G(j\omega_{pc})|$; type-1 systems have $\mathrm{Im}\,G\to-\infty$ with a finite real asymptote as $\omega\to0$.

Why this works. The gain margin measures how much loop gain can grow before the $-180^\circ$ point ($G$ real and negative) lands on $-1$. Because $G(j\omega_{pc})=-1/16$, multiplying the loop gain by $16$ puts that point exactly at $-1$ — the onset of sustained oscillation at $\omega_u=2$. The integrator makes the low-frequency phase start at $-90^\circ$, giving the vertical $\mathrm{Re}=-\tfrac14$ asymptote.

Common pitfall. Taking $\omega=0$ as the phase crossover (it is a $-90^\circ$ asymptote, not $-180^\circ$), or forgetting to invert $|G|$ so GM comes out as $1/16$ instead of $16$. The double pole makes the high-frequency phase $-270^\circ$, so the locus enters the origin from the second quadrant.

Source. Seborg et al., Ch. 14 (Bode/Nyquist, gain and phase margins); Stephanopoulos, Ch. 17–18.

Question 8 of 8: Nyquist Diagram and Gain Margin of $1/(s(s+2)^2)$

Problem 8: Nyquist Diagram and Gain Margin of $1/(s(s+2)^2)$ (20%)