Nyquist stability of the Q1 system: 22-Elec-A2 Systems and Control, May 2013, Q8

Question 8 — Nyquist stability of the Q1 system [8 + 12]

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. $G(j\omega)=\dfrac{100}{(j\omega+2)(j\omega+4)(j\omega+10)}$; the denominator expands to $D(j\omega)=(80-16\omega^2)+j(68\omega-\omega^3)$.

Figure Q8.1 — Nyquist plot of $G(j\omega)$. Real-axis crossings at $\omega=0$ ($+1.25$) and $\omega=8.25$ ($-0.099$); imaginary-axis crossing at $\omega=2.24$ ($-j0.71$). The critical point $-1/K_{crit}$ sits at $-0.099$.

Part A — Graphical Nyquist reading [8]

Encirclement condition. $G(s)$ has no right-half-plane poles ($P=0$), so for closed-loop stability the $K_pG(j\omega)$ locus must not encircle $-1+j0$. Marginal stability occurs when the locus passes through $-1$.
Critical gain. The negative-real-axis crossing is at $-0.099$; scaling by $K_p$ to reach $-1$ gives $K_{crit}=1/0.099=\boxed{10.08}$, safe range $0\lt K_p\lt 10.08$.
Gain margin at $K_p=2$. At $K_p=2$ the crossing is at $2(-0.099)=-0.198$; $$\text{GM}=\frac{1}{0.198}=5.04\ \text{V/V}=20\log_{10}(5.04)=\boxed{14.05\ \text{dB}}.$$

Part B — Analytical axis crossings [12]

Approach. Write $G(j\omega)=100\,\overline{D}/|D|^2$; a real-axis crossing needs $\mathrm{Im}\{D\}=0$, an imaginary-axis crossing needs $\mathrm{Re}\{D\}=0$.

Real-axis crossings ($\mathrm{Im}\,D=0$). $68\omega-\omega^3=0\Rightarrow\omega=0$ or $\omega=\sqrt{68}=8.25\ \text{rad/s}$. At $\omega=0$: $G=100/80=+1.25$ (the DC point). At $\omega=\sqrt{68}$: $D=80-16(68)=-1008$, so $$G=\frac{100}{-1008}=\boxed{-0.099},\qquad K_{crit}=\frac{1}{0.099}=10.08,\quad \omega_{osc}=8.25\ \text{rad/s}.$$
Imaginary-axis crossings ($\mathrm{Re}\,D=0$). $80-16\omega^2=0\Rightarrow\omega=\sqrt{5}=2.24\ \text{rad/s}$. There $D=j(68\sqrt5-\sqrt5^{\,3})=j140.9$, so $$G=\frac{100}{j140.9}=-j0.71.$$
Consistency. The analytical negative-real crossing $-0.099$ and $\omega_{osc}=8.25$ reproduce Part A exactly ($K_{crit}=10.08$, GM $=14.05$ dB at $K_p=2$), and both agree with the root-locus/Bode/Routh results of Q1. All methods are consistent.

Crossing	Frequency	Value
Real axis (DC)	$\omega=0$	$+1.25$
Real axis (critical)	$\omega=\sqrt{68}=8.25$ rad/s	$-0.099$
Imaginary axis	$\omega=\sqrt{5}=2.24$ rad/s	$-j0.71$
Critical gain / margin	$\boxed{K_{crit}=10.08,\ 0\lt K_p\lt 10.08,\ \text{GM}=5.04\ \text{V/V}=14.05\ \text{dB}}$

Concept & Theory

Topic. Nyquist stability criterion and analytical axis crossings.

Key relations. $Z=N+P$ (closed-loop RHP poles $=$ clockwise encirclements of $-1$ $+$ open-loop RHP poles); real crossing $\Leftrightarrow\mathrm{Im}\,D=0$, imaginary crossing $\Leftrightarrow\mathrm{Re}\,D=0$; $K_{crit}=1/|G(j\omega_{pc})|$.

Why this works. With $P=0$, avoiding encirclement of $-1$ guarantees stability. The most negative real-axis crossing is the closest approach to $-1$, so it sets the critical gain and equals the Routh/Bode/root-locus $\omega_{osc}$ — the four methods examine the identical characteristic equation from different angles.

Common pitfall. Mislabeling which axis crossing sets $K_{crit}$ (it is the negative-real crossing, not the imaginary one), or dropping the $\omega=0$ real crossing when listing results.

Source. Nise §10 (Nyquist criterion, gain/phase margins from the polar plot).

Question 8 of 8: Nyquist stability of the Q1 system

Question 8 — Nyquist stability of the Q1 system [8 + 12]

Part A — Graphical Nyquist reading [8]

Part B — Analytical axis crossings [12]