Nyquist Stability Criterion for an Open-Loop-Unstable Plant: 22-Elec-A2 Systems and Control, December 2019, Q5

Question 5: Nyquist Stability Criterion for an Open-Loop-Unstable Plant (20 marks)

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. Unit-feedback loop with $G_{open}(s) = K(1+s)/[s(s-0.5)]$: open-loop poles at $s = 0$ (on the contour) and $s = +0.5$ (right half plane), and one open-loop zero at $s = -1$. The Nyquist path $\Gamma$ is traversed clockwise, so the number of unstable open-loop poles enclosed is $P = 1$.

Find. The polar plot of $G_{open}(j\omega)/K$ including its real- and imaginary-axis crossings and the direction of increasing frequency; and, via the Nyquist criterion, the range of $K$ for closed-loop stability.

Approach. Rationalise $G_{open}(j\omega)/K$ into explicit real and imaginary parts, find where each vanishes and what the low- and high-frequency asymptotes are, sketch the locus and its mirror image, then count encirclements of the $-1$ point against the requirement $Z = N + P = 0$ and confirm the answer independently with the Routh criterion.

Separate real and imaginary parts. Putting $s = j\omega$, $$\frac{G_{open}(j\omega)}{K} = \frac{1+j\omega}{j\omega(j\omega-0.5)} = \frac{1+j\omega}{-\omega^2 - j0.5\omega}$$ Multiplying numerator and denominator by the conjugate $-\omega^2 + j0.5\omega$ gives $$\frac{G_{open}(j\omega)}{K} = \frac{-1.5\omega^2 + j\left(0.5\omega - \omega^3\right)}{\omega^4 + 0.25\omega^2}$$
Low-frequency asymptote. As $\omega \to 0^+$ the $\omega^2$ terms dominate the denominator, so $$\text{Re} \to \frac{-1.5\omega^2}{0.25\omega^2} = -6, \qquad \text{Im} \to \frac{0.5\omega}{0.25\omega^2} = \frac{2}{\omega} \to +\infty$$ The plot therefore comes down from infinity asymptotic to the vertical line $\text{Re} = -6$.
Real-axis crossing. The imaginary part vanishes when $0.5\omega - \omega^3 = 0$, i.e. at $$\omega = \sqrt{0.5} = 0.707\ \text{rad/s}$$ Substituting back, $$\text{Re} = \frac{-1.5(0.5)}{(0.5)^2 + 0.25(0.5)} = \frac{-0.75}{0.375} \;\Rightarrow\; \boxed{\,\frac{G_{open}}{K}\bigg|_{\omega=0.707} = -2.0\,}$$ With the gain restored the crossing sits at $-2K$.
Imaginary-axis crossing and high-frequency behaviour. The real part $-1.5\omega^2/(\omega^4+0.25\omega^2)$ is strictly negative for every finite $\omega \gt 0$, so the locus never crosses the imaginary axis. As $\omega \to \infty$, $G_{open}/K \approx j\omega/(-\omega^2) = -j/\omega \to 0$, so the plot spirals into the origin from the $-90^\circ$ direction. The direction of increasing frequency therefore runs from far up the $\text{Re} = -6$ asymptote, rightwards through $-2$ on the real axis, and into the origin.

Figure 5.1 — Polar plot of G_open(jw)/K (solid, increasing frequency arrowed) with the mirror image for negative frequency (dashed) completing the Nyquist contour. The locus crosses the real axis at -2.0 when w = 0.707 rad/s.

Assemble the Nyquist contour. The full contour comprises the $\omega \gt 0$ locus just derived, its mirror image about the real axis for $\omega \lt 0$, and the large arc mapping the infinite semicircle to the origin. The pole at the origin is indented by a small semicircle in the s-plane, which maps to a large arc sweeping through infinity and joins the two branches on the left. The resulting closed curve crosses the negative real axis once, at $-2K$.
Apply the criterion. The Nyquist criterion states $$Z = N + P$$ where $Z$ is the number of closed-loop poles in the right half plane, $N$ the number of clockwise encirclements of the $-1$ point, and $P = 1$ the number of open-loop right-half-plane poles (the pole at $s = +0.5$). For closed-loop stability we need $Z = 0$, hence $$N = -P = -1$$ that is, exactly one counter-clockwise encirclement of $-1$.
Translate the encirclement condition into a gain range. The contour encircles the $-1$ point precisely when the real-axis crossing lies to the left of $-1$: $$|-2K| \gt 1 \;\Longrightarrow\; 2K \gt 1 \;\Longrightarrow\; \boxed{\,K \gt 0.5\,}$$ For $K \lt 0.5$ the crossing at $-2K$ falls between $-1$ and the origin, the curve fails to enclose $-1$, $N = 0$ and therefore $Z = 1$ — one unstable closed-loop pole. At $K = 0.5$ the contour passes exactly through $-1$ and the loop is marginally stable, oscillating at $\omega = 0.707$ rad/s.
Independent confirmation by Routh. The closed-loop characteristic equation is $$s(s-0.5) + K(1+s) = s^2 + (K-0.5)s + K = 0$$ For a quadratic, stability requires every coefficient positive: $K \gt 0$ and $K - 0.5 \gt 0$, i.e. $K \gt 0.5$ — identical to the Nyquist result. At $K = 0.5$ the $s^1$ coefficient vanishes and the roots are $\pm j\sqrt{0.5} = \pm j0.707$, confirming both the marginal gain and the oscillation frequency. Note that this loop is stable only for sufficiently large gain: because the plant is open-loop unstable, the gain margin is a lower bound with no upper limit.

Quantity	Result
Low-frequency asymptote	$\text{Re} \to -6$, $\text{Im} \to +\infty$
Real-axis crossing frequency	$\omega = 0.707$ rad/s
Real-axis crossing (normalised)	$-2.0$; with gain, $-2K$
Imaginary-axis crossing	None (Re $\lt 0$ for all $\omega$)
High-frequency limit	$\to 0$ from $-90^\circ$
Open-loop RHP poles	$P = 1$ (at $s = +0.5$)
Encirclements required	$N = -1$ (one counter-clockwise)
Stable gain range	$K \gt 0.5$
Marginal gain / frequency	$K = 0.5$ at $\omega_{osc} = 0.707$ rad/s
Routh check	$s^2 + (K-0.5)s + K$: stable iff $K \gt 0.5$

Topic. The Nyquist stability criterion applied to an open-loop-unstable plant, where the intuition built on minimum-phase systems fails.

Key relations. $Z = N + P$ with $N$ counted clockwise for a clockwise $\Gamma$; rationalisation of $G(j\omega)$ into real and imaginary parts; real-axis crossings where $\text{Im}\{G\} = 0$; Routh–Hurwitz on $1 + G_{open}(s) = 0$ as the algebraic cross-check.

Why this works. The criterion is the argument principle in disguise: as $s$ traverses a closed contour enclosing the right half plane, the image of $1 + G_{open}(s)$ winds about the origin once for each enclosed zero minus each enclosed pole — and the zeros of $1+G_{open}$ are exactly the closed-loop poles. Shifting attention from the origin to the $-1$ point lets the same count be read off the plot of $G_{open}$ itself. The crucial consequence here is that a plant with $P = 1$ needs an encirclement to be stable, which inverts the usual rule: more gain is stabilising, less gain is destabilising, and the familiar "keep the crossing inside $-1$" instinct gives precisely the wrong answer.

Common pitfall. Applying the minimum-phase habit and concluding $K \lt 0.5$ is the classic error, and it is easy to miss because the arithmetic looks identical. Sign conventions are the other trap: $N$ must be counted in the same rotational sense as the assumed $\Gamma$ path, and a counter-clockwise encirclement then registers as $N = -1$. Always confirm the final range with Routh on the closed-loop polynomial — it takes two lines and settles any sign dispute.

Source. Nise, Control Systems Engineering, 8th ed., §10.3–10.6; Ogata, Modern Control Engineering, 5th ed., §7.4 (Nyquist criterion, unstable plants).

Question 5 of 8: Nyquist Stability Criterion for an Open-Loop-Unstable Plant

Question 5: Nyquist Stability Criterion for an Open-Loop-Unstable Plant (20 marks)