Question 6 of 8: Nyquist and Routh–Hurwitz stability for a RHP-pole plant

Question 6 — Nyquist and Routh–Hurwitz stability for a RHP-pole plant [7 + 8 + 5]

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. $\dfrac{G_{open}}{K}=\dfrac{s+1}{s(s-5)}$: pole at the origin (Type 1) and one RHP pole at $+5$ ($P=1$ open-loop RHP pole), zero at $-1$.

Find. Polar-plot crossovers; the stable $K$ range by Nyquist; and the Routh confirmation.

Part 1 — Polar plot and crossovers [7]

Approach. Evaluate $\dfrac{G_{open}(j\omega)}{K}=\dfrac{1+j\omega}{-\omega^2-5j\omega}$ and rationalise.

1. Rationalise. Multiplying by the denominator conjugate, $\dfrac{G_{open}}{K}=\dfrac{-6\omega^2+j\,\omega(5-\omega^2)}{\omega^2(\omega^2+25)} =\dfrac{-6}{\omega^2+25}+j\,\dfrac{5-\omega^2}{\omega(\omega^2+25)}$.
2. Real-axis crossover. Imaginary part $=0$ when $\omega^2=5$, i.e. $\omega=\sqrt5=2.236$ rad/s; there $\mathrm{Re}=\dfrac{-6}{5+25}=\boxed{-0.2}$.
3. Imaginary-axis crossover. $\mathrm{Re}=-6/(\omega^2+25)\lt0$ for all finite $\omega$, so the locus never crosses the imaginary axis; as $\omega\to0^+$ it has a vertical asymptote at $\mathrm{Re}=-6/25=-0.24$ with $\mathrm{Im}\to+\infty$, and as $\omega\to\infty$ it approaches the origin from the third quadrant.
4. Direction. Increasing $\omega$ runs from the top of the asymptote (second quadrant) down through the real-axis point $-0.2$ and into the origin — a clockwise sweep for the normalized plot.

Part 2 — Nyquist stability [8]

Approach. Use a clockwise (CW) $\Gamma$ contour that indents to the right of the origin pole. With $P=1$ open-loop RHP pole, closed-loop stability ($Z=0$) requires $Z=N+P=0$, i.e. $N=-1$ (one counter-clockwise encirclement of $-1$).

1. Locate $-1$ relative to the crossing. Scaling by $K$, the real-axis crossing sits at $-0.2K$. The $-1$ point is enclosed appropriately only when the crossing lies to its left, i.e. $-0.2K\lt-1$.
2. Encirclement count. For $-0.2K\lt-1$ (large gain) the mapped contour makes exactly one CCW encirclement of $-1$, giving $N=-1$ and $Z=N+P=0$ — stable. For $-0.2K\gt-1$ (small gain) the $-1$ point is not properly encircled, $N=+1$, $Z=2$ — two closed-loop RHP poles, unstable.
3. Range. $-0.2K\lt-1\Rightarrow\boxed{K\gt5}$ for a stable closed loop; the marginal case $K=5$ puts the crossing exactly at $-1$ ($\omega_{osc}=\sqrt5=2.24$ rad/s).

Part 3 — Routh–Hurwitz check [5]

1. Characteristic polynomial. $s(s-5)+K(s+1)=s^2+(K-5)s+K=0$.
2. Routh conditions. $s^2:1,K$; $s^1:(K-5)$; $s^0:K$. A positive first column needs $K\gt0$ and $K-5\gt0$, i.e. $\boxed{K\gt5}$ — identical to the Nyquist result.
3. Marginal frequency. At $K=5$, $s^2+5=0\Rightarrow s=\pm j\sqrt5$, so $\omega_{osc}=\sqrt5=2.24$ rad/s, matching the polar-plot real-axis crossing.