Question 8 of 8: Nyquist stability (Part A) and state-space response (Part B)

Question 8 — Nyquist stability (Part A) and state-space response (Part B) [10 + 10]

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.

Part A(1) — polar plot coordinates

1. Normalized response. $\dfrac{G_{open}}{K}=\dfrac{1+j\omega}{j\omega(j\omega-5)}$; separating parts, $\text{Re}=\dfrac{-6}{\omega^2+25}$, $\text{Im}=\dfrac{5-\omega^2}{\omega(\omega^2+25)}$.
2. Real-axis crossover. $\text{Im}=0\Rightarrow\omega=\sqrt5=2.24$ rad/s, where $\text{Re}=-6/30=\boxed{-0.20}$.
3. Imaginary-axis behaviour. $\text{Re}=-6/(\omega^2+25)\lt0$ for all $\omega$, so the plot never crosses the imaginary axis; as $\omega\to0^+$ it runs off to $+j\infty$ asymptotic to the vertical line $\text{Re}=-6/25=-0.24$, and as $\omega\to\infty$ it approaches the origin from the third quadrant ($\angle\to-90^\circ$).

Part A(2) — Nyquist criterion

1. Open-loop RHP poles. $P=1$ (the pole at $+5$); the pole at the origin is handled by an infinitesimal right indentation of the $\Gamma$-path.
2. Encirclement requirement. For closed-loop stability $Z=N+P=0$ needs $N=-1$ (one counter-clockwise encirclement of $-1$).
3. Critical gain. The real-axis crossing sits at $-0.2K$; it reaches $-1$ when $K=1/0.2=5$. For $K\gt5$ the point $-1$ is enclosed to give the required CCW encirclement, so the loop is $\boxed{\text{stable for }K_p\gt5}$ (unstable for $K_p\lt5$).
4. Cross-check (Routh). $s(s-5)+K(1+s)=s^2+(K-5)s+K=0$ is stable iff $K-5\gt0$ and $K\gt0$, i.e. $K\gt5$ — identical.

Part B — analytical state-space solution

1. Transfer function. $\det(sI-A)=(s+1)(s+5)+3=(s+2)(s+4)$; $Y/U=C(sI-A)^{-1}B=\dfrac{3}{(s+2)(s+4)}$.
2. Step response transform. $Y(s)=\dfrac{3}{s(s+2)(s+4)}$; partial fractions give residues $\tfrac{3}{8},-\tfrac{3}{4},\tfrac{3}{8}$ at $s=0,-2,-4$.
3. Time response. $\boxed{y(t)=0.375-0.75\,e^{-2t}+0.375\,e^{-4t}},\ t\ge0$; check $y(0)=0$ and $y(\infty)=0.375$.