Question 8 of 8: Nyquist Criterion and State-Space Step Response

Question 8: Nyquist Criterion and State-Space Step Response (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. Part A: $G_{open}(s)/K=(1+s)/[s(s-5)]$, an open-loop plant with ONE pole at the origin and ONE pole in the right half-plane ($s=+5$). Part B: state matrices $A=\begin{bmatrix}-1&1\\-3&-5\end{bmatrix}$, $B=\begin{bmatrix}-1\\0\end{bmatrix}$, $C=\begin{bmatrix}0&1\end{bmatrix}$, unit step input, zero initial state.

Find. Part A: the polar-plot crossovers and the stabilizing range of $K_p$. Part B: the closed-form $y(t)$.

Approach. Part A: evaluate $G_{open}(j\omega)/K$ directly to find where it crosses the real and imaginary axes and its limiting behaviour as $\omega\to0^{+},\infty$; since the open loop has one RHP pole ($P=1$), the Nyquist criterion needs exactly one CCW encirclement of $-1/K$ for stability, which is cross-checked against the Routh–Hurwitz array on the same characteristic equation. Part B: solve $Y(s)=C(sI-A)^{-1}Bu(s)$ and invert the Laplace transform.

1. Part A.1) — polar-plot geometry. With $s=j\omega$, $$\frac{G_{open}(j\omega)}{K}=\frac{1+j\omega}{j\omega(j\omega-5)}=\underbrace{\frac{-6}{\omega^2+25}}_{\text{Re}} +j\underbrace{\frac{5-\omega^2}{\omega(\omega^2+25)}}_{\text{Im}}.$$ The REAL part is negative for every $\omega\ne0$ and never reaches zero, so the plot never crosses the imaginary axis (except in the limits below). The IMAGINARY part vanishes at $\omega=\sqrt5=2.236$ rad/s, where $\text{Re}=-6/(5+25)=\boxed{-0.20}$ — the plot's one crossing of the real axis. As $\omega\to0^{+}$, $G_{open}/K\to j/(5\omega)\to\infty\angle{+}90^{\circ}$ (the plot departs upward along the positive imaginary axis); as $\omega\to\infty$, $G_{open}/K\to1/(j\omega)\to0\angle{-}90^{\circ}$ (the plot spirals into the origin from the negative-imaginary direction). Frequency increases from $0^{+}$ to $+\infty$ moving CLOCKWISE around this path, passing through the real-axis crossing at $-0.20$ on the way.
2. Part A.2) — Nyquist stability range. The open-loop plant has ONE pole in the right half-plane ($s=+5$) plus one pole at the origin needing the usual small indentation in the $\Gamma$ contour, so $P=1$; for closed-loop stability $Z=N+P=0$ requires $N=-1$, i.e. exactly one COUNTER-clockwise encirclement of the critical point $-1/K_p$. The real-axis crossing at $-0.20$ found above is the only point where the normalized plot can straddle the critical point as $K_p$ varies, so the encirclement count changes exactly there: solving the SAME characteristic equation directly via Routh–Hurwitz, $$s(s-5)+K_p(1+s)=0\ \Rightarrow\ s^2+(K_p-5)s+K_p=0,$$ requires both $K_p-5>0$ and $K_p>0$ for stability, i.e. $\boxed{K_p>5}$ — consistent exactly with $-1/K_{crit}=-1/5=-0.20$, the crossing point identified in Part A.1, confirming the encirclement analysis.
3. Part B) — state-space step response. $$Y(s)=C(sI-A)^{-1}B\,\frac{1}{s}=\frac{3}{s(s^2+6s+8)}=\frac{3}{s(s+2)(s+4)}.$$ Partial fractions give $Y(s)=\dfrac{3/8}{s}-\dfrac{3/4}{s+2}+\dfrac{3/8}{s+4}$, so $$\boxed{y(t)=\frac{3}{8}-\frac{3}{4}e^{-2t}+\frac{3}{8}e^{-4t}}\ ,\quad t\ge0,$$ which correctly satisfies $y(0)=3/8-3/4+3/8=0$ (matching $x(0)=0$) and settles at $y_{ss}=\lim_{s\to0}sY(s)=\boxed{3/8=0.375}$.