Question 1 of 8: Frequency-response & Routh stability of a proportional loop (compulsory)

Question 1 — Frequency-response & Routh stability of a proportional loop (compulsory) [20]

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, open loop $L(s)=K_p\,G(s)$ with $G(s)=\dfrac{100}{s(s+5)^2}$: poles at $0,-5,-5$; no finite zeros; one free integrator.

Find. GM, PM, $\omega_{cg}$ (phase-crossover) and $\omega_{cp}$ (gain-crossover); $K_{crit}$, $\omega_{osc}$; stable range of $K_p$; and the Type 1 error constants at $K_p=1$.

[Figure not reproduced: Figure Q1.1 redrawn — proportional unit-feedback loop. See the official exam paper.]

Part 1 — Frequency-response margins & critical gain

Approach. The phase-crossover frequency (where $\angle L=-180^\circ$) fixes the gain margin and the marginal gain; the gain-crossover frequency (where $|L|=1$) fixes the phase margin.

1. Phase-crossover frequency. $\angle L(j\omega)=-90^\circ-2\arctan(\omega/5)$. Setting this to $-180^\circ$ gives $\arctan(\omega/5)=45^\circ$, so $\boxed{\omega_{cg}=5\ \text{rad/s}}$.
2. Gain margin. $|L(j\omega)|=\dfrac{100}{\omega(\omega^2+25)}$; at $\omega=5$, $|L|=\dfrac{100}{5(50)}=0.4$. Hence $\text{GM}=\dfrac{1}{0.4}=\boxed{2.5\ \text{V/V}=7.96\ \text{dB}}$.
3. Critical gain & oscillation frequency. Marginal stability occurs when $K_p|L|=1$ at $\omega_{cg}$: $K_{crit}=\text{GM}=\boxed{2.5}$ with $\boxed{\omega_{osc}=5\ \text{rad/s}}$.
4. Gain-crossover frequency & phase margin. Solving $|L|=1\Rightarrow\omega^3+25\omega-100=0$ gives $\omega_{cp}=2.96$ rad/s. There $\angle L=-90^\circ-2\arctan(2.96/5)=-151.3^\circ$, so $\text{PM}=180^\circ-151.3^\circ=\boxed{28.7^\circ}$.
5. Stable range. The loop is stable for gains below the critical value: $\boxed{0\lt K_p\lt2.5}$.

Part 2 — Routh–Hurwitz verification

Approach. Form the closed-loop characteristic polynomial, build the Routh array, and set the $s^1$ row to zero; the $s^2$-row auxiliary equation gives $\omega_{osc}$.

1. Characteristic polynomial. With $s(s+5)^2=s^3+10s^2+25s$, $1+K_pG=0$ gives $$s^3+10s^2+25s+100K_p=0.$$
2. Routh array. $$\begin{array}{c|cc}s^3&1&25\\ s^2&10&100K_p\\ s^1&\frac{250-100K_p}{10}&0\\ s^0&100K_p&\end{array}$$
3. Marginal stability. The $s^1$ entry vanishes when $250-100K_p=0$, i.e. $\boxed{K_{crit}=2.5}$.
4. Oscillation frequency. The $s^2$-row auxiliary equation $10s^2+100K_{crit}=0$ gives $s^2=-25$, so $\boxed{\omega_{osc}=5\ \text{rad/s}}$.
5. Comparison. Identical to Part 1 — both methods give $K_{crit}=2.5$, $\omega_{osc}=5$ rad/s and the range $0\lt K_p\lt2.5$.

Part 3 — System Type & steady-state errors ($K_p=1$)

Approach. The single open-loop integrator makes the system Type 1; evaluate the three static error constants.

1. System Type. $L(s)$ has one pole at the origin $\Rightarrow$ $\boxed{\text{Type 1}}$.
2. Error constants. $K_{pos}=\lim_{s\to0}L=\infty$; $K_v=\lim_{s\to0}sL=\dfrac{100}{25}=\boxed{4.0}$; $K_a=\lim_{s\to0}s^2L=0$.
3. Steady-state errors. $e_{ss}(\text{step})=\dfrac{1}{1+K_{pos}}=0$; $e_{ss}(\text{ramp})=\dfrac{1}{K_v}=\boxed{0.25}$; $e_{ss}(\text{parabola})=\dfrac{1}{K_a}=\infty$.