Question 1 of 6: DC Gain and Final Value of an Unstable Second-Order System

Question 1: DC Gain and Final Value of an Unstable Second-Order System (25 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. A single-input single-output plant with transfer function $G(s)=3/(s^{2}+2s-3)$, driven by a unit step $r(t)=1$, $R(s)=1/s$, from rest. No feedback is applied — the transfer function as written is the system.

Find. (a) the DC (zero-frequency) gain of $G$, and (b) the final value of the output $c(t)$ when the input is a step, stating whether the Final Value Theorem may legitimately be used.

Approach. Factor the denominator to locate the poles, evaluate $G(0)$ for the DC gain, then test the Final Value Theorem's validity condition against those poles before quoting any limit — and, because that test fails here, obtain $c(t)$ explicitly by partial fractions.

1. Factor the characteristic polynomial and locate the poles. The denominator is a plain quadratic, so it factors by inspection:$$s^{2}+2s-3=(s+3)(s-1)\quad\Longrightarrow\quad p_{1}=-3,\qquad p_{2}=+1$$The constant term is negative, which is already a red flag: a stable second-order polynomial $s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}$ must have all three coefficients strictly positive. Here $p_{2}=+1$ lies in the right half-plane, so the plant is open-loop unstable.
2. Evaluate the DC gain. The DC gain is the transfer function evaluated at zero frequency, $s=j0$:$$G(0)=\left.\frac{3}{s^{2}+2s-3}\right|_{s=0}=\frac{3}{-3}$$so that$$\boxed{\;G(0)=-1\;}$$The gain is negative because the two poles sit on opposite sides of the origin, making the product of the pole distances change sign. Numerically it is a perfectly well-defined number; what it does not mean here is "the steady output for a unit step", and part (b) is precisely the trap that separates the two ideas.
3. Test the Final Value Theorem before using it. For a unit step $R(s)=1/s$ the response transform is$$C(s)=G(s)R(s)=\frac{3}{s\,(s+3)(s-1)}$$The Final Value Theorem $c(\infty)=\lim_{s\to 0}sC(s)$ is valid only if every pole of $sC(s)$ lies strictly in the open left half-plane (a single pole at the origin being permitted for $C(s)$ itself, but not for $sC(s)$). Here$$sC(s)=\frac{3}{(s+3)(s-1)}$$retains the pole at $s=+1$. The validity condition is violated, so the theorem may not be applied.
4. Invert the transform explicitly to see what really happens. Expanding $C(s)$ in partial fractions,$$C(s)=\frac{A_{0}}{s}+\frac{A_{1}}{s+3}+\frac{A_{2}}{s-1}$$with residues obtained by the cover-up rule:$$A_{0}=\frac{3}{(3)(-1)}=-1,\qquad A_{1}=\frac{3}{(-3)(-4)}=0.25,\qquad A_{2}=\frac{3}{(1)(4)}=0.75$$Transforming term by term gives the closed-form response$$\boxed{\;c(t)=-1+0.25\,e^{-3t}+0.75\,e^{+t}\;}$$A free check: at $t=0$ the three terms give $-1+0.25+0.75=0$, which is correct because $G$ is strictly proper and the system starts from rest.
5. State the final value. The term $0.75e^{+t}$ grows without bound, so$$\lim_{t\to\infty}c(t)=+\infty$$i.e. no finite final value exists. Had the Final Value Theorem been applied blindly it would have returned $\lim_{s\to0}sC(s)=3/[(3)(-1)]=-1$, i.e. the DC gain — a number the output never approaches and, in fact, moves away from. In practice the output runs off in the positive direction while the "DC gain" is negative, so even the sign of the naive answer is wrong.