Question 1 of 7: 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 system with transfer function $G(s)=\dfrac{3}{s^{2}+2s-3}$, driven from rest by a unit step $r(t)=1(t)$, i.e. $R(s)=1/s$. Note the negative constant term in the denominator.

Find. (a) the DC (zero-frequency) gain of the system, and (b) the final value approached by the step response — or a demonstration that no such value exists.

Approach. Factor the denominator to locate the poles, evaluate $G(0)$ for the DC gain, then test whether the Final Value Theorem is legally applicable before using it — and confirm the verdict by inverting $C(s)$ term by term.

1. Factor the denominator and locate the poles. The characteristic polynomial factors over the reals:

One pole is comfortably in the left half plane, but the second sits at $s=+1$, in the right half plane. The system is therefore open-loop unstable. This single observation governs both parts of the question, and it is only visible because the constant term $-3$ is negative — a polynomial with a negative coefficient can never have all its roots in the left half plane.

2. Compute the DC gain from $G(0)$. The DC gain is the steady-state ratio of output to input for a constant input, which for a transfer function is simply its value at $s=0$:

This is a well-defined piece of algebra: $s=0$ is not a pole of $G$, so $G(0)$ exists and equals $-1$. Physically it is the gain the system would show at zero frequency if it were stable, and it is exactly what a frequency-response (Bode) magnitude plot would report as $\omega\to0$: $20\log_{10}|{-1}|=0$ dB.

3. Test the Final Value Theorem before applying it. The theorem $\lim_{t\to\infty}c(t)=\lim_{s\to0}sC(s)$ is valid only if every pole of $sC(s)$ lies strictly in the left half plane (a simple pole at the origin is permitted). Here

The step’s pole at the origin cancels as intended, but the pole at $s=+1$ survives in $sC(s)$. The hypothesis of the theorem fails, so the Final Value Theorem must not be used, and the limit $\lim_{s\to0}sC(s)=3/(3)(-1)=-1$ that it would produce is meaningless.

4. Invert $C(s)$ to settle the question directly. Expanding in partial fractions, $C(s)=\dfrac{A}{s}+\dfrac{B}{s+3}+\dfrac{C}{s-1}$, the residues are

A free arithmetic check is available: because $C(s)$ is strictly proper and the system starts from rest, $c(0^{+})=0$, so the residues must sum to zero — and indeed $-1+0.25+0.75=0$. Transforming back with pairs (1) and (15) of the supplied table,

5. Read off the behaviour as $t\to\infty$. The $e^{-3t}$ term decays and the constant stays put, but the $0.75\,e^{+t}$ term grows without bound. Hence

It is worth stating plainly how badly the naive answer fails. Applying the Final Value Theorem regardless would return $-1$: not merely the wrong magnitude, but the wrong sign, since the true response runs off to $+\infty$. The response never goes anywhere near $-1$: differentiating, $\dot{c}(t)=0.75\left(e^{t}-e^{-3t}\right)\ge0$ for every $t\ge0$, so $c(t)$ climbs monotonically away from the origin and never becomes negative, despite the negative DC gain. What it does do is start slowly — $\dot{c}(0)=0$ and $c(0.2)=0.053$ — which is exactly the kind of short-record behaviour that can mislead a simulation stopped too soon; $c(t)$ only passes $1$ at $t=0.974$ s, and by $t=1.6$ s it has reached $2.72$ and is growing like $e^{t}$.