22-Mec-A3 System Analysis and Control · May 2016

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Mec-A3 System Analysis and Control, National Exams May 2016 — 3 hours, closed book (Casio or Sharp approved calculator; semi-log graph paper is the only permitted aid). The rubric states that any four (4) questions constitute a complete paper and that all questions are of equal value, so each question is worth 25 marks of the 100 marked. Because this document is a study resource rather than an exam script, all six questions are solved in full.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Pearson) — Ch. 3–7; N. S. Nise, Control Systems Engineering, 8th ed. (Wiley) — Ch. 4, 6, 8, 10; R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson) — Ch. 5–8; G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson) — Ch. 3–6. A table of Laplace transform pairs is appended to the examination paper (pages 3–4 of the original), together with two sheets of 3-cycle semi-logarithmic graph paper for the Bode question.

Question 1: DC Gain and Final Value of a 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.

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.
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.
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.
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.
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.

Figure 1.1 — Unit-step response of $G(s)=3/(s^{2}+2s-3)$. Because the plant is strictly proper and starts from rest, the response leaves the origin flat, $c(0)=\dot c(0)=0$ with $\ddot c(0)=+3$, and then rises monotonically as the right-half-plane pole at $s=+1$ takes over and the output diverges to $+\infty$. It never approaches the DC gain of $-1$ — indeed it never goes negative at all, so the naive final-value answer has the wrong sign as well as the wrong magnitude. No finite final value exists.

Check: the question as printed asks only for "the final value to a step input" and gives no amplitude, so a unit step is assumed throughout. Because the system is linear, a step of amplitude $R_{0}$ simply scales every term above by $R_{0}$; the conclusion (unbounded response, no final value) is unchanged for any non-zero step size.

Quantity	Result
Poles of $G(s)$	$s=-3$ and $s=+1$ (one RHP pole → open-loop unstable)
(a) DC gain $G(0)$	$-1$ (dimensionless)
Final Value Theorem applicable?	No — $sC(s)$ retains the pole at $s=+1$
Unit-step response	$c(t)=-1+0.25e^{-3t}+0.75e^{+t}$
(b) Final value $c(\infty)$	Does not exist; $c(t)\to+\infty$
Value the FVT would wrongly give	$-1$ (equal to the DC gain, and not attained)

Topic. DC gain, the Final Value Theorem and its region-of-convergence condition.

Key relations. DC gain $=G(0)$; Final Value Theorem $\lim_{t\to\infty}c(t)=\lim_{s\to0}sC(s)$, valid only when all poles of $sC(s)$ have negative real parts; residue (cover-up) rule $A_{k}=\left[(s-p_{k})C(s)\right]_{s=p_{k}}$ for simple poles.

Why this works. The Laplace transform of a signal exists only on a half-plane to the right of that signal's rightmost pole. The Final Value Theorem is really a statement about the behaviour of $C(s)$ as $s\to0$ from inside that region of convergence; if a pole with positive real part lies between the origin and the region of convergence, the point $s=0$ is not reachable and the limit computed there describes nothing physical. The DC gain, by contrast, is purely algebraic: it is the ratio of output to input amplitude for a sustained sinusoid of zero frequency, which presupposes that a steady state exists. For an unstable plant it does not, so the two quantities part company. This is exactly why the examiner pairs the two parts — part (a) is a one-line substitution and part (b) is a test of whether the candidate checks the theorem's hypotheses.

Common pitfall. Quoting $c(\infty)=G(0)=-1$ because "the final value equals the DC gain". That identity holds only for stable systems. A second, subtler slip is reading $s^{2}+2s-3$ as a standard second-order form and reporting $\omega_{n}=\sqrt{-3}$ or $\zeta=2/(2\sqrt{-3})$; the $\zeta$–$\omega_{n}$ parameterisation presumes a pole pair in the left half-plane and simply does not apply when a coefficient is negative.

Source. Ogata, Modern Control Engineering 5th ed., §2-3 and §5-2 (final value theorem and step response); Nise, Control Systems Engineering 8th ed., §4.2; Franklin, Powell & Emami-Naeini, 8th ed., §3.1.6.

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