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

Question 1 of 6: Impulse Response of a Repeated-Pole and of an Underdamped 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 December 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. 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 (page 5 of the original).

Question 1: Impulse Response of a Repeated-Pole and of an Underdamped 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. Two transfer functions driven by a unit impulse, $r(t)=\delta(t)$, so $R(s)=1$ and the response transform is simply $C(s)=G(s)$. Part (a) is third order with a repeated real pole; part (b) is second order and written with a normalised constant term rather than a normalised leading coefficient.

Find. The time functions $g(t)=\mathcal{L}^{-1}\{G(s)\}$ for both systems.

Approach. Because the impulse transform is unity, the impulse response is just the inverse Laplace transform of the transfer function itself: expand (a) in partial fractions with the extra term the repeated pole demands, and force (b) into the standard underdamped second-order form before reading off the damped sine pair from the transform table.

Part (a)

Set up the partial-fraction expansion with a repeated root. The pole at $s=-5$ appears twice, so it contributes two terms — one for each power of the repeated factor:$$G(s)=\frac{100}{(s+2)(s+5)^{2}}=\frac{A}{s+2}+\frac{B}{s+5}+\frac{C}{(s+5)^{2}}$$Omitting the $B$ term is the classic error; the expansion would then be unable to match $G(s)$ at any second point.
Evaluate the simple-pole residue by the cover-up rule. Multiply by $(s+2)$ and set $s=-2$:$$A=\left.\frac{100}{(s+5)^{2}}\right|_{s=-2}=\frac{100}{9}=11.111$$
Evaluate the highest repeated-pole coefficient the same way. Multiply by $(s+5)^{2}$ and set $s=-5$:$$C=\left.\frac{100}{s+2}\right|_{s=-5}=\frac{100}{-3}=-33.333$$
Evaluate the remaining coefficient by differentiating. For a double pole the lower coefficient comes from the first derivative of the covered-up function:$$B=\left.\frac{d}{ds}\!\left[\frac{100}{s+2}\right]\right|_{s=-5}=\left.\frac{-100}{(s+2)^{2}}\right|_{s=-5}=-\frac{100}{9}=-11.111$$Note the free check that falls out immediately: $A+B=0$. That must happen, because $G$ has relative degree three, so $g(0^{+})=0$ and the two simple exponential terms have to cancel at $t=0$.
Invert term by term. Using $1/(s+a)\to e^{-at}$ and $1/(s+a)^{2}\to t\,e^{-at}$ from the appended table,$$\boxed{\;g(t)=\frac{100}{9}e^{-2t}-\frac{100}{9}e^{-5t}-\frac{100}{3}\,t\,e^{-5t}\qquad (t\ge 0)\;}$$
Sanity-check the shape. At $t=0$ the expression gives $100/9-100/9-0=0$, as required. For large $t$ the $e^{-5t}$ terms die four times faster than $e^{-2t}$, so the tail is governed by the slow pole and decays as $11.11\,e^{-2t}$. A numerical sweep puts the maximum at $g=1.817$ at $t=0.540\ \text{s}$.

Impulse response of $G(s)=100/[(s+2)(s+5)^{2}]$. The $t\,e^{-5t}$ term holds the response near zero at the start; the slow $e^{-2t}$ pole governs the tail.

Part (b)

Normalise the leading coefficient. The denominator is written with a unit constant term, which hides the natural frequency. Divide numerator and denominator by 0.04:$$G(s)=\frac{1}{0.04s^{2}+0.08s+1}=\frac{25}{s^{2}+2s+25}$$
Read off the standard second-order parameters. Comparing with $\omega_{n}^{2}/(s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2})$:$$\omega_{n}=\sqrt{25}=5\ \text{rad/s},\qquad 2\zeta\omega_{n}=2\;\Longrightarrow\;\zeta=\frac{2}{2(5)}=0.20$$Since $\zeta<1$ the system is underdamped and the poles are complex.
Compute the damped natural frequency. $$\omega_{d}=\omega_{n}\sqrt{1-\zeta^{2}}=5\sqrt{1-0.04}=4.899\ \text{rad/s}$$so the poles sit at $s=-1\pm j4.899$.
Match the transform pair and invert. The table pair is $\omega_{d}/[(s+\sigma)^{2}+\omega_{d}^{2}]\to e^{-\sigma t}\sin\omega_{d}t$ with $\sigma=\zeta\omega_{n}=1$. Writing the numerator to suit,$$G(s)=\frac{25}{(s+1)^{2}+4.899^{2}}=\frac{25}{4.899}\cdot\frac{4.899}{(s+1)^{2}+4.899^{2}}$$so that$$\boxed{\;g(t)=5.103\,e^{-t}\sin(4.899\,t)\qquad (t\ge 0)\;}$$
Check the result. $g(0)=0$, correct for a system of relative degree two. The decay envelope is $\pm 5.103\,e^{-t}$, giving a time constant of 1 s and 2 % settling in about $4\tau=4\ \text{s}$. The first peak occurs at $t=\tan^{-1}(\omega_{d}/\sigma)/\omega_{d}=0.280\ \text{s}$, where $g=3.78$.

Impulse response of $G(s)=25/(s^{2}+2s+25)$: a lightly damped ($\zeta=0.20$) sine inside an $e^{-t}$ envelope.

Question 1 — final results
Quantity	Value
(a) residues $A,\;B,\;C$	$100/9,\;-100/9,\;-100/3$
(a) impulse response	$g(t)=11.111e^{-2t}-11.111e^{-5t}-33.333\,t\,e^{-5t}$
(b) $\omega_{n}$, $\zeta$, $\omega_{d}$	$5\ \text{rad/s}$, $0.20$, $4.899\ \text{rad/s}$
(b) impulse response	$g(t)=5.103\,e^{-t}\sin(4.899t)$

Topic. Inverse Laplace transformation: partial-fraction expansion with repeated poles, and the standard underdamped second-order form.

Key relations. $g(t)=\mathcal{L}^{-1}\{G(s)\}$ because $\mathcal{L}\{\delta(t)\}=1$; for a double pole, $F(s)=\cdots+B/(s+p)+C/(s+p)^{2}$ with $C=[(s+p)^{2}F]_{s=-p}$ and $B=\frac{d}{ds}[(s+p)^{2}F]_{s=-p}$; $\omega_{n}^{2}=a_{0}/a_{2}$, $2\zeta\omega_{n}=a_{1}/a_{2}$, $\omega_{d}=\omega_{n}\sqrt{1-\zeta^{2}}$.

Why this works. The impulse is the one input whose transform is unity, so the impulse response and the transfer function are a Laplace pair — every dynamic property of the system is visible in $g(t)$ without solving a differential equation. Partial fractions decompose the response into the natural modes of the system, one per pole. A repeated pole is a degenerate case in which two modes have collapsed onto the same exponential rate; the system compensates by producing a mode $t\,e^{-pt}$ that grows algebraically before the exponential wins. In the underdamped case the complex-conjugate pole pair produces a single real sinusoid whose frequency is the imaginary part and whose decay rate is the real part — which is why designers think in terms of $\zeta$ and $\omega_{n}$ rather than in terms of pole coordinates.

Common pitfall. Two recur. First, dropping the $B/(s+5)$ term for the repeated pole, which makes the expansion wrong everywhere except at the two evaluation points. Second, in part (b), reading $\omega_{n}^{2}=1$ straight off the printed denominator: the leading coefficient must be normalised to unity first, or $\omega_{n}$ comes out as 1 rad/s instead of 5.

Source. Ogata, Modern Control Engineering, 5th ed., §2.4 (partial-fraction expansion, repeated poles) and §5.3 (second-order systems); Nise, Control Systems Engineering, 8th ed., Ch. 4.

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