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

Question 2 of 6: Inverse Laplace Transforms by Partial-Fraction Expansion

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 2: Inverse Laplace Transforms by Partial-Fraction Expansion (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. Three strictly proper rational transforms:

Part	$F(s)$	Pole structure
(a)	$2/[s(s+2)]$	Two distinct real poles: $0,\;-2$
(b)	$10/[s(s+1)(s+10)]$	Three distinct real poles: $0,\;-1,\;-10$
(c)	$(3s+2)/(s^{2}+4s+20)$	Complex conjugate pair: $-2\pm j4$

Find. The corresponding time functions $f(t)$ for $t\ge0$, obtained by partial-fraction expansion as the question directs.

Approach. Parts (a) and (b) have simple real poles, so the cover-up (residue) rule applies directly. Part (c) has an irreducible quadratic, so instead of expanding into complex residues we complete the square and match the transform to the standard damped-sine and damped-cosine pairs on the supplied table.

Part (a)

Write the expansion and evaluate the residues. With simple poles at $s=0$ and $s=-2$,$$\frac{2}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2},\qquad A=\left.\frac{2}{s+2}\right|_{s=0}=1,\qquad B=\left.\frac{2}{s}\right|_{s=-2}=-1$$
Invert term by term. Using $1/s\leftrightarrow1$ and $1/(s+a)\leftrightarrow e^{-at}$,$$\boxed{\;f(t)=1-e^{-2t},\qquad t\ge0\;}$$This is the familiar first-order step response with time constant $\tau=0.5\text{ s}$; $f(0)=0$ and $f(\infty)=1$, both consistent with the initial- and final-value theorems applied to $F(s)$.

Part (b)

Expand over the three simple poles.$$\frac{10}{s(s+1)(s+10)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+10}$$and apply the cover-up rule at each pole:$$A=\frac{10}{(1)(10)}=1,\qquad B=\frac{10}{(-1)(9)}=-\frac{10}{9},\qquad C=\frac{10}{(-10)(-9)}=\frac{1}{9}$$
Assemble the time function.$$\boxed{\;f(t)=1-\tfrac{10}{9}e^{-t}+\tfrac{1}{9}e^{-10t}=1-1.1111\,e^{-t}+0.1111\,e^{-10t}\;}$$Checking at $t=0$: $1-10/9+1/9=0$, as required for a transform whose numerator degree is three less than the denominator degree (both $f(0)$ and $\dot f(0)$ vanish). The mode at $s=-10$ decays ten times faster than the one at $s=-1$, so after roughly $0.4\text{ s}$ the response is essentially the single dominant exponential $1-1.111e^{-t}$ — the standard justification for dominant-pole reduction.

Part (c)

Complete the square in the denominator. The quadratic has no real roots ($\Delta=16-80<0$), so write it in the damped-oscillator form:$$s^{2}+4s+20=(s+2)^{2}+16=(s+\sigma)^{2}+\omega_{d}^{2},\qquad \sigma=2,\;\omega_{d}=4$$The poles are therefore $s=-2\pm j4$, i.e. $\omega_{n}=\sqrt{20}=4.472\text{ rad/s}$ and $\zeta=2/\sqrt{20}=0.447$.
Split the numerator about $(s+\sigma)$. The table pairs are $\dfrac{s+\sigma}{(s+\sigma)^{2}+\omega_{d}^{2}}\leftrightarrow e^{-\sigma t}\cos\omega_{d}t$ and $\dfrac{\omega_{d}}{(s+\sigma)^{2}+\omega_{d}^{2}}\leftrightarrow e^{-\sigma t}\sin\omega_{d}t$, so rewrite $3s+2$ in terms of $(s+2)$:$$3s+2=3(s+2)-4$$giving$$F(s)=3\,\frac{s+2}{(s+2)^{2}+4^{2}}-\frac{4}{4}\cdot\frac{4}{(s+2)^{2}+4^{2}}$$
Invert and combine into a single damped sinusoid.$$\boxed{\;f(t)=e^{-2t}\left(3\cos 4t-\sin 4t\right)\;}$$Equivalently, with $\sqrt{3^{2}+1^{2}}=\sqrt{10}=3.1623$ and $\phi=\arctan(1/3)=18.43^{\circ}$,$$f(t)=3.1623\,e^{-2t}\cos\!\left(4t+18.43^{\circ}\right)$$The check $f(0)=3$ agrees with the initial-value theorem, $\lim_{s\to\infty}sF(s)=3$.

Part	$f(t)$, $t\ge0$	Character
(a)	$1-e^{-2t}$	First-order rise, $\tau=0.5\text{ s}$
(b)	$1-\tfrac{10}{9}e^{-t}+\tfrac{1}{9}e^{-10t}$	Overdamped, dominant pole $s=-1$
(c)	$e^{-2t}(3\cos4t-\sin4t)=3.1623\,e^{-2t}\cos(4t+18.43^{\circ})$	Damped oscillation, $\zeta=0.447$, $\omega_{d}=4\text{ rad/s}$

Topic. Partial-fraction expansion and inverse Laplace transformation for simple real poles and for complex-conjugate pairs.

Key relations. Cover-up rule $A_{k}=[(s-p_{k})F(s)]_{s=p_{k}}$; transform pairs $1/s\leftrightarrow1$, $1/(s+a)\leftrightarrow e^{-at}$, $(s+\sigma)/[(s+\sigma)^{2}+\omega_{d}^{2}]\leftrightarrow e^{-\sigma t}\cos\omega_{d}t$, $\omega_{d}/[(s+\sigma)^{2}+\omega_{d}^{2}]\leftrightarrow e^{-\sigma t}\sin\omega_{d}t$; initial- and final-value theorems as free checks.

Why this works. Every strictly proper rational transform is a sum of terms whose inverses are tabulated, and the residue at a simple pole is unique, so the expansion is exact rather than approximate. For a complex pair one could compute two complex residues and add the conjugates, but completing the square and splitting the numerator about $(s+\sigma)$ keeps every quantity real and lands directly on the two table entries — markedly faster under closed-book exam conditions and far less error-prone. The residues also carry physical meaning: in part (b) the residue magnitudes $10/9$ and $1/9$ show immediately that the fast mode contributes only about one-tenth as much as the slow mode, which is what licenses the dominant-pole approximation.

Common pitfall. Forgetting the $\omega_{d}$ scaling on the sine pair — writing $-4/[(s+2)^{2}+16]$ as $-4e^{-2t}\sin4t$ instead of $-1\cdot e^{-2t}\sin4t$ — which multiplies that half of the answer by four. The second frequent error is dropping a sign in the cover-up rule for the middle pole of part (b): $(s+10)$ evaluated at $s=-1$ is $+9$, not $-9$, while $s$ itself is $-1$.

Source. Ogata, Modern Control Engineering 5th ed., §2-5 (partial-fraction expansion with MATLAB) and Table 2-1; Nise, Control Systems Engineering 8th ed., §2.2; Dorf & Bishop, Modern Control Systems 13th ed., §2.4.

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