22-Elec-B2 Advanced Control Systems · December 2017

Question 3 of 5: Partial fractions and the inverse Laplace transform

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

Notes on this paper

Paper format. 16-Elec-B2 Advanced Control Systems, December 2017 — a three-hour open-book examination; any non-communicating calculator is permitted. The cover page states “Any four questions constitute a complete paper. Only the first four questions as they appear in your answer book will be marked” and “All questions are of equal value (25%)”. The sitting prints five questions, so each carries 25 marks and a candidate answers four. Question 1 is a seventeen-item multiple-choice block whose per-item weights are printed in the margin as [1], [2] or [3] and total exactly 25. A table of inverse Laplace transforms and a table of Laplace/z transforms are appended. All five questions are worked below, because this set is a study resource rather than a timed sitting.

Reference texts. N. S. Nise, Control Systems Engineering, 8th ed. (Ch. 4 time response, percent overshoot, peak time and settling time; Ch. 6 Routh–Hurwitz stability; Ch. 7 steady-state error and system type; Ch. 8 root-locus sketching rules including break-away/break-in points and angles of departure; Ch. 10 frequency response and stability margins); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Ch. 5, 6, 7, 9); K. Ogata, Modern Control Engineering, 5th ed. (Ch. 5 transient response, Ch. 6 root-locus analysis, Ch. 7 frequency-response analysis); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Ch. 3, 5, 6). These are the references listed for exam code 16-Elec-B2 in the Engineers Canada syllabus, and this paper’s vocabulary (“break-in point”, “intercept of the asymptotes”, “angle of departure”) follows Nise closely.

Check — three source readings checked against the printed paper.

(1) Question 1, item 2. The printed sentence reads “where $\theta(s)$, $T(s)$, $J$ and $K$ represent torque, angular displacement, moment of inertia and spring constant, respectively”. The paper really does print the first two descriptions in the wrong order. The drawing beside it settles the physics — $T(t)$ is the applied torque acting on the inertia $J$, and $\theta$ is the resulting angular displacement, which is also the only reading under which $\theta(s)/T(s)$ has the units of a compliance. The answer does not depend on the naming at all, because only the denominator is used.

(2) Question 1, item 16. The margins are read off a printed Bode pair, so they are inherently approximate. The printed Bode pair gives a gain crossover near $\omega \approx 3$ rad/s where the phase curve sits near $-116^\circ$, and a phase crossover near $\omega \approx 10$ rad/s where the magnitude curve is near $-21$ dB. Both readings point at the same option. The graph is coarse enough that the honest statement is “approximately”, exactly as the question words it.

(3) Question 2. The pole-zero map prints its two crosses to the right of the imaginary axis, at $s = 1 \pm j1$, and its two circles on the negative real axis at $-2$ and $-3$. This is the whole point of the question, because an open-loop plant with two right-half-plane poles is unstable until enough gain is applied, which is the opposite of the usual root-locus habit.

Question 3: Partial fractions and the inverse Laplace transform (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 strictly proper rational function with three distinct real poles, one of which is repeated.

Given data
Quantity	Value
Function	$F(s) = 4(s+2)\,/\,[\,s(s+1)(s+3)^{2}\,]$
Simple poles	$s = 0$, $s = -1$
Repeated pole	$s = -3$, multiplicity 2
Finite zero	$s = -2$
Relative degree	$4 - 1 = 3$

Find. (a) the partial-fraction expansion, and (b) $f(t)$ for $t \ge 0$.

Approach. Residues at the simple poles come from the cover-up rule; the repeated pole needs both the cover-up value and its derivative; the time function then follows term by term from the inverse-transform table appended to the paper.

Part (a) — write down the form of the expansion. A double pole contributes two terms, one for each power, so $$F(s) = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+3} + \frac{D}{(s+3)^{2}}$$ There is no polynomial part because $F$ is strictly proper.
Residue at the pole at the origin. Multiply by $s$ and evaluate at $s = 0$: $$A = \left.\frac{4(s+2)}{(s+1)(s+3)^{2}}\right|_{s=0} = \frac{4(2)}{(1)(9)} = \boxed{\frac{8}{9} = 0.8889}$$
Residue at $s = -1$. Multiply by $(s+1)$ and evaluate at $s = -1$: $$B = \left.\frac{4(s+2)}{s(s+3)^{2}}\right|_{s=-1} = \frac{4(1)}{(-1)(4)} = \boxed{-1}$$
Highest-power coefficient of the repeated pole. The cover-up rule applies directly to the $(s+3)^{2}$ term: $$D = \left.(s+3)^{2}F(s)\right|_{s=-3} = \left.\frac{4(s+2)}{s(s+1)}\right|_{s=-3} = \frac{4(-1)}{(-3)(-2)} = \boxed{-\frac{2}{3} = -0.6667}$$
Lower-power coefficient of the repeated pole. This one needs a derivative, because covering up $(s+3)^{2}$ and setting $s = -3$ would divide by zero: $$C = \left.\frac{d}{ds}\!\left[(s+3)^{2}F(s)\right]\right|_{s=-3} = \left.\frac{d}{ds}\!\left[\frac{4(s+2)}{s^{2}+s}\right]\right|_{s=-3}$$ $$= \left.\frac{4\left(s^{2}+s\right) - 4(s+2)(2s+1)}{\left(s^{2}+s\right)^{2}}\right|_{s=-3} = \frac{4(6) - 4(-1)(-5)}{6^{2}} = \frac{24-20}{36} = \boxed{\frac{1}{9} = 0.1111}$$
Check the expansion before using it. Because $F$ falls off as $4/s^{3}$, the coefficients of all the simple $1/(s+a)$ terms must sum to zero: $A+B+C = \tfrac{8}{9} - 1 + \tfrac{1}{9} = 0$. Recombining the four terms over a common denominator reproduces $4(s+2)/[s(s+1)(s+3)^{2}]$ identically, so $$\boxed{F(s) = \frac{8/9}{s} - \frac{1}{s+1} + \frac{1/9}{s+3} - \frac{2/3}{(s+3)^{2}}}$$
Part (b) — invert term by term. The table appended to the paper gives $\mathcal{L}^{-1}\{A/(s+\alpha)\} = Ae^{-\alpha t}$ and $\mathcal{L}^{-1}\{A/(s+\alpha)^{n+1}\} = At^{n}e^{-\alpha t}/n!$. With $n = 1$ the factorial is 1, so the double-pole term simply picks up a factor $t$: $$\boxed{f(t) = \frac{8}{9} - e^{-t} + \frac{1}{9}e^{-3t} - \frac{2}{3}\,t\,e^{-3t}, \qquad t \ge 0}$$
Verify the answer at both ends of the time axis. At $t = 0$ the expression gives $\tfrac{8}{9} - 1 + \tfrac{1}{9} - 0 = 0$, which matches the initial-value theorem because $\lim_{s\to\infty} sF(s) = 0$. As $t \to \infty$ the three exponential terms vanish and $f(\infty) = 8/9 = 0.8889$, which matches the final-value theorem $\lim_{s\to 0} sF(s) = 4(2)/[(1)(9)] = 8/9$. The $t\,e^{-3t}$ term peaks at $t = 1/3$ s and is negligible after about a second, so the visible dynamics are dominated by the slow $e^{-t}$ mode.

Question 3(b): the inverse transform. The response starts from zero, rises with the $e^{-t}$ mode and settles on the final value $8/9 = 0.889$.

Question 3 — results
Part	Quantity	Result
(a)	$A$ (pole at $s = 0$)	$8/9 = 0.8889$
(a)	$B$ (pole at $s = -1$)	$-1$
(a)	$C$ (term $1/(s+3)$)	$1/9 = 0.1111$
(a)	$D$ (term $1/(s+3)^{2}$)	$-2/3 = -0.6667$
(b)	Inverse transform	$f(t) = \tfrac{8}{9} - e^{-t} + \tfrac{1}{9}e^{-3t} - \tfrac{2}{3}te^{-3t}$
check	$f(0)$ and $f(\infty)$	$0$ and $8/9 = 0.8889$

Topic. Partial-fraction expansion with a repeated real pole, and inversion through the standard transform table.

Key relations. Simple pole: $A_i = \left.(s+p_i)F(s)\right|_{s=-p_i}$. Repeated pole of multiplicity $r$: $A_{r-k} = \dfrac{1}{k!}\left.\dfrac{d^{k}}{ds^{k}}\left[(s+p)^{r}F(s)\right]\right|_{s=-p}$. Inverse transforms $1/(s+\alpha) \to e^{-\alpha t}$ and $(s+\alpha)^{-(n+1)} \to t^{n}e^{-\alpha t}/n!$. Initial and final value theorems $f(0^{+}) = \lim_{s\to\infty}sF(s)$, $f(\infty) = \lim_{s\to 0}sF(s)$.

Why this works. Every strictly proper rational function is a finite sum of elementary transforms, each of which has a known inverse; the residue formulas are just Taylor coefficients of the analytic part of $F$ around each pole. The derivative in the repeated-pole formula is what recovers the coefficient hidden underneath the highest power.

Common pitfall. Applying the plain cover-up rule to the $1/(s+3)$ term of a double pole, which silently returns the wrong number; and dropping the $1/n!$ in the $t^{n}e^{-\alpha t}$ inverse, which matters from the triple pole onward. The $A+B+C = 0$ test costs one line and catches both.

Source. Nise, Control Systems Engineering, 8th ed., §2.2 (partial-fraction expansion, Case 3: repeated real roots); Ogata, Modern Control Engineering, 5th ed., §2.4; the inverse-transform table printed on page 22 of this paper.

← Question 2 Next: Question 4 →

Part	Quantity	Result
(a)	\(A\) (pole at \(s = 0\))	\(8/9 = 0.8889\)
(a)	\(B\) (pole at \(s = -1\))	\(-1\)
(a)	\(C\) (term \(1/(s+3)\))	\(1/9 = 0.1111\)
(a)	\(D\) (term \(1/(s+3)^{2}\))	\(-2/3 = -0.6667\)
(b)	Inverse transform	\(f(t) = \tfrac{8}{9} - e^{-t} + \tfrac{1}{9}e^{-3t} - \tfrac{2}{3}te^{-3t}\)
check	\(f(0)\) and \(f(\infty)\)	\(0\) and \(8/9 = 0.8889\)

Quantity	Value
Function	\(F(s) = 4(s+2)\,/\,[\,s(s+1)(s+3)^{2}\,]\)
Simple poles	\(s = 0\), \(s = -1\)
Repeated pole	\(s = -3\), multiplicity 2
Finite zero	\(s = -2\)
Relative degree	\(4 - 1 = 3\)