Analytical Step Response by Partial Fractions: 17-Phys-B5 Systems and Control, May 2017, Q3

Question 3: Analytical Step Response by Partial Fractions (20 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. $G(s)=Y(s)/R(s)=\dfrac{7s(s+3)}{(s+0.5)(s^2+1.6s+16)}$, a stable LTI system (all poles in the open LHP: $s=-0.5$ and $s=-0.8\pm j3.919$) driven by a unit step $r(t)=1(t)$, i.e. $R(s)=1/s$.

Find. A closed-form $y(t)$ for $t\ge0$.

Approach. Form $Y(s)=G(s)R(s)=G(s)/s$; the explicit factor of $s$ in $G(s)$'s numerator cancels the $1/s$ from the step input, so $Y(s)$ has NO pole at the origin and the response decays to zero (no steady-state offset). Expand $Y(s)$ in partial fractions (one real pole, one complex-conjugate pair) and invert term-by-term using the Laplace table on page 2.

Step 1 — form $Y(s)$ and note the pole-at-origin cancellation. $$Y(s)=\frac{G(s)}{s}=\frac{7s(s+3)}{s(s+0.5)(s^2+1.6s+16)}=\frac{7(s+3)}{(s+0.5)(s^2+1.6s+16)}.$$ Because $G(s)$ itself carries an explicit zero at the origin, the step input's own pole at $s=0$ cancels exactly — the response is a pure transient with $y(\infty)=0$ (confirmed by the final-value theorem: $\lim_{s\to0}sY(s)=0$).
Step 2 — partial-fraction expansion. $$Y(s)=\frac{A}{s+0.5}+\frac{Bs+C}{s^2+1.6s+16}.$$ The residue at the real pole: $A=\left.(s+0.5)Y(s)\right|_{s=-0.5}=\dfrac{7(2.5)}{0.5^2-0.8+16} =\dfrac{350}{309}\approx1.133$. Matching the remaining numerator by subtraction gives $B=-\tfrac{350}{309}\approx-1.133$, $C=\tfrac{1778}{309}\approx5.754$: $$Y(s)=\frac{1.133}{s+0.5}+\frac{-1.133\,s+5.754}{s^2+1.6s+16}.$$
Step 3 — complete the square and invert term-by-term. $s^2+1.6s+16=(s+0.8)^2+3.919^2$ (so $\sigma=0.8$, $\omega_d=3.919$ rad/s). Writing $-1.133s+5.754=-1.133(s+0.8)+6.660$ isolates the $\cos$/$\sin$ Laplace-table pairs ($e^{-\sigma t}\cos\omega_dt \leftrightarrow \tfrac{s+\sigma}{(s+\sigma)^2+\omega_d^2}$, $e^{-\sigma t}\sin\omega_dt \leftrightarrow \tfrac{\omega_d}{(s+\sigma)^2+\omega_d^2}$): $$y(t)=1.133\,e^{-0.5t}+e^{-0.8t}\left[-1.133\cos(3.919t)+1.699\sin(3.919t)\right],\quad t\ge0.$$ Combining the sinusoidal terms into a single magnitude-phase form ($M=\sqrt{1.133^2+1.699^2}=2.042$, $\phi=123.7^\circ$): $$\boxed{y(t)=1.133\,e^{-0.5t}+2.042\,e^{-0.8t}\cos(3.919t-123.7^\circ)\ \text{V},\quad t\ge0.}$$

Final results — Question 3
Item	Result
Partial-fraction residues	$A=1.133$ (real pole $-0.5$); $B=-1.133,\ C=5.754$ (quadratic)
Complex-pole parameters	$\sigma=0.8$, $\omega_d=3.919$ rad/s
$y(t)$	$1.133e^{-0.5t}+2.042e^{-0.8t}\cos(3.919t-123.7^\circ)$ V
$y(\infty)$	$0$ (no pole at origin in $Y(s)$)

Topic: Analytical time response of an LTI system via Laplace-domain partial-fraction expansion.

Key relations: $Y(s)=G(s)R(s)$; residue at a simple real pole $p$: $A=(s-p)Y(s)|_{s=p}$; complex-pair inversion via completing the square, $\tfrac{Bs+C}{(s+\sigma)^2+\omega_d^2}\to e^{-\sigma t}[B\cos\omega_dt+\tfrac{C-B\sigma}{\omega_d}\sin\omega_dt]$.

Why this works: every strictly-proper rational $Y(s)$ with distinct poles decomposes uniquely into first- and second-order terms whose inverse transforms are read directly off a standard table — there is no need to simulate or numerically integrate when the transfer function and input are both rational functions of $s$.

Common pitfall: forgetting that a zero at the origin in $G(s)$ cancels the step input's own pole at $s=0$, and instead expecting (incorrectly) a nonzero steady-state term $D/s$ in the partial-fraction expansion.

Source: Nise, Control Systems Engineering, Ch. 2 (Laplace transforms and partial-fraction expansion); Ogata, Modern Control Engineering, Ch. 2.

Question 3 of 8: Analytical Step Response by Partial Fractions

Question 3: Analytical Step Response by Partial Fractions (20 marks)