Impulse Response of Two Transfer Functions: 23-Mechatronics-A1 Systems Dynamics and Controls, Undated paper, Q2

Question 2: Impulse Response of Two Transfer Functions

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) $G(s)=100/[(s+2)(s+5)^2]$, a strictly-proper transfer function with a simple pole at $s=-2$ and a repeated pole at $s=-5$. (b) $G(s)=1/(0.04s^2+0.08s+1)$, a normalized 2nd-order transfer function.

Find. The impulse response $g(t)=\mathcal{L}^{-1}[G(s)]$ for each system (the output when the input is a unit impulse $\delta(t)$, since $\mathcal{L}[\delta(t)]=1$).

Approach. Expand $G(s)$ into partial fractions (part a) or standard underdamped 2nd-order form (part b), then invert term-by-term with the Laplace transform table supplied with the exam.

Part (a) — partial-fraction setup. $$\frac{100}{(s+2)(s+5)^2}=\frac{A}{s+2}+\frac{B}{s+5}+\frac{C}{(s+5)^2}$$
Part (a) — solve the residues. $$A=\left.\frac{100}{(s+5)^2}\right|_{s=-2}=\frac{100}{9}=11.11,\qquad C=\left.\frac{100}{s+2}\right|_{s=-5}=\frac{100}{-3}=-33.33$$ $$B=\left.\frac{d}{ds}\!\left[\frac{100}{s+2}\right]\right|_{s=-5} =\left.\frac{-100}{(s+2)^2}\right|_{s=-5}=\frac{-100}{9}=-11.11$$
Part (a) — invert term by term (table entries: $e^{-at} \leftrightarrow 1/(s+a)$, $te^{-at}\leftrightarrow 1/(s+a)^2$): $$\boxed{g_a(t)=11.11\,e^{-2t}-11.11\,e^{-5t}-33.33\,t\,e^{-5t}\ \ (t\ge 0)}$$
Part (b) — normalize to standard 2nd-order form. Dividing by $0.04$: $G(s)=25/(s^2+2s+25)$, so $\omega_n^2=25\Rightarrow\omega_n=5$ rad/s, $2\zeta\omega_n=2\Rightarrow\zeta=0.2$, and $\omega_d=\omega_n\sqrt{1-\zeta^2}=4.899$ rad/s.
Part (b) — apply the standard pair $\omega_n^2/(s^2+2\zeta\omega_ns+\omega_n^2)\leftrightarrow (\omega_n/\sqrt{1-\zeta^2})e^{-\zeta\omega_nt}\sin\omega_dt$: $$\boxed{g_b(t)=5.103\,e^{-t}\sin(4.899\,t)\ \ (t\ge 0)}$$

Part	Impulse response $g(t)$, $t\ge 0$
(a)	$11.11e^{-2t}-11.11e^{-5t}-33.33te^{-5t}$
(b)	$5.103\,e^{-t}\sin(4.899t)$

Concept & Theory

Topic: Inverse Laplace transform via partial fractions and the standard underdamped pair.

Key relations: a repeated real pole expands as $A/(s-p_1)+B/(s-p_2)+C/(s-p_2)^2$ with $C$ the direct residue and $B$ its derivative residue; the standard underdamped pair $\omega_n^2/(s^2+2\zeta\omega_ns+\omega_n^2)\leftrightarrow(\omega_n/\sqrt{1-\zeta^2}) e^{-\zeta\omega_nt}\sin\omega_dt$.

Why this works: the impulse response is the inverse transform of $G(s)$ itself, so any strictly-proper $G(s)$ decomposes into a sum of first- and second-order modes whose individual transforms are tabulated.

Common pitfall: forgetting the extra factor of $t$ that a repeated real pole introduces (using $Ce^{-5t}$ instead of the correct $Cte^{-5t}$), and forgetting to re-normalize a transfer function whose $s^2$ coefficient is not 1 before reading $\omega_n,\zeta$ off it.

Source: Ogata, Modern Control Engineering, Ch. 3 (Laplace transform); Nise, Control Systems Engineering, §4.4 (2nd-order systems).

Question 2 of 6: Impulse Response of Two Transfer Functions

Question 2: Impulse Response of Two Transfer Functions