22-Mec-A3 System Analysis and Control · Undated paper

Question 2 of 6: Impulse response of two transfer functions

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

Question 2: Impulse response of two transfer functions (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: (a) a third-order all-pole system with a simple pole at $s=-2$ and a repeated pole at $s=-5$; (b) a second-order system written with a non-unity leading coefficient, $0.04s^2+0.08s+1$.

Find. $g(t)=\mathcal{L}^{-1}\{G(s)\}$ in each case, in closed form.

Approach. The Laplace transform of a unit impulse is $1$, so $C(s)=G(s)\cdot 1=G(s)$ and the impulse response is simply the inverse transform of the transfer function itself — no differential equation need be solved. Part (a) is a partial-fraction expansion with a repeated pole; part (b) is normalised to standard second-order form before the transform pair is applied.

1. Set up the expansion for (a). A double pole needs two terms, so $$G(s)=\frac{100}{(s+2)(s+5)^2}=\frac{A}{s+2}+\frac{B}{s+5}+\frac{C}{(s+5)^2}$$
2. Evaluate the residues. The simple-pole residue and the highest-order repeated term follow from direct substitution, while $B$ requires one derivative: $$A=\left.\frac{100}{(s+5)^2}\right|_{s=-2}=\frac{100}{9},\qquad C=\left.\frac{100}{s+2}\right|_{s=-5}=-\frac{100}{3}$$ and $$B=\left.\frac{d}{ds}\!\left(\frac{100}{s+2}\right)\right|_{s=-5}=\left.\frac{-100}{(s+2)^2}\right|_{s=-5}=-\frac{100}{9}$$
3. Check the residues before inverting. The relative degree of $G$ is three, so $g(0^+)$ must vanish; that requires $A+B=0$, and indeed $100/9-100/9=0$. This one-line test catches almost every algebraic slip in a repeated-pole expansion.
4. Invert term by term. Using pairs (4) and (5) of the supplied table, $1/(s+a)\to e^{-at}$ and $1/(s+a)^2\to te^{-at}$: $$\boxed{g_a(t)=\frac{100}{9}e^{-2t}-\frac{100}{9}e^{-5t}-\frac{100}{3}\,t\,e^{-5t}\quad (t\ge 0)}$$

The response starts from zero, rises to a peak of $g=1.817$ at $t=0.540$ s and then decays with the slow $e^{-2t}$ mode, the $t e^{-5t}$ term contributing only during the first half second. Turning to part (b), the leading coefficient must be removed before the standard form can be recognised.

5. Normalise (b) to monic form. Dividing numerator and denominator by $0.04$: $$G(s)=\frac{1}{0.04s^2+0.08s+1}=\frac{25}{s^2+2s+25}$$ Comparing with $\omega_n^2/(s^2+2\zeta\omega_n s+\omega_n^2)$ gives $\omega_n=5$ rad/s and $2\zeta\omega_n=2$, hence $\zeta=0.2$. The damped natural frequency is $\omega_d=\omega_n\sqrt{1-\zeta^2}=5\sqrt{0.96}=4.899$ rad/s.
6. Match the sine transform pair. Completing the square puts $G$ in the form of pair (8), $\omega/[(s+a)^2+\omega^2]\to e^{-at}\sin\omega t$, with $a=1$ and $\omega=\omega_d$: $$G(s)=\frac{25}{(s+1)^2+4.899^2}=\frac{25}{4.899}\cdot\frac{4.899}{(s+1)^2+4.899^2}$$ so that $$\boxed{g_b(t)=5.103\,e^{-t}\sin(4.899\,t)\quad (t\ge 0)}$$
7. Sanity-check the second-order result. At $t=0$ the response is zero, as the relative degree of two demands. The envelope decays as $e^{-\zeta\omega_n t}=e^{-t}$, so the $2\%$ settling time is $4/(\zeta\omega_n)=4$ s, and successive zero crossings are spaced by $\pi/\omega_d=0.641$ s. With $\zeta=0.2$ the system is lightly damped and the impulse response rings for roughly six cycles.