Step response by Laplace transform: 22-Elec-A1 Circuits, May 2014, Q6

Question 6: Step response by Laplace transform [6 + 6 + 8]

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.

Element	Value
Input	$V_{in}=5\,u(t)$ V
Series resistor	$2\ \Omega$
Shunt capacitor	$0.2$ F
Series (branch) resistor	$4\ \Omega$
Inductor ($V_{out}$ across it)	$0.5$ H
Initial conditions	$i_L(0^{+})=0,\ v_c(0^{+})=0$

Find. The $s$-domain circuit, $V_{out}(s)$, and $V_{out}(t)$.

Figure 6 — Time-domain network: $2\ \Omega$ in series feeding a $0.2$ F capacitor in parallel with a $4\ \Omega + 0.5$ H branch; $v_{out}$ is across the inductor.

Approach. Because both initial conditions are zero, transform each element to its impedance ($R\to R$, $C\to 1/sC$, $L\to sL$) with no added sources, then use two voltage dividers to reach $V_{out}(s)$ and invert.

(i) Laplace-transformed circuit. With zero ICs: $V_{in}(s)=5/s$, the $2\ \Omega$ stays, the capacitor becomes $1/(0.2s)=5/s$, the $4\ \Omega$ stays and the inductor becomes $0.5s$ — no initial-condition sources appear.

Figure 6($s$) — The $s$-domain equivalent. Node $V_x(s)$ is the top of the parallel section; $V_{out}(s)$ is across the $0.5s$ inductor impedance.

Reduce the parallel section. The capacitor branch $Z_C=5/s$ is in parallel with the RL branch $Z_{RL}=0.5s+4$: $$Z_{p}=\frac{Z_C Z_{RL}}{Z_C+Z_{RL}}=\frac{(5/s)(0.5s+4)}{(5/s)+0.5s+4}=\frac{2.5s+20}{0.5s^{2}+4s+5}.$$
First divider — node $V_x$. Across the $2\ \Omega$ and $Z_p$: $$V_x(s)=\frac{5}{s}\cdot\frac{Z_p}{2+Z_p}=\frac{5(2.5s+20)}{s\,(s^{2}+10.5s+30)}.$$
(ii) Second divider — output across the inductor. Within the RL branch the inductor takes the fraction $0.5s/(0.5s+4)=s/(s+8)$ of $V_x$; since $2.5s+20=2.5(s+8)$ the $(s+8)$ factors cancel: $$V_{out}(s)=V_x(s)\cdot\frac{0.5s}{0.5s+4}=\boxed{\dfrac{12.5}{s^{2}+10.5s+30}}.$$
Pole location. $s^{2}+10.5s+30=0\Rightarrow s=-5.25\pm j1.561$ (complex → under-damped). Complete the square: $$V_{out}(s)=\frac{12.5}{(s+5.25)^{2}+(1.561)^{2}}.$$
(iii) Invert. Using $\mathcal{L}^{-1}\{\omega_d/[(s+a)^{2}+\omega_d^{2}]\}=e^{-at}\sin\omega_d t$ with $a=5.25$, $\omega_d=1.561$: $$V_{out}(t)=\frac{12.5}{1.561}\,e^{-5.25t}\sin(1.561\,t)=\boxed{8.01\,e^{-5.25t}\sin(1.561\,t)\ \text{V}},\quad t\ge0.$$

Quantity	Result
$V_{out}(s)$	$\dfrac{12.5}{s^{2}+10.5s+30}$
Poles	$-5.25\pm j1.561$ (under-damped)
$V_{out}(t)$	$\boxed{8.01\,e^{-5.25t}\sin(1.561t)\ \text{V}}$
Checks	$V_{out}(0^{+})=0$, $V_{out}(\infty)=0$ (inductor short at DC)

Concept & Theory

Topic. Transient (step) analysis of a second-order network by the Laplace-transform / impedance-divider method.

Key relations. Zero-IC element impedances $R,\ 1/sC,\ sL$; voltage division in the $s$-domain; inverse pair $\dfrac{\omega_d}{(s+a)^{2}+\omega_d^{2}}\leftrightarrow e^{-at}\sin\omega_d t$.

Why this works. With no stored energy at $t=0$, the transformed circuit is purely algebraic — no impulse sources for $L$ or $C$ — so the response is just the product of the input transform and the network’s divider ratios. The denominator $s^{2}+10.5s+30$ is the circuit’s characteristic polynomial; complex roots give a damped sinusoid.

Common pitfall. Adding initial-condition sources when the ICs are zero, or mismatching the numerator when completing the square (the sine transform needs $\omega_d$ in the numerator, hence the $12.5/\omega_d$ scale factor).

Source. Sadiku §15.3–15.5 (circuit element models in the $s$-domain and Laplace circuit analysis).

Question 6 of 6: Step response by Laplace transform

Question 6: Step response by Laplace transform [6 + 6 + 8]