Laplace (s-domain) step response: 22-Elec-A1 Circuits, May 2017, Q6

Question 6: Laplace (s-domain) step response [8 + 12]

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. Step source $15u(t)\text{ V}$, series $5\,\Omega$ to node $X$; at $X$ an inductor $L=2\text{ H}$ (current $i_L$ down) in parallel with a capacitor $C=0.3\text{ F}$ (voltage $v_c$, + up). Initial conditions $i_L(0^+)=0.5\text{ A}$, $v_c(0^+)=4\text{ V}$.

Find. the s-domain equivalent and the closed-form $v_c(t)$, $t\ge0$.

[Figure not reproduced: Figure 6 — redrawn. The output node $X$ carries $v_c$; $L$ and $C$ are in parallel across it, fed through $5\,\Omega$ from the step source. See the official exam paper.]

Approach. Replace each element by its s-domain model carrying its initial condition, write one nodal equation at $v_c$, solve for $V_c(s)$, then invert by completing the square to a decaying sinusoid pair.

(a) s-domain elements. Source $\to 15/s$; resistor $\to5\,\Omega$; inductor $\to$ impedance $sL=2s$ carrying its current so its branch current is $\dfrac{V_c}{2s}+\dfrac{i_L(0)}{s}=\dfrac{V_c}{2s}+\dfrac{0.5}{s}$; capacitor $\to$ current $sC\,V_c-C\,v_c(0)=0.3sV_c-1.2$.
(a) Nodal equation at $v_c$. KCL (currents leaving $=0$): $$\frac{V_c-15/s}{5}+\Big(\frac{V_c}{2s}+\frac{0.5}{s}\Big)+\big(0.3sV_c-1.2\big)=0.$$
(a) Solve for $V_c(s)$. Multiplying out and collecting $V_c$: $$\boxed{V_c(s)=\frac{12s+25}{3s^{2}+2s+5}.}$$ (Initial-value check: $\lim_{s\to\infty}sV_c=12/3=4\text{ V}=v_c(0)$; final-value $\lim_{s\to0}sV_c=0$.)
(b) Complete the square. $3s^2+2s+5=3\big[(s+\tfrac13)^2+\tfrac{14}{9}\big]$, so with $\omega_d=\dfrac{\sqrt{14}}{3}=1.247\text{ rad/s}$ and numerator $\tfrac{12s+25}{3}=4(s+\tfrac13)+7$: $$V_c(s)=\frac{4\,(s+\tfrac13)}{(s+\tfrac13)^2+\omega_d^{2}}+\frac{7}{(s+\tfrac13)^2+\omega_d^{2}}.$$
(b) Invert term-by-term. Using $e^{-\alpha t}\cos\omega_d t\leftrightarrow\frac{s+\alpha}{(s+\alpha)^2+\omega_d^2}$ and the matching sine pair (with $7/\omega_d=5.6125$): $$\boxed{v_c(t)=e^{-t/3}\big[\,4\cos(1.247t)+5.6125\sin(1.247t)\,\big]\text{ V},\qquad t\ge0.}$$
Sanity. $v_c(0)=4\text{ V}$ ✓ the response decays ($\alpha=\tfrac13$) toward $v_c(\infty)=0$ (the inductor shorts node $X$ at dc); a single-amplitude form is $v_c(t)=6.89\,e^{-t/3}\cos(1.247t-54.5^\circ)\text{ V}$.

Quantity	Value
$V_c(s)$	$\dfrac{12s+25}{3s^{2}+2s+5}$
Damping / damped freq.	$\alpha=\tfrac13\text{ s}^{-1},\ \omega_d=1.247\text{ rad/s}$
$v_c(t),\ t\ge0$	$e^{-t/3}[\,4\cos1.247t+5.6125\sin1.247t\,]\text{ V}$

Topic: Laplace-transform analysis of a second-order circuit with initial conditions.

Key relations: element models $sL$ with series $Li_L(0)$ (or parallel $i_L(0)/s$), $1/(sC)$ with the $v_c(0)/s$ / $Cv_c(0)$ sources; nodal KCL in $s$; inversion pairs $e^{-\alpha t}\cos\omega_dt$ and $e^{-\alpha t}\sin\omega_dt$ after completing the square.

Why this works: The Laplace transform turns the circuit’s differential equations into algebra and folds the initial conditions directly into the element models as extra sources, so a single KCL yields $V_c(s)$ with no separate homogeneous/particular bookkeeping. A quadratic denominator with complex roots inverts to a damped sinusoid; completing the square exposes the damping $\alpha$ and damped frequency $\omega_d$ that appear in the time answer. The real part of the roots ($-\tfrac13$) sets the decay; the imaginary part ($\pm1.247$) sets the ringing.

Common pitfall: dropping or mis-signing the initial-condition sources (the $0.5/s$ and $1.2$ terms), or mislabeling $\omega_d$; also forgetting that at dc the inductor shorts the output so $v_c(\infty)=0$, which the final-value theorem confirms.

Source: Sadiku, Fundamentals of Electric Circuits, §16.3–16.5 (circuit-element models in the s-domain and Laplace circuit analysis).

Question 6 of 6: Laplace (s-domain) step response

Question 6: Laplace (s-domain) step response [8 + 12]