Laplace-domain step-like response: 22-Elec-A1 Circuits, December 2016, Q6

Question 6: Laplace-domain step-like response [10 + 10]

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. Source $50e^{-2t}\text{ V}$ ($+$ up) in series with $2\,\Omega$, feeding the parallel combination of a $1\text{ H}$ inductor (with $i_L(0^-)=0$) and a $5\,\Omega$ resistor; the output $v_o$ is the voltage across the $5\,\Omega$.

Element	Laplace (s-domain)
Source $50e^{-2t}$	$\dfrac{50}{s+2}$
Series $2\,\Omega$	$2$
$1\text{ H}$, $i_L(0^-)=0$	$sL=s$ (no IC source)
$5\,\Omega$	$5$

Find. $V_o(s)$ and its inverse $v_o(t)$ for $t\ge0$.

[Figure not reproduced: Figure 6 — redrawn. In the s-domain the source is $50/(s+2)$, the inductor is an impedance $s$ (zero initial current means no series/parallel IC source), and $v_o$ is taken across the $5\,\Omega$. See the official exam paper.]

Approach. Transform each element, form the parallel impedance of $sL\|5$, apply a voltage divider against the series $2\,\Omega$, then invert by partial fractions.

(a) s-domain network. Source $V_s(s)=\dfrac{50}{s+2}$; series $2\,\Omega$; load $Z_p=sL\,\|\,5=\dfrac{5s}{s+5}$ (the $1\text{ H}$ becomes $s$, no initial-condition source because $i_L(0^-)=0$).
Voltage divider for $V_o$. $$V_o(s)=V_s(s)\,\frac{Z_p}{2+Z_p}=\frac{50}{s+2}\cdot\frac{5s/(s+5)}{2+5s/(s+5)}=\frac{50}{s+2}\cdot\frac{5s}{7s+10}.$$
(a) result. $$\boxed{V_o(s)=\frac{250\,s}{(s+2)(7s+10)}}.$$
(b) Partial fractions. Write $\dfrac{250s}{(s+2)(7s+10)}=\dfrac{A}{s+2}+\dfrac{C}{7s+10}$. Residues: $A=\dfrac{250(-2)}{7(-2)+10}=125$ and $C=\dfrac{250(-10/7)}{(-10/7)+2}=-625$.
Rewrite the second term. $\dfrac{-625}{7s+10}=\dfrac{-625/7}{s+10/7}$, so $$V_o(s)=\frac{125}{s+2}-\frac{625/7}{s+10/7}.$$
Inverse transform. Using $\mathcal{L}^{-1}\{1/(s+a)\}=e^{-at}u(t)$, $$\boxed{v_o(t)=125\,e^{-2t}-89.29\,e^{-1.4286\,t}\ \text{V},\quad t\ge0.}$$ Check: $v_o(0^+)=125-89.29=35.71\text{ V}=50\cdot\dfrac{5}{2+5}$ (inductor open at $t=0^+$) and $v_o(\infty)=0$ (inductor shorts, source decays) ✓.

Quantity	Value
$V_o(s)$	$\dfrac{250s}{(s+2)(7s+10)}$
$v_o(0^+)$	$35.71\text{ V}$
$v_o(t)$, $t\ge0$	$125e^{-2t}-89.29e^{-1.4286t}\text{ V}$

Topic: Laplace (s-domain) analysis of a source-driven RL circuit.

Key relations: $L\to sL$ (series IC source $Li_L(0)$, zero here), $C\to1/sC$; divider $V_o=V_s\,Z_p/(Z_{series}+Z_p)$; inversion by partial fractions and the $1/(s+a)$ pair.

Why this works: Because the inductor starts with zero current, no initial-condition source appears, and the network is a plain divider of impedances in $s$. The excitation $50e^{-2t}$ contributes a pole at $s=-2$, and the RL network contributes the natural pole at $s=-10/7$; each maps to a decaying exponential. The two initial/final checks confirm the algebra without re-solving.

Common pitfall: forgetting that $i_L(0^-)=0$ removes the IC source (a spurious $Li_L(0)$ term derails the divider), or mis-handling the $7s+10$ factor when reading off the exponential rate ($10/7\approx1.4286$, not $10$).

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-domain step-like response

Question 6: Laplace-domain step-like response [10 + 10]