Question 6 of 6: Second-order transient via the Laplace transform

Question 6: Second-order transient via the Laplace transform [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. $V_s=12\text{ V}$, $R=5\,\Omega$, $L=2\text{ H}$, $C=1\text{ F}$; $V_c(0)=4\text{ V}$, $i_L(0)=1\text{ A}$. For $t\ge0$ the source feeds $R$ into a parallel $L$–$C$ pair; $V_c$ is the voltage at that node.

Find. The s-domain model and the capacitor voltage $V_c(t)$.

[Figure not reproduced: Figure 6 — redrawn ($t\ge0$). The step source $V_s/s$ drives $R$; the capacitor becomes $1/(sC)$ with initial charge, and the inductor $sL$ with its initial current. See the official exam paper.]

(a) Laplace equivalent. Replace the step source by $V_s/s=12/s$; the resistor stays $R=5$. Model the capacitor as an admittance $sC$ in parallel with an initial-condition current source $C\,V_c(0)=4\text{ A}$, and the inductor as an admittance $1/(sL)$ in parallel with an initial-condition current source $i_L(0)/s=1/s$. (Equivalently, series-source forms: $V_c(0)/s$ in series with $1/(sC)$, and $L\,i_L(0)=2\text{ V}$ in series with $sL$.)

Approach. Write one nodal equation at the $V_c$ node in the s-domain, solve for $V_c(s)$, then invert.

1. Nodal equation at the capacitor node. Current in through $R$ equals current into $C$ plus current into $L$: $$\frac{12/s-V}{5}=\big(sV-4\big)+\Big(\frac{V}{2s}+\frac{1}{s}\Big),\qquad V\equiv V_c(s).$$
2. Solve for $V_c(s)$. Multiplying by $10s$ and collecting terms, $$V\,(10s^{2}+2s+5)=40s+14\;\Rightarrow\;\boxed{V_c(s)=\dfrac{40s+14}{10s^{2}+2s+5}}.$$
3. Locate the poles. Dividing through by 10, $s^{2}+0.2s+0.5=0$ gives $$s=-0.1\pm j0.7,$$ an underdamped pair ($\alpha=0.1\text{ Np/s}$, $\omega_d=0.7\text{ rad/s}$).
4. Match to the standard transform pairs. Writing the numerator about the shifted pole, $$V_c(s)=\frac{4s+1.4}{(s+0.1)^{2}+0.7^{2}}=\frac{4(s+0.1)}{(s+0.1)^2+0.7^2}+\frac{1.0}{(s+0.1)^2+0.7^2}.$$
5. Invert. Using $\dfrac{s+a}{(s+a)^2+\omega^2}\!\to\! e^{-at}\cos\omega t$ and $\dfrac{\omega}{(s+a)^2+\omega^2}\!\to\! e^{-at}\sin\omega t$, $$\boxed{V_c(t)=e^{-0.1t}\big[\,4\cos0.7t+1.429\sin0.7t\,\big]\text{ V},\quad t\ge0}$$ or, in single-sinusoid form, $V_c(t)=4.248\,e^{-0.1t}\cos(0.7t-19.65^\circ)\text{ V}.$
6. Sanity checks. At $t=0$: $V_c=4\text{ V}$ (matches $V_c(0)$). Initial slope $\dot V_c(0^+)=-0.1(4)+1.429(0.7)=0.6\text{ V/s}$, i.e. $i_C(0^+)=C\dot V_c=0.6\text{ A}=\frac{12-4}{5}-1$ ✓. As $t\to\infty$, $V_c\to0$ (the inductor short-circuits the node at DC) ✓.