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Find V0(t) in this circuit assuming zero initial conditions.

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I solved it (refer image above) using frequency-domain analysis using Laplace and found the answer, but I am not able to get the same answer using time-domain analysis. How to solve it in thr time domain using transient analysis?

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  • \$\begingroup\$ Simulators use transient analysis and would easily solve this problem. Maybe have a quick word with Alfred? \$\endgroup\$
    – Andy aka
    Sep 12 at 13:57
  • \$\begingroup\$ @Andyaka Alfred's asking, which one? ;P \$\endgroup\$ Sep 12 at 14:02
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    \$\begingroup\$ Tell Alfred that I use micro-cap 12: zip file or link. \$\endgroup\$
    – Andy aka
    Sep 12 at 15:53

2 Answers 2

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Using nodal analysis at the \$\small V_x\$ node: \$\Sigma \small currents\:away\:from \:node =0 \$

\$\large\frac{(V_x -1)}{1}\: +\large\frac{1}{3}\frac{dV_x}{dt}+\large\frac{1}{5}\left(\small V_x +\large\frac{dV_x}{dt}+\frac{1}{3}\frac{d^2V_x}{dt^2}\right)\small=0\$

Simplifying,

\$\large\frac{d^2 V_x}{dt^2}\small+8\large\frac{dV_x}{dt}\small +18V_x=15 \$

Auxiliary equation:

\$\small m^2 +8m +18=0\$

\$\small m=-4\pm j\sqrt{2}\$

Complementary function:

\$\small V_{xcf}(t)=e^{-4t}(A cos\sqrt{2}t+Bsin\sqrt{2}t) \$

Supplement, in response to comment:

\$\small i_L=\large \frac{1}{5}\small (V_x-V_0)=\large \frac{1}{5}\small \left(V_x-\large\frac{d}{dt}\small i_L\right) =\large \frac{1}{5}\small \left(V_x-\large\frac{d}{dt}\small(i-i_C)\right) =\large \frac{1}{5}\small \left(V_x-\large\frac{di}{dt}+ \frac{di_C}{dt}\right)\$

where \$ i\$ is the source current, \$i_L \$ is the inductor current, and \$\small i_C\$ is the capacitor current.

Thus,

\$ i_L=\large \frac{1}{5}\small \left(V_x+\large\frac{dV_x}{dt}+\frac{1}{3} \frac{d^2V_x}{dt^2}\right)\$

Also, the other two currents are,

\$\small i=1-V_x \$, and \$ i_C=\large\frac{1}{3}\frac{dV_x}{dt}\$

Finally, node current balance,

\$i+i_C +i_L=0\$

gives,

\$\large\frac{d^2 V_x}{dt^2}\small+8\large\frac{dV_x}{dt}\small +18V_x=15 \$

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  • \$\begingroup\$ Could you care to explain how you got 1/5(Vx+dVx/dt...) that equation? \$\endgroup\$ Sep 14 at 10:26
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    \$\begingroup\$ See supplement. \$\endgroup\$
    – Chu
    Sep 14 at 13:33
  • \$\begingroup\$ Got it! Tysm dude. Cheers. \$\endgroup\$ Sep 15 at 14:03
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You've already used Kirchhoff for the Laplace domain, why not use it for the time domain, too? I'll use the loop:

$$\begin{align} &\begin{cases} 1\cdot i_1(t)+\dfrac{1}{\frac13}\int{[i_1(t)-i_2(t)]\mathrm{d}t}&=\theta(t) \\ \dfrac{1}{\frac13}\int{[i_1(t)-i_2(t)]\mathrm{d}t}&=5\cdot i_2(t)+1\cdot \dfrac{\mathrm{d}i_2(t)}{\mathrm{d}t} \end{cases} \\ \Rightarrow \\ &\begin{cases} 1\cdot \dfrac{\mathrm{d}i_1(t)}{\mathrm{d}t}+\dfrac{i_1-i_2}{\frac13}&=\delta(t) \\ \dfrac{i_1-i_2}{\frac13}&=5\cdot \dfrac{\mathrm{d}i_2(t)}{\mathrm{d}t}+1\cdot \dfrac{\mathrm{d}^2i_2(t)}{\mathrm{d}^2t} \end{cases} \tag{1} \\ \Rightarrow \\ i_2(t)&=\dfrac16-\dfrac13\mathrm{e}^{-4t}\left(\sqrt2\sin(\sqrt2 t)+\cos(\sqrt2 t)\right) \tag{2} \end{align}$$

\$i_1(t)\$ can be discarded. Since the voltage across the inductor is the derivative of the current through it:

$$\dfrac{\mathrm{d}i_2(t)}{\mathrm{d}t}=\dfrac{3}{\sqrt2}\mathrm{e}^{-4t}\sin(\sqrt2 t) \tag{3}$$

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