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schematic

simulate this circuit – Schematic created using CircuitLab

This is a simple RC circuit with an AC voltage source, \$E = V_0 \cos(\omega t)\$. In the sinusoidal state, \$Y_{\text{resistor}} = G\$ and \$Y_{\text{capacitor}} = C\omega j\$. The capacitor is not charged at all at \$t = 0\$. According to the KCL at each node, we get a system of equations that can be represented in terms of matrices \$\mathbf{Y}\mathbf{V} = \mathbf{I}\$:

\begin{gather} \begin{bmatrix} G & -G & 0 \\ -G & G + C\omega j & -C\omega j \\ 0 & -C\omega j & C\omega j \end{bmatrix} \begin{bmatrix} v_a \\ v_b \\ v_c \end{bmatrix} = \begin{bmatrix} -i_a \\ 0 \\ i_a \end{bmatrix}\text{.} \end{gather}

Setting node \$v_c\$ as the reference node (\$v_c = 0\$) and node \$v_a\$ as the super node (\$v_a = E\$) gives this matrix equation, which is the same as modified nodal analysis:

\begin{gather} \begin{bmatrix} G & -G & 1 \\ -G & G + C\omega j & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_a \\ v_b \\ i_a \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ E \end{bmatrix}\text{.} \end{gather}

Solving the equation (and substituting \$E\$ with \$V_0\angle 0\$) yields below:

\begin{gather} \begin{bmatrix} v_a \\ v_b \\ i_a \end{bmatrix} = \begin{bmatrix} V_0 \\ \frac{V_0}{G^2 + C^2{\omega}^2} \left( G^2 - GC{\omega}j \right) \\ \frac{-V_0}{G^2 + C^2{\omega}^2} \left( GC^2\omega^2 + G^2C{\omega} j \right) \end{bmatrix}\text{.} \end{gather}

Substituting \$V_0 = G = C = 1\$ and \$ s = \omega j\$, and solving for \$v_{resistor}\$ and \$v_{capacitor}\$ yields below:

$$ v_{\text{resistor}}(s) = v_a - v_b = 1 - \frac{1-\omega j}{1+\omega ^2} = 1 - \frac{1}{s+1}\text{,} $$ $$ v_{\text{capacitor}}(s) = v_b - v_c = v_b = \frac{1-\omega j}{1+\omega ^2} = \frac{1}{s+1}\text{.} $$

The problem begins when I apply the inverse Laplace transform and get the solution in \$t\$: $$ v_{\text{resistor}}(t) = \mathcal{L}^{-1} \left\{v_{\text{resistor}}(s)\right\} = \mathcal{L}^{-1} \left\{1 - \frac{1}{s+1}\right\} = \delta(t) - e^{-t} \text{,} $$ $$ v_{\text{capacitor}}(t) = \mathcal{L}^{-1} \left\{v_{\text{capacitor}}(s)\right\} = \mathcal{L}^{-1} \left\{\frac{1}{s+1}\right\} = e^{-t} \text{.} $$

I have several questions: Why do i get an exponentially decaying time function (is this correct), when there was an assumption made about the sinusoidal state which led to \$Y_{\text{capacitor}} = C\omega j\$ and \$ s = \omega j\$ (no exponential decay)? Also, I don't have an exact clue about what these two represent: \$v_{\text{resistor}}(t)\$ looks similar to \$ 1- e^{-t}\$, and this should be for the capacitor voltage, not resistor. I think the two are swapped, but I couldn't find a flaw in the logic. I'd greatly appreciate your help.

Reason for using modified nodal analysis I have very limited knowledge of AC electric circuit analysis/signal processing (Dirac delta function)/time and phase/frequency domain in Laplace transform. My ultimate goal is to perform modified nodal analysis on a system of very large and complex RC circuit network with a DC voltage source (hence multiple nodes) and obtain all nodal voltages with respect to time \$ t\$, probably using Python (Numpy and Scipy). I've been searching for the most efficient and scalable method for the past 3 weeks for the RC circuit analysis. I know how to obtain a differential equation for above circuit, but with many resistors and capacitors, it will lead to solving a differential matrix equation (preferably analytically), which would be my last option.

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    \$\begingroup\$ Fred's answer is solid, but another thing to think about is that your forcing function is not actually \$E(t)=V_0\sin\omega t\$, it's \$u(t)V_0\sin\omega t\$. \$\endgroup\$ Commented Sep 14, 2024 at 18:56
  • \$\begingroup\$ Keep in mind that \$\mathscr{L}\left\{x_t^{'}\right\}=sX_s-x_{_0}\$ ( \$\mathscr{L}\left\{x_t^{'}\right\}\ne sX_s\$.) Also, when you write "most efficient and scalable method" several things bother me. First the word most is quite extreme. Also most and efficient are quite malleable to one's goals of the moment. I might understand rote and scalable better, where rote implies something like "more readily constructed without complexity or thought." But even then... Finally, are you familiar with the more general methods of directed graph theory and how MNA evolves from it? \$\endgroup\$ Commented Sep 14, 2024 at 19:25
  • \$\begingroup\$ @ThePhoton but why is the necessity of adding \$u(t)\$ we are working with \$ \forall t \geq 0 \$ \$\endgroup\$ Commented Sep 14, 2024 at 22:45
  • \$\begingroup\$ @periblepsis I need to compute more than hundred thousands of nodal voltages. Now, I have done the exact one with a resistor-only circuit network with a DC voltage source by algorithmic MNA and sparse matrix representation of the G matrix in Python. The reason I'm doing this, a RC circuit with an AC voltage source with MNA, is because I'm testing my method with trival cases. Using MNA for more than 1000x1000 RC network will be efficient (I assume, if you know any other method; please help), but I just need to know, how I can get those nodal voltages in time t from this equations in phase. \$\endgroup\$ Commented Sep 15, 2024 at 1:37
  • \$\begingroup\$ @ThePhoton Okay, I substituted just \$E(t) = u(t) V_0\$ and actually \$ v_{\text{capacitor}}\$ leads to \$ \mathcal{L}^{-1} \left\{ \frac{1}{s(s+1)} \right\} = 1 - e^{-t} \$ which is the correct solution for the case of a DC voltage source. If \$E(t) = u(t) V_0 \sin{ \omega t}\$, then what should I substitute into the the \$E \$ in the matrix equation? I heard that it should be the phasor equivalent (which is why I put \$V_0\$ but it looks incorrect). Also, if a RC circuit has a DC voltage source where \$t\geq 0\$, can we use \$E(t) = u(t) V_0\$? I'm not sure if I get the difference. Thanks. \$\endgroup\$ Commented Sep 15, 2024 at 2:25

2 Answers 2

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  1. Nodal analysis should begin by choosing a reference node. If you choose one of the voltage source terminals as the reference node there remains only one independent node. And only one node equation. You have made this problem far more difficult than it needs to be.
  2. A Laplace transform solution yields the sum of the natural and forced responses. So it will (almost always) not be equal to the result of phasor analysis, which gives the forced response to a sinusoidal forcing function.
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    \$\begingroup\$ what is a forced response to sinusoidal forcing function? \$\endgroup\$ Commented Sep 14, 2024 at 22:41
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You can use Laplace to show that the output is related to the cosine driving the circuit.

KCL:

$$\begin{align*} C_1\frac{\text{d}}{\text{d}t}v_{cap}+\frac{v_{cap}}{R_1}&=\frac{V_0\cos\left(\omega\,t\right)}{R_1} \\\\ \frac{\text{d}}{\text{d}t}v_{cap}+\frac{v_{cap}}{R_1\,C_1}&=\frac{V_0\cos\left(\omega\,t\right)}{R_1\,C_1} \\\\ \mathscr{L}\left\{\frac{\text{d}}{\text{d}t}v_{cap}+\frac{v_{cap}}{R_1\,C_1}\right\}&=\mathscr{L}\left\{\frac{V_0\cos\left(\omega\,t\right)}{R_1\,C_1}\right\} \\\\ \mathscr{L}\left\{\frac{\text{d}}{\text{d}t}v_{cap}\right\}+\frac{1}{R_1\,C_1}\mathscr{L}\left\{v_{cap}\right\}&=\frac{V_0}{R_1\,C_1}\mathscr{L}\left\{\cos\left(\omega\,t\right)\right\} \\\\ sV_s+v_0+\frac{1}{R_1\,C_1}V_s&=\frac{V_0}{R_1\,C_1}\cdot\frac{s}{s^2+\omega^2} \\\\ \left(s+\frac{1}{R_1\,C_1}\right)V_s&=\frac{V_0}{R_1\,C_1}\cdot\frac{s}{s^2+\omega^2}-v_0, \text{where }v_0=0 \\\\ V_s &= \frac{V_0}{R_1\,C_1}\cdot\frac{s}{s^2+\omega^2}\cdot\frac1{s+\frac{1}{R_1\,C_1}} \\\\ V_s &= \frac{V_0}{\tau}\cdot\frac{s}{s^2+\omega^2}\cdot\frac1{s+\frac{1}{\tau}}, \text{where }\tau=R_1\,C_1 \\\\ \mathscr{L}^{-1}\left\{V_s\right\}&=\mathscr{L}^{-1}\left\{\frac{V_0}{\tau}\cdot\frac{s}{s^2+\omega^2}\cdot\frac1{s+\frac{1}{\tau}}\right\} \\\\ v_{cap(t)} &= \frac{V_0}{\tau^2\omega^2+1}\left[\cos\left(\omega\,t-\tan^{-1}\left(\tau\,\omega\right)\right)-\exp\left(-\frac{t}{\tau}\right)\right] \end{align*}$$

That's a cosine, with a phase shift, and an exponential component that gradually disappears, leaving only the attenuated cosine and phase shift as \$t\to\infty\$.

Compare that with your own work.

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