Band Stop Filter Charging Transient

Let the current through the inductor be $I_1$ and that through the capacitor be $I_2$ , the charge on the capacitor be $Q$ and output voltage be $V$ . The circuit equations are:

$-V_S + L\dot{I}_1 + (I_1+I_2)R=0$ $L\dot{I}_1=\frac{Q}{C}$ $V= (I_1+I_2)R$ $I_1(0) = I_2(0) = Q(0) = 0$

Numerically integrating the above equations gives the required answer of $\approx\boxed{10.002}$ .

I am pasting my solution to the frequency domain problem without revealing the answer below:

The circuit has been analysed in steady-state. Let the magnitude of input voltage be $V$ and its frequency be $\omega$ . The impedance due to the inductor is $Z_1=j\omega L$ and that due to the capacitor is $Z_2=\frac{-1}{j\omega C}$ . Let the steady-state current through the inductor be $I_1$ and that through the capacitor be $I_2$ . The circuit equations are according to Kirchoff's laws as follows.

$Z_1I_1 + (I_1 + I_2)R = V$ $Z_1I_1 = Z_2I_2$

Rearranging these equations gives:

$\left[\begin{matrix}Z_1+R&R\\Z_1&-Z_2\end{matrix}\right]\left[\begin{matrix}I_1\\I_2\end{matrix}\right]=\left[\begin{matrix}V\\0\end{matrix}\right]$

From here, the currents can be computed as such:

$\left[\begin{matrix}I_1\\I_2\end{matrix}\right]=\left[\begin{matrix}Z_1+R&R\\Z_1&-Z_2\end{matrix}\right]^{-1}\left[\begin{matrix}V\\0\end{matrix}\right]$

The matrix is denoted as A where: $A = \left[\begin{matrix}Z_1+R&R\\Z_1&-Z_2\end{matrix}\right]$

The output voltage is: $V_o = R(I_1 + I_2)$

The filter gain is, therefore, the ratio of the magnitude of the complex numbers:

$G(\omega)= \left\lvert \frac{V_o}{V} \right\rvert$

Now, keeping these steps in mind, the=is problem is solved numerically. A plot of the magnitude response can be studied below:

One can see a sharp drop in magnitude ratio at a frequency of $\omega = 1$ . This can be verified by a time-domain simulation where the output voltage indeed converges to zero.

For the computation of the required integral, simulation code is attached below:

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clear all
clc

% Initialisation:
R = 3;
L = 1;
C = 1;

% Frequency range definition:
dw = 1e-4;
w  = dw:dw:5;
N  = length(w);

% Initial value of required integral:
S  = 0;

% Frequency ranges from dw (instead of zero) to 5 rad/sec in steps of dw to prevent any singularities.

% Here, j = sqrt(-1)..

for k = 1:length(w)

    % Computing the matrix A:
    Z1 = j*w(k)*L;
    Z2 = -j/(w(k)*C);
    A  = [Z1+R R;Z1 -Z2];

    % Computing the current column matrix and assuming that the amplitude of input voltage is unity:
    I  = inv(A)*[1;0];

    % Computation of output voltage magnitude which is also the Magnitude ratio in this case:
    G(k) = abs(R*(I(1)+I(2)));

    % Numerical integration of the required function:
    f  = 1 - G(k);
    S  = S + dw*f;
end

% Plotting frequency response:
plot(w,G,'Linewidth',1.5)
grid on
xlabel('Frequency [rad/sec]')
ylabel('G(\omega)')
title('Frequency Response - Magnitude Response')