Op Amp Filter Transient

Consider the diagram as labelled above. The currents and junctions have been assigned a name.

Let the voltages at junctions $P$ , $Q$ , $R$ , $S$ and $T$ respectively be $V_P$ , $V_Q$ , $V_R$ , $V_S$ , $V_T = 0$ . The charges on the capacitors $C_1$ and $C_2$ are $Q_1$ and $Q_2$ respectively. Therefore:

$I_1 = \dot{Q}_1$ $I_2 = \dot{Q}_2$

As per the circuit diagram above, the following equations can be generated.

$-V_{\mathrm{in}} + IR_1 + I_2R_2 + \frac{Q_2}{C_2} = 0$ $V_{\mathrm{in}} -V_P = IR_1$ $V_P - V_R = \frac{Q_1}{C_1}$ $V_Q - V_T = \frac{Q_2}{C_2} \implies V_Q - 0 = \frac{Q_2}{C_2}$

Considering an ideal operational amplifier, I have had no prior experience of working with them, so I had to read up a bit. What I understood was that the input current $I_{\mathrm{in}} = 0$ for the ideal case and that $V_Q = V_S$ .

Using these bits of information gives us the following equations:

$V_R = V_S = V_{\mathrm{out}} = V_Q$ $\implies V_Q = V_{\mathrm{out}} = \frac{Q_2}{C_2}$

Moreover, Kirchoff's current law dictates that:

$I = I_1 + I_2$

Manipulating, rearranging and simplifying the above equations leads to the following system of equations:

$-V_{\mathrm{in}} + I + I_2 + Q_2 =0$ $-V_{\mathrm{in}} - I - V_{\mathrm{out}} =Q_1$ $I_1 = \dot{Q}_1$ $I_2 = \dot{Q}_2$ $I = I_1 + I_2$ $V_{\mathrm{out}} = Q_2$

The above system of equations can be recast into a state-space form and is solved numerically to obtain the output voltage as a function of the input voltage. The required integral is also computed numerically. The numerical details are left out and I will attach if requested. This solution lays out the outline of the problem-solving approach.

The answer is:

$\boxed{\int_{0}^{20} \lvert 1 - V_{\mathrm{out}} \rvert dt \approx 2.933}$

The plot showing input and output voltage is as follows:

It can be seen that the transients die out somewhere between five and six seconds after start of the simulation, and the amplitude of the steady-state output is highly attenuated. I suppose there might be some kind of electronic application where such a filter proves to be useful. Would like to know more.

@Karan Chatrath has provided a nice explanation already. I will show my approach as well, since there are some minor stylistic differences. I also want to point out that the filter obeys the principle of superposition. Since it is a low-pass filter, the DC component passes through without attenutation, while the AC portion is greatly reduced. As it turns out, the frequency of the AC portion is three times the cutoff frequency of this filter. This is known as a Sallen-Key topology.

The state-space model is:

$V = V_{C2} + V_{C1} \,\,* \\ I_{R2} = \frac{V - V_{C2}}{R_2} \\ I_{R1} = \frac{V_{in} - V}{R_1} \\ I_{C2} = I_{R2} \,\,** \\ I_{C1} = I_{R1} - I_{R2} \\ I_{C1} = C_1 \dot{V}_{C1} \\ I_{C2} = C_2 \dot{V}_{C2} \\ * \text{from op amp voltage property} \\ ** \text{from op amp current property}$