Question

For the following circuit, sketch the outputs at $v_{oA}$and $v_{oB}$if the input is initially zero but is switched to a constant positive voltage at time zero.

Short answer

The $v_{oA}$will be $v_{oA}=-\frac{1}{R_i{C}_f}{\int }_0^t{v}_{i}dt$and$v_{oB}$will be $v_{oB}=\frac{R_f{C}_i}{R_i{C}_f}{v}_i$.

Step-by-step solution

Step 1. Given Information

Given two Op-Amps arranged in such a way that the first one is an integrator and the second one is a differentiator.

Step 2. Explanation

The first Op-Amp forms an integrator circuit. It integrates the variable signal

The output of the first part of the circuit is $v_{oA}$.

The output of the integrator is;

$v_{oA}=-\frac{1}{R_i{C}_f}{\int }_0^t{v}_{i}dt$

In the second part of the circuit, the input is $v_{oA}$and the output is localid="1645452359582" $v_{oB}$.

The output of the differentiator is;

$v_{oB}=-R_f{C}_i\frac{d{v}_{oA}}{dt}$

Step 3. Substitution

We have, the net output is;

$v_{oB}=-R_f{C}_i\frac{d{v}_{oA}}{dt}$

Substituting the value of $v_{oA}$in the above equation, we get;

localid="1645771299320" $$\begin{gathered} v_{oB}=-R_f{C}_i\frac{d}{dt}\left(-\frac{1}{R_i{C}_f}{\int }_0^t{v}_{i}dt\right) \\ =\frac{R_f{C}_i}{R_i{C}_f}{v}_i \end{gathered}$$