This question was previously asked in

IPRC (ISRO) Technician B (Electronics) 2018 Official Paper (Held on 22 April 2018)

Option 2 : Rectangular

__Concept:__

1. The basic **operational amplifier differentiator circuit **produces an output signal which is the first derivative of the input signal.

2. The input signal to the differentiator is applied to the capacitor. The capacitor blocks any DC content so there is no current flow to the amplifier summing point, resulting in zero output voltage.

3. The capacitor only allows AC-type input voltage changes to pass through and whose frequency is dependent on the rate of change of the input signal.

__Calculation:__

The circuit is drawn as:

Applying KCL, we can write:

\(\frac{{{V_i} - 0}}{{\frac{1}{{sC}}}} + \frac{{{V_0} - 0}}{R} = 0\)

\(\frac{{{V_0}}}{R} = - sC{V_i}\)

V0 = -sRC Vi

Taking the inverse Laplace transform of the above, we get:

\({V_0}\left( t \right) = - RC \cdot \frac{{d{V_i}}}{{dL}}\)

This indicates a differentiator circuit.

So now if we give a **sawtooth wave as input (Vi)** to this circuit then it will give a rectangular wave as output after differentiating the input.

Hence option(2) is the correct answer.

__Important Points__

Now when we interchange the position of capacitor and feedback resistor in the given circuit then the circuit will be known as an Integrator circuit.

\(\frac{{{V_i}}}{R} + \frac{{{V_0}}}{{1/sC}} = 0\)

\({V_0} = \frac{{ - 1}}{{sCR}}{V_i}\)

\({V_0}\left( t \right) = \frac{{ - 1}}{{RC}}\mathop \smallint \limits_0^t {V_i}\left( t \right)dt\)

This circuit acts as an integrator.