An integrator circuit is also a ________ circuit.

A simple RC integrator circuit is as shown:

Derivation:

Vi = VR + VC

VR = Voltage across the Resistor, given by:

V_R = i×R

V_C = Voltage across the capacitor

Current through the capacitor is given by:

\(i = C~\frac{{d{V_c}}}{{dt}}\)

\({V_i} = {V_c}\left( t \right) + RC\;\frac{{d{V_c}}}{{dt}}\)

For a large time constant RC >> 1, the above equation becomes:

\({V_i} \approx RC~\frac{{d{V_c}}}{{dt}}\)

Vc = V0

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

Obtaining the transfer function in the frequency domain:

The transfer function of the above circuit is simply the ratio of the output voltage to the input voltage and is obtained using voltage division as:

\(V_i(ω ) \frac{\left( \frac{{{X}_{c}}}{j} \right)}{\frac{{{X}_{c}}}{j}+R}={{V}_{o}}\left( ω \right)\)

\(X_C= \frac{{1}}{{ω C}}\)

\(\frac{{{V}_{0}}\left( ω \right)}{{{V}_{i}}\left( ω \right)}=\frac{1}{jω C}\times \frac{1}{\left( \frac{1}{jω C}+R \right)}\)

\(\frac{{{V}_{0}}\left( ω \right)}{{{V}_{i}}\left( ω \right)}=\frac{1}{1+Rjω C}\)

\(H\left( ω \right)=\frac{{{V}_{o}}\left( ω \right)}{{{V}_{i}}\left( ω \right)}=\frac{1}{1+jω RC}\)

The magnitude of the above transfer function is represented as:

\(\left| H\left( ω \right) \right|=\frac{1}{\sqrt{1+{{ω }^{2}}{{R}^{2}}{{C}^{2}}}}\)

For ω = 0

|H(ω)| = 1

For ω = ∞

|H(ω)| = 0

This is the characteristic of a low pass filter.

We, therefore, conclude that the integrator circuit is also a low pass circuit.