If τ is time constant and ω is the applied frequency, a low pass
| If τ is time constant and ω is the applied frequency, a low pass RC filter acts as a pure integrator when
A. ωτ = 0
B. ωτ > > 1
C. ωτ = 1
D. ωτ < < 1
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Low-pass filter is as shown:-
\( \frac{{{V_o}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{1 + sCR}} \)
\(= \frac{1}{{1 + s\tau }} = \frac{1}{{1 + j\omega \tau }}\)
\({V_o}\left( t \right) = {V_o}{e^{ - \frac{t}{\tau }}}\) .
For the capacitor to act as an integrator the capacitor should discharge slowly. And this discharging time depends upon the time constant (τ).
Less the time constant → faster will be the discharge
More the time constant → Slower will be the discharge
So, for the capacitor to act as an integrator,
ωτ ≫ 1
