If τ is time constant and ω is the applied frequency, a low pass RC filter acts as a pure integrator when

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LPSC ISRO Technical Assistant Electronics 23 Feb 2020 Official Paper
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  1. ωτ = 0
  2. ωτ > > 1
  3. ωτ = 1
  4. ωτ < < 1

Answer (Detailed Solution Below)

Option 2 : ωτ > > 1
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Low-pass filter is as shown:-

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\( \frac{{{V_o}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{1 + sCR}} \)

\(= \frac{1}{{1 + s\tau }} = \frac{1}{{1 + j\omega \tau }}\)

\({V_o}\left( t \right) = {V_o}{e^{ - \frac{t}{\tau }}}\) .

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

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