Topic: Analog Circuits
If $\tau$ is the time constant and w is the applied frequency, a low pass RC filter acts as a pure integrator when
- A. $w\tau=0$
- B. $w\tau>>1$
- C. $w\tau=1$
- D. $w\tau<<1$
Answer: B. $w\tau>>1$
Explanation: For a low pass RC filter, the transfer function is $\frac{V_{0}(s)}{V_{i}(s)} = \frac{1}{1+RCs} = \frac{1}{1+j\omega RC}$. For it to act as a pure integrator, the term $j\omega RC$ must dominate the denominator, which requires $\omega\tau >> 1$ (or $\omega >> \frac{1}{RC}$).