Topic: Analog Circuits
Gain of an RC low pass filter having a time constant '$\tau$' and frequency '$\omega$' is
- A. $\sqrt{1+(\omega\tau)^{2}}$
- B. $\frac{1}{\sqrt{1+(\omega\tau)^{2}}}$
- C. $\frac{\omega\tau}{\sqrt{1+(\omega\tau)^{2}}}$
- D. $\frac{\omega\tau}{\sqrt{1-(\omega\tau)^{2}}}$
Answer: B. $\frac{1}{\sqrt{1+(\omega\tau)^{2}}}$
Explanation: For an RC low pass filter, $H(s) = \frac{1}{RCs+1}$. The frequency response is $H(j\omega) = \frac{1}{1+j\omega RC}$. The magnitude or gain is $|H(j\omega)| = \left|\frac{1}{1+j\omega\tau}\right| = \frac{1}{\sqrt{1+\omega^{2}\tau^{2}}}$, where $\tau=RC$ is the time constant.