Topic: Analog Circuits
Phase of the transfer function of the following circuit is (Circuit with a capacitor C in series and a resistor R in parallel/shunt to ground)
- A. $\tan^{-1}(1/\omega RC)$
- B. $\tan^{-1}(\omega RC)$
- C. $\tan^{-1}(RC/\omega)$
- D. $\tan^{-1}(\omega/RC)$
Answer: A. $\tan^{-1}(1/\omega RC)$
Explanation: The circuit acts as a high-pass filter. The transfer function is $V_o/V_i = \frac{R}{R + 1/(j\omega C)} = \frac{j\omega RC}{1 + j\omega RC}$. Alternatively written as $\frac{R}{R - j/\omega C}$, the phase angle is $\tan^{-1}(\frac{1}{\omega RC})$.