A sine wave of angular frequency ω = 1/(RC) is applied to the RC high-pass filter in figure (b). Which of the following correctly describes the amplitude and phase of the output relative to the input?

Answer: B. The amplitude is about 0.71 times, and the output leads the input by 45°

Method 1 (complex numbers): the transfer function is H(jω) = jωRC/(1 + jωRC), and for ωRC = 1, H = j/(1 + j) = (1 + j)/2. Its magnitude is 1/√2 ≈ 0.71 (−3 dB) and its angle is +45°, so the output leads the input by 45°. Method 2 (phasor diagram): the current in the series RC circuit leads the input voltage because the circuit is capacitive, and at ω = 1/(RC), R and 1/(ωC) are equal, so the lead is exactly 45°. The output is the voltage across R, which is in phase with the current, so the output also leads the input by 45°. At the same frequency, the output of the low-pass filter in figure (a) lags by 45°.

• A. The magnitude is right, but a 45° lag applies to the low-pass filter in figure (a).
• B. Correct. H = (1 + j)/2: about 0.71 times with a 45° lead.
• C. This splits the voltage equally between R and 1/(ωC) without considering phase. The two voltages are at right angles, so each is 1/√2 times the input.