In a half-wave rectifier supplied by a sinusoidal ac source of frequency 60 Hz, what is the frequency of the resulting pulsating dc output waveform (expressed in pulses per second, pps)?

Introduction / Context:
Rectifiers convert ac to dc, but the output remains a series of pulses. Understanding the relation between input line frequency and the ripple (pulse) frequency is essential for filter design, ripple calculations, and power-supply sizing.

Given Data / Assumptions:

• Half-wave rectifier using a single diode and a resistive load.
• Input frequency f_in = 60 Hz.
• Ideal diode behavior; no filtering after rectification.

Concept / Approach:
A half-wave rectifier conducts on only one half-cycle (positive half, assuming conventional orientation) and blocks the other half. Therefore, one output pulse is produced for each input cycle. The pulse repetition frequency equals the line frequency.

Step-by-Step Solution:
Input: sinusoid with period T = 1/60 s.Half-wave action: only positive half-cycles pass; negative halves are blocked.Thus, exactly one pulse per full cycle appears at the output.Therefore, pulse frequency f_pulse = 60 pulses per second (pps).

Verification / Alternative check:
Compare with full-wave rectification, which produces two pulses per cycle (120 pps at 60 Hz). The half-wave result is half of that, confirming 60 pps.

Why Other Options Are Wrong:
30 pps: would imply one pulse every two cycles, not correct.

90 pps and 120 pps: correspond to 1.5× and 2× the line frequency; full-wave gives 120 pps, not half-wave.

15 pps: no basis in standard rectification.

Common Pitfalls:
Confusing ripple frequency of half-wave vs full-wave rectifiers. Always remember: half-wave → f_ripple = f_line; full-wave → f_ripple = 2 * f_line.

Final Answer:
60 pps