Cycloconverter frequency reduction concept A bridge-type single-phase cycloconverter reduces the input frequency from f to f/3. During one half-cycle of the output (that is, one half-wave at f/3), how many half-waves of the input waveform are spanned?

Introduction / Context:
Cycloconverters directly convert AC at one frequency to AC at a lower frequency without an intermediate DC link. Understanding the timing relationship between the input frequency and the synthesized output frequency is crucial for timing, gating logic, and harmonic analysis.

Given Data / Assumptions:

• Single-phase bridge cycloconverter.
• Input frequency = f.
• Output frequency = f/3 (one-third of input).
• We consider one output half-cycle (one half-wave) at f/3.

Concept / Approach:
The period T of a waveform is the reciprocal of frequency. If the output frequency is f/3, then T_out = 1/(f/3) = 3 * (1/f) = 3 * T_in. An output half-cycle lasts T_out/2 = 3 * T_in / 2. Each input half-wave lasts T_in/2. Therefore, counting how many input half-waves fit inside one output half-cycle becomes a simple ratio problem.

Step-by-Step Solution:
Let T_in = 1/f and T_out = 3 * T_in.Duration of one output half-wave = T_out / 2 = 3 * T_in / 2.Duration of one input half-wave = T_in / 2.Number of input half-waves in one output half-wave = (3 * T_in / 2) / (T_in / 2) = 3.Thus, one output half-wave at f/3 spans three input half-waves.

Verification / Alternative check:
Another way: since the output frequency is one-third of input, every output cycle lasts three input cycles. A half-cycle of output is therefore one and a half input cycles, which equals three input half-cycles.

Why Other Options Are Wrong:

• 3 full waves of input: that would be a full output cycle, not a half-cycle.
• 6 full waves or 6 half waves: overcounts the input segments for a single output half-cycle.
• 2 half waves: undercounts the required span.

Common Pitfalls:

• Confusing cycles with half-cycles while converting frequencies.
• Forgetting that a bridge cycloconverter stitches selected input half-waves to synthesize the lower-frequency output.

Final Answer:
3 half waves of input