High-frequency period calculation: What is the period (time for one cycle) of a 16 MHz sine wave?

Difficulty: Easy

Correct Answer: 62.5 ns

Explanation:


Introduction:
Converting between frequency and period is fundamental in signal processing, microcontroller timing, and communications. The period is the reciprocal of frequency: higher frequency means shorter period.


Given Data / Assumptions:

  • Frequency f = 16 MHz = 16 * 10^6 Hz.
  • Ideal single-tone sine wave.
  • We need T in seconds (or nanoseconds).


Concept / Approach:

The relationship is T = 1 / f. After computing T in seconds, convert to nanoseconds by multiplying by 10^9. This quick conversion is widely used for clock signals in digital electronics.


Step-by-Step Solution:

Compute T = 1 / f = 1 / (16 * 10^6) s.T = 0.0625 * 10^-6 s.Convert to nanoseconds: T = 62.5 ns.Thus, one cycle of a 16 MHz signal lasts 62.5 ns.


Verification / Alternative check:

Check by multiplication: 16 * 10^6 Hz * 62.5 * 10^-9 s = 1.0 (dimensionless), confirming consistency. Also, 8 MHz would be 125 ns; doubling frequency halves period, aligning with 16 MHz → 62.5 ns.


Why Other Options Are Wrong:

  • 196 ns / 31.25 ns / 19.9 ns / 6.25 ns: Do not equal 1 / 16 MHz; some correspond to other frequencies (e.g., 32 MHz → 31.25 ns).


Common Pitfalls:

  • Misplacing decimal points during unit conversion between microseconds and nanoseconds.
  • Confusing MHz (10^6) with kHz (10^3) or GHz (10^9).


Final Answer:

62.5 ns

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