AD/DA conversion

If a 10 kHz sine wave is sampled at 12 kHz without an anti-aliasing filter, at what frequency does it appear?

  1. A10 kHz
  2. B22 kHz
  3. C2 kHz
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Answer: C. 2 kHz

The Nyquist frequency (half the sampling frequency) is 6 kHz, and 10 kHz exceeds it. Such a signal folds back and appears at |fs − f| = |12 − 10| = 2 kHz (aliasing). Once aliased, it cannot be distinguished from a real signal, so it must be removed by a filter before sampling.

Example of aliasing Time Original 10 kHz signal Samples at 12 kHz Apparent 2 kHz Components above half the sampling rate show up at a lower frequency
  • A. 10 kHz is above the Nyquist frequency (6 kHz) and cannot be represented as is.
  • B. 22 kHz is above the Nyquist frequency and does not appear in the sampled signal.
  • C. Correct. It folds back to 12 kHz − 10 kHz = 2 kHz.
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More AD/DA conversion questions

  1. Which correctly describes the SFDR (spurious-free dynamic range) of an AD converter?
  2. How many comparisons does an N-bit successive-approximation AD converter basically perform per conversion?
  3. Which correctly describes the acquisition time of a sample-and-hold circuit?
  4. What is the technique of deliberately sampling a signal centered at 70 MHz with a 10 MHz bandwidth at a lower sampling frequency, which also performs frequency conversion?
  5. What is the main purpose of deliberately adding a small noise (dither) before AD conversion?