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Digital Signal Processing Basics

A signal sampled at 48 kHz is reduced to 12 kHz by keeping one sample out of every four (decimation by 4). Which of the following frequency components should the low-pass filter placed before the downsampling remove?

  1. AComponents above 6 kHz
  2. BComponents above 12 kHz
  3. CComponents above 24 kHz
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Answer: A. Components above 6 kHz

After downsampling, the sampling frequency is 48 ÷ 4 = 12 kHz and its Nyquist frequency is 6 kHz. If components above 6 kHz remain when the samples are dropped, they fold back under 12 kHz sampling and overlap the band below 6 kHz (for example, a 9 kHz component appears at 12 − 9 = 3 kHz). A digital low-pass filter therefore attenuates everything above 6 kHz sufficiently before the downsampling.

  • A. Correct. Everything above the new Nyquist frequency of 6 kHz is removed.
  • B. This is the new sampling frequency itself. The folding boundary is half of it, 6 kHz.
  • C. This is the Nyquist frequency of the original 48 kHz. The original signal has no components above 24 kHz anyway, so this alone does not prevent aliasing.
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