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Digital Signal Processing Basics

Let k = 0 be the DC bin. When a 1 kHz sine wave sampled at 8 kHz is analyzed with a 256-point DFT, which of the following is the index k of the bin where the peak appears on the positive-frequency side?

  1. A32
  2. B64
  3. C128
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Answer: A. 32

The bin spacing is Δf = fs/N = 8000 ÷ 256 = 31.25 Hz, so 1 kHz falls in bin k = 1000 ÷ 31.25 = 32. 1 kHz is an exact multiple of the bin spacing (exactly 32 cycles in 256 samples), so the peak is concentrated in the single bin k = 32 and there is no spectral leakage. For a real signal, a peak of the same size also appears at k = N − 32 = 224, which corresponds to the negative frequency.

  • A. Correct. k = f × N/fs = 1000 × 256/8000 = 32.
  • B. This divides by the Nyquist frequency of 4 kHz instead of 8 kHz (1000 × 256/4000).
  • C. k = N/2 = 128 is the bin of the Nyquist frequency (4 kHz).
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  1. A 64-point DFT of a real signal x[n] gives X[10] = 3 + 4j. Which of the following is X[54]?
  2. The DFT of the 8-point real signal x[n] = {3, 1, 3, 1, 3, 1, 3, 1} is defined as X[k] = Σx[n]e^(−j2πkn/8). Which of the following is the value of X[4], the bin at the Nyquist frequency?
  3. Suppose an N-point DFT takes about N² operations when computed from its definition and about N log₂N operations with the FFT (fast Fourier transform). For N = 1024, which of the following is closest to the reduction in the number of operations achieved by the FFT?
  4. A 256-point signal is followed by 768 zeros (zero padding) and a 1024-point DFT is computed. Compared with computing the DFT of the 256 points as they are, which of the following is correct?
  5. Which of the following is the main cause of spectral leakage in a DFT?