Short notes

DFT Short Notes

Running a 4-point DFT on the sequence x[n] = {1, 2, 3, 4} is one of the most repeated problems in Anna University DSP papers, and it appears simple until you forget that W_4 = e^(-j2π/4) = e^(-jπ/2) = -j. The DFT takes a finite, discrete-time sequence sampled at, say, 8 kHz and produces a discrete set of complex frequency coefficients — four input samples give exactly four output bins, each corresponding to a frequency of k × (f_s/N).

ECE, EI

How it works

The DFT analysis equation is X[k] = Σ x[n] · W_N^(nk) for n = 0 to N-1, where W_N = e^(-j2πN). Each output bin X[k] represents the correlation of the input with a complex sinusoid at frequency k/N cycles per sample. For a real input of length N = 8 sampled at 8 kHz, bin k=1 corresponds to 1 kHz, bin k=2 to 2 kHz, and so on. The inverse DFT (IDFT) uses W_N^(-nk) and a 1/N scaling factor to recover x[n]. Computing all N points of an N-point DFT directly requires N² complex multiplications, which for N = 1024 means over a million operations.

Key points to remember

An N-point DFT has exactly N complex output bins, but for a real input, bins above N/2 are complex conjugate mirrors of the lower bins — only N/2 + 1 unique values exist. Frequency resolution is Δf = f_s / N; doubling N halves the frequency resolution. DFT assumes the N-point block is one period of a periodic signal — discontinuities at block boundaries cause spectral leakage. Multiplying x[n] by a Hanning or Hamming window before the DFT reduces leakage at the cost of slightly widened spectral peaks. The twiddle factor W_N^(nk) satisfies the periodicity W_N^(nk+N) = W_N^(nk).

Exam tip

Every Anna University DSP exam has at least one 4-point or 8-point DFT calculation — draw the twiddle factor table first, then substitute, and double-check your W_4 = -j and W_8 = (1-j)/√2 values before computing.

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