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8-point DFT by FFT of {1,−1,2,2,1,1,2,2}

Short answer

Use radix-2 DIT FFT with input in bit-reversed order. The 8-point DFT is X(k)={10, −1.4142+j1.4142, −2+j4, 1.4142+j1.4142, 2, 1.4142−j1.4142, −2−j4, −1.4142−j1.4142}.

Find 8-point DFT of sequence x[n] = {1, −1, 2, 2, 1, 1, 2, 2} using Fast Fourier Transform algorithm.

The question does not name the algorithm, so DIT-FFT is used.

Method: radix-2 decimation in time (N = 8, 3 stages)

Input in bit-reversed order: x(0), x(4), x(2), x(6), x(1), x(5), x(3), x(7). Each butterfly: A' = A + W·B, B' = A − W·B, with W₈⁰ = 1, W₈¹ = 0.7071 − j0.7071, W₈² = −j, W₈³ = −0.7071 − j0.7071. Stage 1 uses W₂ (= ±1), stage 2 uses W₄⁰ = 1 and W₄¹ = −j, and stage 3 uses W₈⁰ … W₈³.

x(n) = {1, -1, 2, 2, 1, 1, 2, 2}

PositionInput (bit-reversed)After stage 1After stage 2After stage 3 = X(k)
012610
1100−1.4142 + j1.4142
224-2−2 + j4
32001.4142 + j1.4142
4-1042
51-2-21.4142 − j1.4142
624-4−2 − j4
720-2−1.4142 − j1.4142

Result:

X(0) = 10
X(1) = −1.4142 + j1.4142
X(2) = −2 + j4
X(3) = 1.4142 + j1.4142
X(4) = 2
X(5) = 1.4142 − j1.4142
X(6) = −2 − j4
X(7) = −1.4142 − j1.4142

Check: Σx(n) = 10 = X(0); the result equals the direct 8-point DFT.

Answer: X(k) = {10, −1.4142 + j1.4142, −2 + j4, 1.4142 + j1.4142, 2, 1.4142 − j1.4142, −2 − j4, −1.4142 − j1.4142}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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