- 2069 Chaitra · 7 marks
8-point DIF-FFT of {1,1,2,2,1,1,2,1}
Short answer
Use radix-2 DIF FFT with outputs in bit-reversed order. X(k)={11, −0.7071−j0.7071, −2+j1, 0.7071−j0.7071, 1, 0.7071+j0.7071, −2−j1, −0.7071+j0.7071}.
Find 8-point DFT of sequence x[n] = {1, 1, 2, 2, 1, 1, 2, 1} using Decimation in frequency FFT (DIFFFT) algorithm.
Method: radix-2 decimation in frequency (N = 8, 3 stages)
Each butterfly: A' = A + B, B' = (A − B)·W. Stage 1 pairs n and n + 4 (twiddle W₈ⁿ on the difference), stage 2 pairs n and n + 2 (W₄ⁿ), stage 3 pairs adjacent samples (W₂⁰ = 1). The outputs come out in bit-reversed order.
x(n) = {1, 1, 2, 2, 1, 1, 2, 1}
| Position | Input x(n) | After stage 1 | After stage 2 | After stage 3 |
|---|---|---|---|---|
| 0 | 1 | 2 | 6 | 11 |
| 1 | 1 | 2 | 5 | 1 |
| 2 | 2 | 4 | -2 | −2 + j1 |
| 3 | 2 | 3 | j1 | −2 − j1 |
| 4 | 1 | 0 | 0 | −0.7071 − j0.7071 |
| 5 | 1 | 0 | −0.7071 − j0.7071 | 0.7071 + j0.7071 |
| 6 | 2 | 0 | 0 | 0.7071 − j0.7071 |
| 7 | 1 | −0.7071 − j0.7071 | 0.7071 − j0.7071 | −0.7071 + j0.7071 |
Output positions 0, 1, 2, 3, 4, 5, 6, 7 hold X(0), X(4), X(2), X(6), X(1), X(5), X(3), X(7) (bit-reversed). Rearranging:
Result:
X(0) = 11
X(1) = −0.7071 − j0.7071
X(2) = −2 + j1
X(3) = 0.7071 − j0.7071
X(4) = 1
X(5) = 0.7071 + j0.7071
X(6) = −2 − j1
X(7) = −0.7071 + j0.7071
Check: Σx(n) = 11 = X(0); the result equals the direct 8-point DFT.
Answer: X(k) = {11, −0.7071 − j0.7071, −2 + j1, 0.7071 − j0.7071, 1, 0.7071 + j0.7071, −2 − j1, −0.7071 + j0.7071}.
✓ Answer checked by Claude Opus 5.5 (high effort)
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