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4-point DFT of {0, 1, 2, 3}

Short answer

Using W4 = -j and the DFT matrix, X[0] = 6, X[1] = -2 + 2j, X[2] = -2, X[3] = -2 - 2j. Thus the four-point DFT is {6, -2+2j, -2, -2-2j}.

Compute four-point DFT of four-point sequence: x[n] = {0, 1, 2, 3}.

Formula: For N=4N = 4,

X[k]=∑n=03x[n] W4nk,W4=e−j2π/4=−jX[k] = \sum_{n=0}^{3} x[n]\,W_4^{nk}, \qquad W_4 = e^{-j2\pi/4} = -j

So W40=1W_4^0 = 1, W41=−jW_4^1 = -j, W42=−1W_4^2 = -1, W43=jW_4^3 = j.

Matrix form:

[X[0]X[1]X[2]X[3]]=[11111−j−1j1−11−11j−1−j][0123]\begin{bmatrix} X[0] \\ X[1] \\ X[2] \\ X[3] \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -j & -1 & j \\ 1 & -1 & 1 & -1 \\ 1 & j & -1 & -j \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 2 \\ 3 \end{bmatrix}

Each term:

X[0]=0+1+2+3=6X[1]=0+(1)(−j)+(2)(−1)+(3)(j)=−2+2jX[2]=0+(1)(−1)+(2)(1)+(3)(−1)=−2X[3]=0+(1)(j)+(2)(−1)+(3)(−j)=−2−2j\begin{aligned} X[0] &= 0 + 1 + 2 + 3 = 6 \\ X[1] &= 0 + (1)(-j) + (2)(-1) + (3)(j) = -2 + 2j \\ X[2] &= 0 + (1)(-1) + (2)(1) + (3)(-1) = -2 \\ X[3] &= 0 + (1)(j) + (2)(-1) + (3)(-j) = -2 - 2j \end{aligned}

Answer: X[k]={6,  −2+2j,  −2,  −2−2j}X[k] = \{6,\; -2+2j,\; -2,\; -2-2j\}

kkX[k]X[k]∣X[k]∣\lvert X[k]\rvert∠X[k]\angle X[k]
06660∘0^\circ
1−2+2j-2+2j22=2.8282\sqrt2 = 2.828135∘135^\circ
2−2-22180∘180^\circ
3−2−2j-2-2j2.8282.828−135∘-135^\circ

Checks:

  • X[0]X[0] equals the sum of samples (6).
  • Since x[n]x[n] is real, X[3]=X∗[1]X[3] = X^*[1], which holds.
  • Parseval: ∑∣x[n]∣2=0+1+4+9=14\sum|x[n]|^2 = 0+1+4+9 = 14 and 14∑∣X[k]∣2=14(36+8+4+8)=14\frac{1}{4}\sum|X[k]|^2 = \frac{1}{4}(36+8+4+8) = 14.

✓ Answer checked by Claude Opus 5.5 (high effort)

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