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4-point DFT of {2, 1+j, 1−j}

Short answer

Zero-pad to length 4: x[n]={2, 1+j, 1−j, 0}. Using W4=−j, X[0]=4, X[1]=2, X[2]=2−2j, X[3]=2j. Magnitudes: 4, 2, 2.83, 2; phases: 0°, 0°, −45°, 90°.

Compute 4-point DFT of a signal x[n] = {2, 1+j, 1−j} and plot its magnitude and phase spectrums.

The sequence has only 3 samples, so append one zero for a 4-point DFT: x[n]={2,  1+j,  1−j,  0}x[n] = \{2,\; 1+j,\; 1-j,\; 0\}.

Formula: X[k]=∑n=03x[n]W4nkX[k] = \sum_{n=0}^{3}x[n]W_4^{nk}, with W4=e−jπ/2=−jW_4 = e^{-j\pi/2} = -j, so W40=1W_4^0=1, W41=−jW_4^1=-j, W42=−1W_4^2=-1, W43=jW_4^3=j.

[X[0]X[1]X[2]X[3]]=[11111−j−1j1−11−11j−1−j][21+j1−j0]\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} 2 \\ 1+j \\ 1-j \\ 0 \end{bmatrix}

Working:

X[0]=2+(1+j)+(1−j)+0=4X[1]=2+(1+j)(−j)+(1−j)(−1)+0=2+(1−j)+(−1+j)=2X[2]=2−(1+j)+(1−j)−0=2−2jX[3]=2+(1+j)(j)+(1−j)(−1)+0=2+(−1+j)+(−1+j)=2j\begin{aligned} X[0] &= 2 + (1+j) + (1-j) + 0 = 4 \\ X[1] &= 2 + (1+j)(-j) + (1-j)(-1) + 0 = 2 + (1 - j) + (-1 + j) = 2 \\ X[2] &= 2 - (1+j) + (1-j) - 0 = 2 - 2j \\ X[3] &= 2 + (1+j)(j) + (1-j)(-1) + 0 = 2 + (-1 + j) + (-1 + j) = 2j \end{aligned}

Answer: X[k]={4,  2,  2−2j,  2j}X[k] = \{4,\; 2,\; 2-2j,\; 2j\}

kkX[k]X[k]∣X[k]∣\lvert X[k]\rvert∠X[k]\angle X[k]
04440∘0^\circ
12220∘0^\circ
22−2j2-2j22=2.832\sqrt2 = 2.83−45∘-45^\circ
32j2j290∘90^\circ

Spectra (stem plots):

 |X[k]|
 4 |  |
 3 |  |       |
 2 |  |   |   |   |
 1 |  |   |   |   |
 0 +--+---+---+---+--> k
      0   1   2   3

 phase (degrees)
  90 |              |
  45 |              |
   0 +--o---o---+---+--> k
 -45 |          |
        0   1   2   3

(Here x[n]x[n] is complex, so X[3]≠X∗[1]X[3] \ne X^*[1]; the spectrum has no conjugate symmetry.)

Check (Parseval): ∑∣x[n]∣2=4+2+2+0=8\sum|x[n]|^2 = 4 + 2 + 2 + 0 = 8 and 14∑∣X[k]∣2=14(16+4+8+4)=8\frac{1}{4}\sum|X[k]|^2 = \frac{1}{4}(16 + 4 + 8 + 4) = 8.

✓ Answer checked by Claude Opus 5.5 (high effort)

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