4-point DFT of {3, −2, 1}

The sequence has 3 samples, so for a 4-point DFT it is padded with one zero: $x[n] = \{3, -2, 1, 0\}$ for $n = 0, 1, 2, 3$.

Formula

$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N}, \qquad N = 4, \quad e^{-j\pi kn/2} = (-j)^{kn}$$

The twiddle factor is $W_4 = e^{-j\pi/2} = -j$, with $W^0 = 1$, $W^1 = -j$, $W^2 = -1$, $W^3 = j$.

Matrix form

$$\begin{bmatrix}X[0]\\X[1]\\X[2]\\X[3]\end{bmatrix} = \begin{bmatrix}1 & 1 & 1 & 1\\ 1 & -j & -1 & j\\ 1 & -1 & 1 & -1\\ 1 & j & -1 & -j\end{bmatrix}\begin{bmatrix}3\\-2\\1\\0\end{bmatrix}$$

Each value

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

Result

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

Checks:

• $X[3] = X^*[1]$, as required for a real sequence.
• $X[0] = \sum x[n] = 2$.
• Parseval: $\sum|x[n]|^2 = 9 + 4 + 1 = 14$, and $\frac{1}{4}(4 + 8 + 36 + 8) = 14$.
• Inverse DFT: $x[0] = \frac{1}{4}\sum X[k] = \frac{1}{4}(2 + 2 + 2j + 6 + 2 - 2j) = 3$. Correct.