4-point DFT by linear transformation

Find the sequence

$x[n] = u[n] + u[n-1] - u[n-3] - u[n-4]$. Evaluate it term by term:

So $x[n] = \{1, 2, 2, 1\}$ for $n = 0, 1, 2, 3$.

Linear transformation (matrix) method

$X_N = W_N x_N$, where $W_N^{kn} = e^{-j2\pi kn/N}$. For $N = 4$, $W_4 = e^{-j\pi/2} = -j$, with $W^0 = 1$, $W^1 = -j$, $W^2 = -1$ and $W^3 = j$:

$$W_4 = \begin{bmatrix}1 & 1 & 1 & 1\\ 1 & -j & -1 & j\\ 1 & -1 & 1 & -1\\ 1 & j & -1 & -j\end{bmatrix}$$ $$\begin{bmatrix}X[0]\\X[1]\\X[2]\\X[3]\end{bmatrix} = \begin{bmatrix}1 & 1 & 1 & 1\\ 1 & -j & -1 & j\\ 1 & -1 & 1 & -1\\ 1 & j & -1 & -j\end{bmatrix}\begin{bmatrix}1\\2\\2\\1\end{bmatrix}$$

Row by row:

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

Answer: $X[k] = \{6,\ -1 - j,\ 0,\ -1 + j\}$.

Check: $X[3] = X^*[1]$, as expected for a real sequence. By Parseval, $\sum|x[n]|^2 = 1 + 4 + 4 + 1 = 10$, and $\frac{1}{4}\sum|X[k]|^2 = \frac{1}{4}(36 + 2 + 0 + 2) = 10$.