信号: x[n] = {1, 1} (N=2)

1. DTFT

$$ X(e^{j\omega}) = \sum_{n=0}^{1} x[n]e^{-j\omega n} = x[0]e^{-j\omega(0)} + x[1]e^{-j\omega(1)} = (1)(1) + (1)e^{-j\omega} = 1 + e^{-j\omega} $$

2. ZT

$$ X(z) = \sum_{n=0}^{1} x[n]z^{-n} = x[0]z^{-0} + x[1]z^{-1} = (1)(1) + (1)z^{-1} = 1 + z^{-1}, \quad (\text{ROC: } |z| > 0) $$

3. DFT

步骤 1: 构建变换矩阵 W 并进行计算

$$ \begin{bmatrix} X(0) \\ X(1) \end{bmatrix} = \begin{bmatrix} W_2^{0} & W_2^{0} \\ W_2^{0} & W_2^{1} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} (1)(1) + (1)(1) \\ (1)(1) + (-1)(1) \end{bmatrix} = \begin{bmatrix} 1 + 1 \\ 1 - 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} $$

最终结果: X(k) = {2, 0}

4. 3D 可视化

信号: x[n] = {1, 1, 1, 1} (N=4)

1. DTFT

$$ X(e^{j\omega}) = \sum_{n=0}^{3} e^{-j\omega n} = e^{-j\omega(0)} + e^{-j\omega(1)} + e^{-j\omega(2)} + e^{-j\omega(3)} = \frac{1 - (e^{-j\omega})^4}{1 - e^{-j\omega}} = \frac{1 - e^{-j4\omega}}{1 - e^{-j\omega}} $$

2. ZT

$$ X(z) = \sum_{n=0}^{3} z^{-n} = z^{-0} + z^{-1} + z^{-2} + z^{-3} = \frac{1 - (z^{-1})^4}{1 - z^{-1}} = \frac{1 - z^{-4}}{1 - z^{-1}}, \quad (\text{ROC: } z \neq 0) $$

3. DFT

步骤 1: 构建变换矩阵 W 并进行计算

$$ \begin{bmatrix} X(0) \\ X(1) \\ X(2) \\ X(3) \end{bmatrix} = \begin{bmatrix} W_4^{0} & W_4^{0} & W_4^{0} & W_4^{0} \\ W_4^{0} & W_4^{1} & W_4^{2} & W_4^{3} \\ W_4^{0} & W_4^{2} & W_4^{4} & W_4^{6} \\ W_4^{0} & W_4^{3} & W_4^{6} & W_4^{9} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -j & -1 & j \\ 1 & -1 & 1 & -1 \\ 1 & j & -1 & -j \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(1)(1)+(1)(1)+(1)(1) \\ (1)(1)+(-j)(1)+(-1)(1)+(j)(1) \\ (1)(1)+(-1)(1)+(1)(1)+(-1)(1) \\ (1)(1)+(j)(1)+(-1)(1)+(-j)(1) \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$

最终结果: X(k) = {4, 0, 0, 0}

4. 3D 可视化