Digital Signal Processing Sanjit K Mitra 3rd Edition Solution Manual -

2.1 (a) The even part of the signal $x[n] = \cos(0.5\pi n)$ is $x_e[n] = \cos(0.5\pi n)$.

$$X[k_1, k_2] = \begin{bmatrix} 10 & -2 \ -2 & -2 \end{bmatrix}$$

3.2 The FFT of the sequence $x[n] = 1, 2, 3, 4$ is:

$$X[k] = \begin{bmatrix} 10 & -2+j2 & -2 & -2-j2 \end{bmatrix}$$

The impulse response of the filter is:

$$X[k] = \begin{bmatrix} 10 & -2+j2 & -2 & -2-j2 \end{bmatrix}$$

$$H(z) = \frac{1}{1 - 0.5z^{-1} - 0.2z^{-2}}$$