The impulse response of the filter is:
$$X[k] = \begin{bmatrix} 10 & -2+j2 & -2 & -2-j2 \end{bmatrix}$$ The impulse response of the filter is: $$X[k]
2.1 (a) The even part of the signal $x[n] = \cos(0.5\pi n)$ is $x_e[n] = \cos(0.5\pi n)$. The impulse response of the filter is: $$X[k]
$$H(z) = \frac{1}{1 - 0.5z^{-1} - 0.2z^{-2}}$$ The impulse response of the filter is: $$X[k]
$$y[n] = x[2n]$$
$$H(z) = 1 + 2z^{-1} + 3z^{-2}$$