$$X[k] = \begin{bmatrix} 10 & -2+j2 & -2 & -2-j2 \end{bmatrix}$$
$$H(z) = \frac{1}{1 - 0.5z^{-1}}$$
2.1 (a) The even part of the signal $x[n] = \cos(0.5\pi n)$ is $x_e[n] = \cos(0.5\pi n)$. $$X[k] = \begin{bmatrix} 10 & -2+j2 & -2
The impulse response of the filter is:
(b) The maximum and minimum values that can be represented by 12-bit unsigned binary numbers are 4095 and 0, respectively. $$X[k] = \begin{bmatrix} 10 & -2+j2 & -2
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