H(e^(jω)) and h[n] for y[n] = 0.5y[n−1] + x[n]

Given: $y[n] = 0.5\,y[n-1] + x[n]$, i.e. $y[n] - 0.5\,y[n-1] = x[n]$ (system initially at rest).

Frequency response

Take the DTFT of both sides, using the shift property $x[n-1] \leftrightarrow e^{-j\omega}X(e^{j\omega})$:

$$\begin{aligned} Y(e^{j\omega}) - 0.5\,e^{-j\omega}Y(e^{j\omega}) &= X(e^{j\omega}) \\ H(e^{j\omega}) = \frac{Y(e^{j\omega})}{X(e^{j\omega})} &= \frac{1}{1 - 0.5\,e^{-j\omega}} \end{aligned}$$

Writing $e^{-j\omega} = \cos\omega - j\sin\omega$, the denominator is $(1 - 0.5\cos\omega) + j\,0.5\sin\omega$.

Magnitude response:

$$|H(e^{j\omega})| = \frac{1}{\sqrt{(1-0.5\cos\omega)^2 + (0.5\sin\omega)^2}} = \frac{1}{\sqrt{1.25 - \cos\omega}}$$

Phase response:

$$\angle H(e^{j\omega}) = -\tan^{-1}\left(\frac{0.5\sin\omega}{1 - 0.5\cos\omega}\right)$$

The gain is largest at $\omega = 0$ and smallest at $\omega = \pi$, so the system is a low pass filter. $H(e^{j\omega})$ is periodic with period $2\pi$.

Impulse response

Using the DTFT pair $a^n u[n] \leftrightarrow \frac{1}{1 - a e^{-j\omega}}$ for $|a| < 1$, with $a = 0.5$:

$$h[n] = (0.5)^n\,u[n]$$

Check by recursion ($x[n] = \delta[n]$, $h[-1] = 0$):

This matches $(0.5)^n$. The system is causal ($h[n] = 0$ for $n < 0$), stable ($\sum|h[n]| = \frac{1}{1-0.5} = 2 < \infty$), and IIR (infinite-length $h[n]$). Note $\sum h[n] = 2 = H(e^{j0})$, as expected.