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

Given: $y[n] = 0.5\,y[n-1] + x[n]$, initially at rest.

Frequency response

DTFT of both sides:

$$\begin{aligned} Y(e^{j\omega})\left(1 - 0.5e^{-j\omega}\right) &= X(e^{j\omega}) \\ H(e^{j\omega}) &= \frac{1}{1 - 0.5e^{-j\omega}} = \frac{1}{(1 - 0.5\cos\omega) + j0.5\sin\omega} \end{aligned}$$

Magnitude:

$$|H(e^{j\omega})| = \frac{1}{\sqrt{(1 - 0.5\cos\omega)^2 + 0.25\sin^2\omega}} = \frac{1}{\sqrt{1.25 - \cos\omega}}$$

Phase:

$$\theta(\omega) = -\tan^{-1}\frac{0.5\sin\omega}{1 - 0.5\cos\omega}$$

Magnitude plot

 |H|
 2.0 |          *
     |        .' '.
 1.36|      *       *
     |    .'         '.
 0.89|  *               *
 0.67|*                   *
     +--+---+---+---+---+--> w
     -pi   -pi/2  0  pi/2  pi

The magnitude is even in $\omega$, has its maximum 2 at $\omega = 0$ and its minimum $2/3$ at $\omega = \pm\pi$, and repeats every $2\pi$. So the system is a low pass filter. The phase is odd in $\omega$ (maximum lag 30° at $\omega = \pm\pi/3$).

Impulse response

Using $a^n u[n] \leftrightarrow \frac{1}{1 - ae^{-j\omega}}$ with $a = 0.5$:

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

 h[n]
 1.0 |   o
     |   |
 0.5 |   |  o
0.25 |   |  |  o
     |   |  |  |  o  o  .
   0 +-o-+--+--+--+--+--+--> n
      -1  0  1  2  3  4  5

The system is causal, stable ($\sum h[n] = 2$) and IIR.