Frequency response of y[n] = 0.5x[n] − 0.5x[n−1]

Given: $y[n] = 0.5\,x[n] - 0.5\,x[n-1]$, so $h[n] = 0.5\,\delta[n] - 0.5\,\delta[n-1] = \{\underset{\uparrow}{0.5}, -0.5\}$ (a two-tap FIR system: a scaled first difference).

Frequency response

$$H(e^{j\omega}) = \sum_n h[n]e^{-j\omega n} = 0.5 - 0.5e^{-j\omega}$$

Factor out $e^{-j\omega/2}$:

$$\begin{aligned} H(e^{j\omega}) &= 0.5\,e^{-j\omega/2}\left(e^{j\omega/2} - e^{-j\omega/2}\right) \\ &= 0.5\,e^{-j\omega/2}\,(2j\sin\tfrac{\omega}{2}) \\ &= j\sin\left(\tfrac{\omega}{2}\right)e^{-j\omega/2} = \sin\left(\tfrac{\omega}{2}\right)e^{j\left(\frac{\pi}{2} - \frac{\omega}{2}\right)} \end{aligned}$$

Magnitude: $|H(e^{j\omega})| = \left|\sin\frac{\omega}{2}\right|$

Phase: $\theta(\omega) = \frac{\pi}{2} - \frac{\omega}{2}$ for $0 < \omega < \pi$ (and $-\frac{\pi}{2} - \frac{\omega}{2}$ for $-\pi < \omega < 0$): linear phase, apart from a jump of $\pi$ at $\omega = 0$, where $\sin(\omega/2)$ changes sign.

 |H|
  1 |*                   *
    | '.               .'
0.7 |   *             *
    |     '.       .'
  0 +--------'-*-'--------> w
   -pi         0         pi

Interpretation: DC ($\omega = 0$) is blocked completely and the highest frequency ($\omega = \pi$) passes with gain 1, so this is a simple high pass filter (a difference/differentiator-like system). Being FIR, it is always stable.