Impulse and frequency response of y[n] = 0.5y[n−1] + x[n] + x[n+1]

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

Frequency response

Take the DTFT, using $x[n-n_0]\leftrightarrow e^{-j\omega n_0}X(e^{j\omega})$ (so $x[n+1]\leftrightarrow e^{j\omega}X$):

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

Magnitude and phase: since $1+e^{j\omega}=2\cos(\omega/2)e^{j\omega/2}$,

$$|H(e^{j\omega})|=\frac{2|\cos(\omega/2)|}{\sqrt{1.25-\cos\omega}},\qquad \angle H=\frac{\omega}{2}-\tan^{-1}\frac{0.5\sin\omega}{1-0.5\cos\omega}$$

It is a low pass response (gain 4 at dc, zero at $\omega=\pi$).

Impulse response

Write $H=\frac{1}{1-0.5e^{-j\omega}}+e^{j\omega}\frac{1}{1-0.5e^{-j\omega}}$. Using $(0.5)^n u[n]\leftrightarrow\frac{1}{1-0.5e^{-j\omega}}$ and the time-advance property ($e^{j\omega}$ means $n\to n+1$):

$$h[n]=(0.5)^n u[n]+(0.5)^{n+1}u[n+1]$$

Simplified: at $n=-1$, $h[-1]=1$; for $n\ge0$, $h[n]=(0.5)^n+(0.5)^{n+1}=1.5(0.5)^n$. So

$$h[n]=\delta[n+1]+1.5(0.5)^n u[n]$$

Values: $h[-1]=1,\ h[0]=1.5,\ h[1]=0.75,\ h[2]=0.375,\ h[3]=0.1875,\dots$ (checked by running the recursion with $x[n]=\delta[n]$).

Answer: $H(e^{j\omega})=\dfrac{1+e^{j\omega}}{1-0.5e^{-j\omega}}$, $h[n]=(0.5)^n u[n]+(0.5)^{n+1}u[n+1]$. Since $h[-1]\ne0$ the system is non-causal; it is stable because $\sum|h|=1+3=4<\infty$.