I'm trying to compute the impulse response for the following system:
$$ y(n) = \frac{1}{2} \left( x(n) + x(n-1) \right) + \frac{1}{2} y(n-1) $$
and am told to assume that $y(-1)=0$.
My solution: Letting $x(n) = \delta(n)$, we have
$$ h(n) = \frac{1}{2}\left(\delta(n) + \delta(n-1)\right) + \frac{1}{2}h(n-1). $$
Plugging in a few values for $n$ (and assuming that $h(-1)=0$), I've obtained:
$$ h(0) = \frac{1}{2}, $$
$$ h(1) = \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} $$
and
$$ h(2) = \frac{1}{2} + \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}. $$
From this, I've observed that
$$ h(n) = \begin{cases} 0, & n<0 \\ \frac{1}{2}, & n=0 \\ \frac{1}{2} + \left(\frac{1}{2}\right)^n, & n > 0 \end{cases} $$
Does this seem correct? I feel like I made a mistake somewhere and that the expression for $h(n)$ should be in a simpler form.
Note: I computed the impulse response by taking the Fourier Transform and then plugging in values for $n$ and got something different than this. However, I believe that this discrepancy is due to the initial condition $y(-1)=0$.