Well, i transform g and x into the frequency domain.
u[n] = 1, n ≥ 0
u[n] = 0, n < 0
\begin{aligned}
x[n] & = u[n] \\
h_1[n] & = (\frac{1}{2})^n u[n] \\
g[n] & = (\frac{1}{2})^n u[n] \\
G(e^{j*\phi}) & = \frac{1}{1-0.5 * e^{-j*\phi}} \\
X(e^{j*\phi}) & = \frac{1}{1- e^{-j*\phi}} \\
H(e^{j*\phi}) & = \frac{G(e^{j*\phi})}{X(e^{j*\phi})}
\end{aligned}
But i don't know how to go on.
Best Answer
PART 1: I think you nearly got everything. You know your transmission function in the Fourier domain is:
$$H(\omega)=\frac{G(\omega)}{X(\omega)}=\frac{1- e^{-j\omega}}{1-0.5 e^{-j\omega}} \; .$$
The Fourier transform of $u[n]$ being $X(\omega)$, the Fourier transform of $u[n-1]$ is then
$$\frac{e^{-j\omega}}{1- e^{-j\omega}} \; .$$
Combining everything you get that the Fourier transform of $\delta[n]=u[n]-u[n-1]$ is
$$\frac{1}{1- e^{-j\omega}}-\frac{e^{-j\omega}}{1- e^{-j\omega}} = 1$$
and after passing through your system it will become
$$\frac{1-e^{-j\omega}}{1-0.5 e^{-j\omega}} \; .$$
Working out the components of the series expansion should give you the inverse Fourier transform.
PART 2: As I suggested before, you can solve the exercise without ever doing a Fourier transform. Since the system acts linearly on any input, and $\delta[n]=u[n]-u[n-1]$, it immediately follows that the output is
$$\left(\frac{1}{2}\right)^n u[n] - \left(\frac{1}{2}\right)^{n-1} u[n-1] \; .$$