LTI: How to derive the impulse response of this system?

Question

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.

Answer 1

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

$$\left(\frac{1}{2}\right)^n u[n] - \left(\frac{1}{2}\right)^{n-1} u[n-1] \; .$$