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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.

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$x$ is nowhere defined in your text. Neither are $h_1$ and $h_2$. You'll have to provide more background. Maybe this is standard notation in system/signal theory, but unless a specialist passes by, you'll have to wait for a long time before getting an answer if you leave the question as it is now. – Raskolnikov Apr 16 '11 at 11:11
If I understand the question correctly, you feed the system a Heaviside step function, which is here symbolized by $u[n]$ and as a result you get $g_2[n]$. They ask what the response is when you feed an impulse $\delta[n]$ to the system. But since $\delta[n]=u[n]-u[n-1]$ do you really need to go to the Fourier transform? – Raskolnikov Apr 16 '11 at 11:20
@Raskolnikov i added x[n], h2 i need to calculate. I think u are on the right way, yes. But i am new to signal processing, so i don't know what the best way is to get the solution. – madmax Apr 16 '11 at 12:42
A couple more remarks, at some point, you use $\phi$ where you actually mean $\omega$. The $\delta$ in your transform of $x$ is not necessary I think. I'm not sure where you get that from. The rest should be okay. – Raskolnikov Apr 16 '11 at 12:57
There was a fourier transform table added to this. And for x[n] the transform was that fraction plus the δ. So just dividing should get me to my solution? – madmax Apr 16 '11 at 13:07
up vote 1 down vote accepted

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

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

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Ok, thank you. But i don't understand where i need the δ[n]=u[n]−u[n−1]. Can't i just try to make the inverse Fourier transform out of H(ω)? – madmax Apr 16 '11 at 13:24
@madmax, in the problem, they ask you to compute the impulse response, which is the effect of the system $H(\omega)$ on a pulse $\delta[n]$. You may be right about that being just $H(\omega)$ since the Fourier transform of a $\delta$ pulse is just $1$. I must have made some mistake there. Can you find the exact definition of the Heaviside step function $u[n]$? I must have made my mistake there. – Raskolnikov Apr 16 '11 at 13:35
@madmax, never mind, I found my mistake. It was in the Fourier transform of $u[n-1]$. I edited my answer. – Raskolnikov Apr 16 '11 at 13:37
hmm, i thought the output is the step response g[n] in this case. Isn't just asked for h[n]? Sorry, but it is not that easy to understand for me right now. – madmax Apr 16 '11 at 13:56
The step response is given, but they ask for the impulse response, which as you say is $h[n]$. The output in the case of input $u[n]$ is what is called step response $g[n]$, the output in the case of input $\delta[n]$ is what is called impulse response $h[n]$. – Raskolnikov Apr 16 '11 at 14:04

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