Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

exercise

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.

share|improve this question
    
$x$ 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

1 Answer 1

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

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.