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I understand how to find the output from the input with an impulse response, but how can I go about finding the input if given the other two?

I have $y[n] = [-1, -1, 11, -3, 30, 28, 48]$ and $h[n] = [-1, 2, 3, 4]$

How can I go about finding $x[n]$?

I know very little about signal processing, so if you don't mind giving an easy explanation, then I appreciate it. Or, if it's possible to do an example, that's better.

EDIT: I think that if I know $y[n]$ I can guess what $x[n]$ would be by multiplying and summing the results, but I'm not sure how to figure out the length of $x[n]$.

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Wow, your edits make so many changes that my answer, which you have accepted, is to a completely different problem and now looks like complete nonsense. – Dilip Sarwate Feb 24 '13 at 14:57
up vote 1 down vote accepted

This answer was given in response to the ORIGINAL version of the question in which $y[n]$ and $h[n]$ were vastly different sequences. Click on the edited n hours ago link that appears below the question to view the original version of the question.

Define polynomials $$\begin{align} y(\alpha) &= -\alpha^6+11\alpha^5 +2\alpha^4+\cdots + 4\alpha+3,\\ h(\alpha) &= -\alpha^3+2\alpha^2+\alpha+3. \end{align}$$ Divide $y(\alpha)$ by $h(\alpha)$. Assuming $h(\alpha)$ divides $y(\alpha)$ evenly, meaning that the remainder is $0$, the quotient tells you the input sequence $x[n]$. If the remainder is not $0$, whoever told you that the output is $y[n]$ is mistaken. I have a sneaking suspicion that the latter is likely to be true.

Some people will insist that you replace $\alpha$ by $z$ in the above formulas, and then divide $z^{-6}y(z)$ by $z^{-3}h(z)$, but you will arrive at the same end result if you follow their method.

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Tried it, and it works. Thanks! – Chris Harris Feb 23 '13 at 20:26
@ChrisHarris You are welcome. But, could you very carefully proof-read, and if necessary, correct, your question? I don't think you can get an answer using $$y = y[n] = [-1, 11, 2, 4, 5, 4, 3].$$ That $11$ is either a $1$ or a $1,1$ – Dilip Sarwate Feb 24 '13 at 3:44

Mathematically, here is the demostration for what you want to do: y[n] = x[n]*h[n] (This is a convolution), Applying convolution theorem you can prove that: Y[f] = X[f]*H[f] (this is an ordinary product), Then: X[f] = Y[f]/H[f], Now if we apply fourier inverse transform to X[f] you can have x[n]. Implementing this in matlab can be pretty straightforward now. Cheers

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So, I need to do the fourier transform for both y[n] and h[n]? – Chris Harris Feb 23 '13 at 20:09
This answer is way too complicated. There is no need for Fourier transforms and the like. – Dilip Sarwate Feb 23 '13 at 20:17
@Daniel If you can show the fourier transforms, and inverse transforms, then that would be helpful. – Chris Harris Feb 23 '13 at 20:28
@Chris Harris: Yes Harry, you have to transform into frequency domain.This is a very straightforward solution although Dilip says otherwise – Daniel Conde Feb 23 '13 at 20:47
@DanielConde You might want to give some thought to what one should do if $H(f_0) = 0$ for some frequency $f_0$ while $Y(f_0) \neq 0$. What value should be ascribed to $X(f_0)$? – Dilip Sarwate Feb 24 '13 at 4:14

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