Wednesday, January 1, 2020

deconvolution - How to Estimate the Input for a Convolution Given the Filter (Impulse Response) and the Output of the Convolution


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 $x[n]$.



Answer




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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