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I need to find out the recurrence relation and the impulse response of the following system.

The relation seems to be $$y[n] = \frac{2}{16} x[n] + \frac{3}{16} x[n-1] + \frac{6}{16} x[n-2] + \frac{3}{16} x[n-3] + \frac{2}{16}x[n-4]$$

Since there is no recursion the impulse response $h[n]$ would contain only 5 values (below) being the coefficients?

[2/16, 3/16, 6/16, 3/16, 2/16]

System

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  • $\begingroup$ yes you are right. $\endgroup$ – Fat32 Dec 19 '18 at 23:59
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It looks like you answered your own question but to be complete..The impulse response $H(z)=\frac{Y(z)}{X(z)}$ can be found from the difference equation which you have found from the block diagram. So taking the z-transform of both sides of the difference equation we have: $Y(z)=X(z)(\frac{2}{16}+\frac{3}{16}z^{-1}+\frac{6}{16}z^{-2}+\frac{3}{16}z^{-3}+\frac{2}{16}z^{-4})$ and so we have $H(z)=\frac{2}{16}+\frac{3}{16}z^{-1}+\frac{6}{16}z^{-2}+\frac{3}{16}z^{-3}+\frac{2}{16}z^{-4}$.

Now taking the inverse z-transform you get the impulse response to be $h[n]=\frac{2}{16}+\frac{3}{16}\delta[n-1]+\frac{6}{16}\delta[n-2]+\frac{3}{16}\delta[n-3]+\frac{2}{16}\delta[n-4]$, which has the coefficients that you listed.

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  • $\begingroup$ Ok great. The next task is to use Matlab's filter and conv function to convolute a sinus signal of [1, 3, 4] kHz and a length of 2 seconds with the impulse responce. I find it strange to convolute a vector with 32k~ values with my impulse responce of 5 values. Is there something I'm missing? $\endgroup$ – Pulz Dec 20 '18 at 0:25
  • $\begingroup$ Where did you get 32k values from? $\endgroup$ – BryanEhlers Dec 20 '18 at 0:54
  • $\begingroup$ The sinus is 2 seconds long with a sampling rate of 16kHz. My sinus vector is basically sin(2 * Pi * 1000Hz * 0:1/16kHz:2) $\endgroup$ – Pulz Dec 20 '18 at 1:40
  • $\begingroup$ Ok so you may convolute the input with the impulse response, nothing wrong with that. $\endgroup$ – BryanEhlers Dec 20 '18 at 2:55

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