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So I had this question (f) at my last exam that I failed and I'm now studying for my re-exam, the problem is that I still don't know what the answer is supposed to be in this question (f):

Question

My attempt

Do you have any suggestions how to solve it?

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  • $\begingroup$ Hint: you do understand that $*$ is convolution, not multiplication? $\endgroup$ – Peter K. Aug 20 '15 at 18:34
  • $\begingroup$ Yes, but the convolution is supposed to be distributive so the first step should be correct. The second assumption is wrong but now looking at it I would say that the answer should be X(n)-X(n-1) based on that the dirac delta only giving a value to X(n) when n=0. Am I missing something? $\endgroup$ – ricksson Aug 20 '15 at 18:52
  • $\begingroup$ Well $\delta(n)*x(n)*u(n)$ doesn't equal $x(n)$. Apart from that, you are correct. $\endgroup$ – Peter K. Aug 20 '15 at 18:55
  • $\begingroup$ Sorry but I don't know what it would equal then :/ δ(n)∗x(n) from what I understand is x(n) but I don't understand what effect that d(n) would have except for limiting the possible answer to n>0 $\endgroup$ – ricksson Aug 20 '15 at 19:35
  • $\begingroup$ OK. So $u(n)$ is the unit step. $x(n)u(n)$ limits $x(n)$ to times $n>0$. However, we are dealing with $x(n)*u(n)$ (convolution not multiplication). $\endgroup$ – Peter K. Aug 20 '15 at 20:05
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HINT:

The only thing you need to know is that any sequence convolved with a (shifted) delta impulse is a shifted version of itself:

$$f[n]*\delta[n-k]=f[n-k]$$

Apply this to the expression $u[n]*(\delta[n]-\delta[n-1])$, and realize that what you're left with is a very simple sequence. (The result of (a) will be helpful.)

Then convolve this very simple sequence with $x[n]$ to get the final result.

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  • $\begingroup$ Too much information!!! ;-) (And +1) $\endgroup$ – Peter K. Aug 20 '15 at 20:22
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    $\begingroup$ Thanks for the help. u(n)-u(n-1) = d(n) and convoluted by x(n) gives x(n). I think both of you guys helped widening my perception of the convolution! $\endgroup$ – ricksson Aug 20 '15 at 20:32
  • $\begingroup$ @ricksson Glad to hear it! :-) $\endgroup$ – Peter K. Aug 20 '15 at 20:36
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x(n)*[d(n)-d(n-1)]=x(n)*d(n)-x(n)*d(n-1)
                              = x(n)-x(n-1)  
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