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This question already has an answer here:

If I have a LTI system which has impulse response $h(n) = \left(\frac{1}{2}\right)^nu(n)$ and the entry is $x[n]=u[n]-u[n-1]$ where $u[n]$ is the unit step function. In order to find the output I have to convolute the $h[n]$ with $u[n]$ and knowing $x(n)=u[n]-u[n-1]\Longrightarrow y(n)=h(n)+h(n-1)$ take the result. Is my thought right? Because I saw somewhere some examples solving differently.

for the above example the result is $y[n]=\left(\frac{1}{2}\right)^nu(n)-\left(\frac{1}{2}\right)^{n-1}u(n-1) $

(there is a similar question link but I have confused from the discussion.)

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marked as duplicate by Dilip Sarwate, jojek, MBaz, lennon310, datageist Jun 17 '15 at 0:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Is $x[n]$ now equal to $u[n]-u[n-1]$ or $u[n]+u[n-1]$? Both occur in your question. $\endgroup$ – Matt L. Jun 16 '15 at 9:37
  • $\begingroup$ I am sorry, my fault, the correct is,t: $x[n]=u[n]−u[n−1]$. Is the answer right? $\endgroup$ – Eric P. Jun 16 '15 at 9:42
  • $\begingroup$ You should correct this in your question to avoid further misunderstandings. Have a look at my answer below. $\endgroup$ – Matt L. Jun 16 '15 at 10:02
  • $\begingroup$ @DilipSarwate: That's right, and the user reminded me of that other question. So let's delete one of them ... $\endgroup$ – Matt L. Jun 16 '15 at 15:03
  • $\begingroup$ @MattL. So you could vote to close this one (or flag it for deletion).... $\endgroup$ – Dilip Sarwate Jun 16 '15 at 15:06
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If your input signal is

$$x[n]=u[n]-u[n-1]$$

then it is in fact a unit impulse: $x[n]=\delta[n]$, which has only one non-zero value for $n=0$ ($x[0]=1$). All other values are zero. So the response to this input signal is the system's impulse response, because that's the definition of the impulse response: $y[n]=h[n]$

Note that $u[n]$ is the unit step, and the response to a unit step is not the impulse response but the step response. But for this example you don't need to compute the step response, because, as already mentioned, the input signal is simply a unit impulse.

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  • $\begingroup$ so $y[n]=\left(\frac{1}{2}\right)^nu(n)-\left(\frac{1}{2}\right)^{n-1}u(n-1) $ is correct. right? the output is directly obtained from $y(n)=h(n)-h(n-1)$ $\endgroup$ – Eric P. Jun 16 '15 at 10:18
  • $\begingroup$ @EricP.: No, as I said in my answer: $y[n]=h[n]$. If the input is an impulse, the output must be the impulse response. $\endgroup$ – Matt L. Jun 16 '15 at 10:20
  • $\begingroup$ Firstly, thanks for your time. I am sorry I meant, (my fault again forgot to change the symbols in copy-paste the latex formula): $y[n]=\left(\frac{1}{2}\right)^nh[n]-\left(\frac{1}{2}\right)^{n-1}h[n-1] $ $\endgroup$ – Eric P. Jun 16 '15 at 10:23
  • $\begingroup$ @EricP.: No, $y[n]=h[n]$, I don't know how to make it any clearer. The input is an impulse, so the output is the impulse response given by $h[n]$. So no need to compute anything etc. $\endgroup$ – Matt L. Jun 16 '15 at 10:28
  • $\begingroup$ Well, I have a LTI system which has: Impulse response $h(n) = \left(\frac{1}{2}\right)^nu(n)$ and an entry $x[n]=u[n]-u[n-1]$ where $u[n]$ unit step function. The output $y[n]=\left(\frac{1}{2}\right)^nu(n)$ ? I have confused I must search for some theory about the topic :( $\endgroup$ – Eric P. Jun 16 '15 at 10:37

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