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Is it possible to compute the output from one linear time invariant system which unit impulse sequence is:

$$h(n) = \left(\frac{1}{3}\right)^nu(n)$$

and the entry in the system is:

$$x(n)=u(n)-u(n-2)$$

where $u(n)$ is the unit step sequence?

Or I need more information (so something missing)

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  • $\begingroup$ LTI systems are completely characterized by their impulse response, so it's possible to compute the output from what is given. Try to simplify your input signal (it's actually zero almost everywhere) and try to see that the output signal is a simple combination of the impulse response and a shifted version of the impulse response. $\endgroup$ – Matt L. Jun 27 '14 at 14:39
  • $\begingroup$ Matt thanks for your answer I truly appreciate, can you explain me, why is zero everywhere ? because as I know unit step sequence is 1 for n bigger or equal to zero $\endgroup$ – Aris Chrisak Jun 27 '14 at 14:58
  • $\begingroup$ Yes, but you subtract two steps. The first one goes up and starts at $n=0$, and the second one goes down (because of the negative sign) and starts at $n=2$, so your input signal can be written as $x(n)=\delta(n)+\delta(n-1)$. $\endgroup$ – Matt L. Jun 27 '14 at 15:01
  • $\begingroup$ I Think I am confused, for n=0: x(0)= u(o)-u(-2)=1-0=1 right? because u(n) is the unit step sequence $\endgroup$ – Aris Chrisak Jun 27 '14 at 15:13
  • $\begingroup$ Yes, and the same for $x(1)$. For all other values of $n$, $x(n)$ is zero. $\endgroup$ – Matt L. Jun 27 '14 at 16:07
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HINT:

$$x(n)=\delta(n)+\delta(n-1)\Longrightarrow y(n)=h(n)+h(n-1)$$

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