In the following question and answer I kind of get it how it is calculated at surface but I don't get it fully. Can someone please explain the answer in depth?enter image description here

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    $\begingroup$ Do you know what zero state response means? $\endgroup$
    – Batman
    Mar 1 '14 at 6:22
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    $\begingroup$ The input to the system consists of $2$ at time $n = 0$ and $-3$ at time $n = 2$. The impulse response is $[1\quad 1\quad 4]$. Write out a table as shown in this answer to a related question and verify that you get what is given as the answer. $\endgroup$ Mar 1 '14 at 15:21
  • $\begingroup$ As Dilip as stated the impulse response is $[1,1 ,4]$, so the input $2\delta(n)$ will produce $2[1,1 ,4]$, similarly the input $-3\delta(n-2)$ will produce $-3[0,0,1,1 ,4]$. The total output is then give by summing these two to give $[2,2,5,-3,-12]$. You may want to look at this answer $\endgroup$
    – David
    Mar 4 '14 at 14:41

The answer to your question is essentially contained in Dilip's comment, but I guess as a beginner you need some more help. First, it's important to realize that the response of an LTI system to a scaled and shifted impulse $$x[n]=a\delta(n-k)$$ is simply a scaled and shifted version of its impulse response:


Furthermore, the response to a sum of signals equals the sum of the responses to the individual components of the input signal. Consequently, the response to the given input signal




Now you just have to express the impulse response $h[n]$ in terms of $\delta$-impulses. By noting that $$u[n]-u[n-3]=\delta[n]+\delta[n-1]+\delta[n-2]$$ you get

$$h[n]=\delta[n]+\delta[n-1]+4\delta[n-2] \tag{2}$$

Plugging (2) into (1) gives the desired result.


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