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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. $\endgroup$ – Dilip Sarwate 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
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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:

$$y[n]=ah[n-k]$$

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

$$x[n]=2\delta[n]-3\delta[n-2]$$

is

$$y[n]=2h[n]-3h[n-2]\tag{1}$$

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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