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Came across this example in class but I'm not sure how the expression $x(n)$ was derived.

enter image description here

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  • $\begingroup$ I don't think $x(n)$ was "derived" -- it's a signal that corresponds to the drawing. Can you clarify what you are confused about? $\endgroup$ – MBaz Aug 28 '18 at 16:40
  • $\begingroup$ I’m confused about how it corresponds to the drawing $\endgroup$ – Carpetfizz Aug 28 '18 at 16:41
  • $\begingroup$ OK. My suggestion is, go to your instructor's office hours. $\endgroup$ – MBaz Aug 28 '18 at 16:43
  • $\begingroup$ $x(n)$ consists of $3$ scaled impulses, so you can express it in terms of weighted and shifted deltas. What exactly don't you understand? $\endgroup$ – Matt L. Aug 28 '18 at 16:46
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The plot that shows the sequence $x[n]$ is also called as the lollipop plot and it shows the individual samples of $x[n]$ with their weights and locations.

Each lollipop represents a discrete-time unit-impulse signal, $\delta[n]$, which is located at the position it's shown and weighted by the value of the signal there. The superposition sum of all those impulses compose the signal $x[n]$.

Now, your signal has 3 lollipops located at $n = \{-1,0,1\}$ and weighted by $\{1,2,3\}$ respectively.

Since an impulse located at the position $n=d$, and weighted by $a$, is given by $$ a \cdot \delta[n-d]$$

by using this, you can compose the signal $x[n]$ from its lollipops as:

$$ x[n] = 1 \cdot \delta[n-(-1)] + 2 \cdot \delta[n-(0)] + 3 \cdot \delta[n-(1)] $$

$$ x[n] = 1 \cdot \delta[n+1] + 2 \cdot \delta[n] + 3 \cdot \delta[n-1] $$

Note that a generalization of this also yields the sifting-sum decomposition of any signal $x[n]$ as

$$x[n] = \sum_{k=-\infty}^{\infty} x[k] \delta[n-k] $$

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  • $\begingroup$ I don't get it, could you maybe explain it with Mars bars? $\endgroup$ – Matt L. Aug 28 '18 at 20:10
  • $\begingroup$ what's a Mars bar ? $\endgroup$ – Fat32 Aug 28 '18 at 20:11
  • $\begingroup$ loooool.. I clicked on the link !!! @MattL. $\endgroup$ – Fat32 Aug 28 '18 at 20:14
  • $\begingroup$ Thanks @Fat32 . Where does $\alpha \cdot \delta [n - d]$ come from and why is it defined that way? $\endgroup$ – Carpetfizz Aug 28 '18 at 20:34
  • $\begingroup$ So $a \cdot \delta[n-d]$ represents a unit-impulse $\delta[n]$ shifted to location $n=d$ and multiplied by (weighted by) a constant $a$. The specific values are obtained from the plot as you have provided. $\endgroup$ – Fat32 Aug 28 '18 at 20:36

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