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Hi sorry this question might sound silly but,

a signal $$x[n]=[ 6 \: 4\: 0\: 2]$$ what is $$x[n-2]$$ then?

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  • $\begingroup$ why is the six bold? $\endgroup$ – Marcus Müller Feb 8 '20 at 20:20
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    $\begingroup$ Typically: [0 0 6 4 0 2] $\endgroup$ – Hilmar Feb 8 '20 at 23:35
  • $\begingroup$ the 6 is bold because its the first value of the signal ie x[0]. $\endgroup$ – J Leo Feb 9 '20 at 22:50
  • $\begingroup$ Not silly. A potential for confusion for the beginners who make mistakes shifting the origin to $x[n]=[6\;4\;\textbf{0}\; 2]$ $\endgroup$ – jomegaA Feb 10 '20 at 9:30
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The problem with representing a discrete time just as a list like: $$x[n]=[6, 4, 0, 2]$$ is that you don't have a time reference.

The first index $n=0$ or the sample in $\textbf{bold}$ or $\underline{\mathrm{underline}}$ is usually taken as the start of the sequence.

For example, $$x[n]=[\mathbf{6}, 4, 0, 2]$$ or $$x[n]=[\underline{6}, 4, 0, 2]$$

So $x[n-2]$ is a delay of two samples, and $x[0]$ now happens two samples later: $$x[n]=[\underline{0}, 0, 6, 4, 0, 2]$$

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