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The discrete signal is provided in the following table:

$$\begin{array}{|c|ccccccc|}\hline n& -3 & -2 & -1 & 0 & 1 & 2 & 3\\ \hline x[n] & 3 & 2 & 1 & 0 & 1 & 2 & 3 \\ \hline \end{array}$$

The book is asking to determine $x[3n-1]$. This involves shifting by one and then scaling by $3$. My answer to this problem is:

$$\begin{array}{|c|cc|}\hline 3n-1 & 0 & 1\\ \hline x[3n-1]& 1 & 2 \\ \hline \end{array}$$

However the answer in the book is:

$$\begin{array}{|c|ccc|}\hline 3n-1 & -1 & 0 & 1\\ \hline x[3n-1]& 0 & 1 & 2 \\ \hline \end{array}$$

Which answer is correct?

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The sequence $x[n]$ is only non-zero in the range $-3\le n\le 3$. So if you change the index to $3n-1$, in order to remain inside that region, $n$ can only take the values $n=0$ (giving $3n-1=-1$), and $n=1$ (giving $3n-1=2$). This gives

$$x[3n-1]=\begin{cases}x[-1]=1,&n=0\\x[2]=2,&n=1\\ 0,&\text{otherwise}\end{cases}$$

So if the exercise is as you stated it, then your solution is correct, apart from the fact that it is not $3n-1$ but $n$ that takes the values $0$ and $1$.

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  • $\begingroup$ I meant by $3n-1$ the new index which only takes two values but you are right. $\endgroup$ – CroCo Feb 27 '16 at 12:56

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