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enter image description here

If I want to do the following transform $$x[(n-1)^2]$$

I have thought about substituting each value of $n$ in $x[n]$ according to $x[(n-1)^2]$

i.e at $n=0, x[0]=1$ to put it in $x[(n-1)^2]$

$x[(0-1)^2] = 1$ and so on...

Am I right ? and if I am at $n=-4$ it will be replaced at $n=25$

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    $\begingroup$ at $n=4$ you should transform it to $n'=(4-1)^2 = 9$, so that $y[4]=x[9]$ This is effectively a nonuniform sampling performed on $x[n]$ to produce $y[n] = x[(n-1)^2]$, on the other hand at $n=-2$ the new index will be $n'=(-2-1)^2=9$ so that $y[-2]=x[9]$ again, i.e. $y[n]$ will be even symmetric about $n=1$, based on its definition. $\endgroup$ – Fat32 May 25 '16 at 2:28
  • $\begingroup$ that means at $n=-4$ $x[(-4-1)^2] = 25$. Is that corerct? $\endgroup$ – Hossam Houssien May 25 '16 at 14:52
  • $\begingroup$ No , that is quite not the case, for the sample index $n=-4$, you have the following relationship between $y[n]$ and $x[n]$ as: $y[-4] = x[(-4-1)^2]=x[25]$. Hope this is enough to see what it means. (i.e. index $n=-4$ of y[n] is transformed into index n'=(-4-1)^2=25 of x[n']) $\endgroup$ – Fat32 May 25 '16 at 15:11
  • $\begingroup$ $x[n]$ is the transformed signal, while $y[n]$ is the original signal , is that what you meant? and do you meant what happens at $y[-4]=-1$ happens at $x[25] =-1$ $\endgroup$ – Hossam Houssien May 25 '16 at 15:27
  • $\begingroup$ $x[n]$ is the original signal and $y[n]$ is new signal which is defined based on $x[n]$, by transforming the argument $n$ of $x[n]$ into $(n-1)^2$. So in effect $x[n]$ is transformed into $y[n]$. Your second cliam is right. $\endgroup$ – Fat32 May 25 '16 at 16:00
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This is what I came up for using the table below. IF anyone can confirm?**

$$ \begin{array}{c|c|c} n & (n-1)^2 & x\left[(n-1)^2\right] \\ \hline -2& (-2-1)^2=9\ & x[9]=0 \\ -1& (-1-1)^2=4\ &x[4]=0\\ 0& (0-1)^2=1& x[1]=1\\ 1&(1-1)^2=0& x[0]=1\\ 2& (2-1)^2=1& x[1]=1\\ 3& (3-1)^2=4& x[4]=0\\ 4&(4-1)^2=9& x[9]=0 \end{array} $$

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  • $\begingroup$ confirmed ! :)) $\endgroup$ – Fat32 May 25 '16 at 22:54
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as addition for Houssam's answer, we can see that for every "n" value from the graph P1.22, we replace $n$ in order to find $N$ by the relation $N=(n-1)^2 \implies x[(n-1)^2]= x[N]$ ,so we find that for $n=0,1,2 \implies x[N]=1$. And $0$ elsewhere (depends on figure p1.22) so now we represent $x[(n-1)^2]$ as :

[x[(n-1)²] graph [1

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