# Simple discrete time transformation $x[(n-1)^2]$

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 $n=25$

• 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. – Fat32 May 25 '16 at 2:28
• that means at $n=-4$ $x[(-4-1)^2] = 25$. Is that corerct? – Hossam Houssien May 25 '16 at 14:52
• 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']) – Fat32 May 25 '16 at 15:11
• $x[n]$ 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 $x[25] =-1$ – Hossam Houssien May 25 '16 at 15:27
• $x[n]$ 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. – Fat32 May 25 '16 at 16:00

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