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How to sketch the following discrete-time signal:
\begin{gather*} x[(n-1)^2] \end{gather*} for the signal given below: enter image description here

Any help would be much appreciated.

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I give you some hints and then you can solve this homework.

  • Your $x[n]$ has only $8$ nonzero values. Figuring out what happens to them (in an exhaustive way) is not difficult.

  • Consider $(n-1)^2$ and change $n$ over $(-\infty, \infty)$. You can see that it is always non-negative. It means that we never can have $(n-1)^2=-1$. Hence the samples of $x[n]$ for $n=-3,-2,-1$ are never used. We can only have samples of $x[n]$ at $n=0,1,2,3$.

  • Solve $(n-1)^2=0$, $(n-1)^2=1$,$(n-1)^2=2$, $(n-1)^2=3$. Only accept those results that lead to an integer $n$. Then you can see at what $n$ in $x[(n-1)^2]$ we have what value of $x[n]$. For instance, $(n-1)^2=\color{red}1\Rightarrow n=0,2$. It means in $x[(n-1)^2]$ the samples at $n=0$ and $n=2$ are equal to $x[\color{red}1]=1$.

  • Except at the points mentioned above, the transformed signal would be zero.

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    $\begingroup$ Thank you very much for the explanations. I solved the equations and got n=1 for (n-1)^2=0 , n=0,2 for (n-1)^2=1. For the last two equations I get no integer values. Does that mean that my new signal has a value of 1 at points n=0,2 and a value of 0 at point n=1 (and for the other points)? @LaurentDuval $\endgroup$ – Crypted_39 Jul 17 '17 at 11:55
  • $\begingroup$ Almost correct. You only have three nonzero terms, but $x[0]$ is also equal to one. Not zero. So your transformed signal has only three nonzero samples equal to one at $n = 0,1,2$. $\endgroup$ – msm Jul 17 '17 at 13:21
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The set of indices where $x$ is non-zero is $(-4,\ldots,3)$. You can rephrase the question as:

which $n$ (possibly real) are such as $(n-1)^2$ is an integer taking values in $(-4,\ldots,3)$

In other words, which integers in $(-4,\ldots,3)$ are (non-negative) squares?

Here, this is quite simple, only $0$ and $1$ are perfect squares, attained when $n=1$ (for $0^2=0$) and $n=0$ or $n=2$ (for $1^2=1$).

So if $y[n] =x\left[(n-1)^2\right]$:

  • $y[0] =x[1] $,
  • $y[1] =x[0] $,
  • $y[2] =x[1] $.
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