Signal compression I believe I have an extremely simple question but I can't seem to figure it out.
This image shows $x[n]$ 
 
and I have to draw $y[n]=x[2n-4]$ by first doing a compression and then a time shift.
The image from the compression $y[n] = x[2n]$ is this:

I understand how the time shift is done but can't seem to figure how from the first image, the compression is done resulting in the second image. Can someone explain me how its made?
 A: The second graph is $x(2n-4),$ not $x(2n)$.  To obtain the correct graph of $x(2n)$ note that $x(n)$ is nonzero only when $n$ lies in the interval $[-4,2].$  Therefore $x(2n)$ will be nonzero only when $2n$ lies in the interval $[-4,2],$ which is to say, when $n$ lies in the interval $[-2,1].$  
Let $z(n)=x(2n).$  To compute an example, $z(-1)=x(2\cdot(-1))=x(-2)=0.$  In general, the value that $z$ takes at $n$ is the same as the value that $x$ takes at $2n.$  Geometrically, this is saying that if the horizontal distance to the vertical axis of every point on the graph of $x(n)$ is halved, then one has the graph of $z(n)=x(2n).$  This results in a horizontal compression of the graph.
The graph of $y(n)=x(2n-4)$ is the the graph of $z(n)=x(2n),$ shifted four units to the right.  This is what is seen in your second picture.  So the rising portion of the graph of $x(n)$ occurs between $n=-4$ and $n=2.$  The corresponding rising portion in the graph of $z(n)$ occurs between $n=-2$ and $x=1.$  Finally the rising portion of $y(n)$ occurs between $n=2$ and $n=5.$
