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As I understand it, shifted output means first getting the output and then shifting it. We feed in $x[n]$, and get our output as $x[2n]$. Then we shift it by $k$ to get $x[2n-k]$. This is the shifted output.

Now, for output of the shifted input, we shift the input to get $x[n-k]$, then feed it as input to get output as $x[2(n-k)]=x[2n-2k]$

However, the text in the image has these answers reversed. Is there something I'm not getting?


This is all about consequently substituting variables. Shifting by $t_0$ means replacing the variable $t$ by $t-t_0$. If $y(t)=x(2t)$, then $y(t-t_0)=x(2(t-t_0))=x(2t-2t_0)$.

If you have a shifted input $x_2(t)=x(t-t_0)$, then the output is $y(t)=x_2(2t)$, i.e., the variable $t$ is multiplied by $2$. Replacing $t$ by $2t$ in $x(t-t_0)$ results in $x(2t-t_0)$.

So the slide is correct.

EDIT: Concerning shifting the output, imagine a triangular input $x(t)$ with its apex at $t=0$. The output is a compressed version of the input, i.e., it is also triangular with its apex at $t=0$. If the output is shifted by $t_0$ the apex is shifted to $t=t_0$. The function $x(2t-t_0)$ has its apex at $t$ satisfying $2t-t_0=0$, i.e., at $t=t_0/2$, so the correct formula for the shifted output is $x(2t-2t_0)$.

  • $\begingroup$ I still don't get it. First we get get the output x[2n] and then shift to get the shifted output: x[2n-k]. Second, we first shift the input to get x[n-k] and then provide this as input to get x[2(n-k)] as output $\endgroup$ – Ryder Rude Nov 9 '19 at 11:53
  • $\begingroup$ @RyderRude: I added an example concerning the shifted output to show that your argument is wrong. If you don't see it, just draw a simple example and it should become obvious. $\endgroup$ – Matt L. Nov 9 '19 at 13:03

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