enter image description here

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.