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I'm confused with the time-shifting operation post the time-reversal operation when performing convolution. Let's say we were to convolve $x(t)$ and $h(t)$, so I would have the term $x(k)$ and $h(t-k)$ to solve for. In $h(t-k)$, we first reverse $h(k)$ to $h(-k)$ and then shift it to $t$ units. Ideally, we should shift t units to the left but I find examples where they have moved $t$ units to the right instead.

I'm confused because, for any $x(t+n)$, we shift $n$ units to the left of $t$, so for $x(-t+n)$, why we are moving n units to the right of $-t$?

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Just think of the value of $t$ where $x(0)$ appears. For $x(t+t_0)$ it is at $t=-t_0$, which corresponds to a left shift if $t_0>0$. However, for $x(-t+t_0)$ the value $x(0)$ occurs at $t=t_0$, which is to the right of its original position if $t_0>0$.

You can think of deriving $x(-t+t_0)$ from $x(t)$ in two different ways:

  1. invert the time axis: $x(-t)$
  2. replace $t$ by $t-t_0$, i.e., shift to the right if $t_0>0$: $x(-(t-t_0))=x(-t+t_0)$

Or

  1. shift $x(t)$ to the left (for $t_0>0$): $x(t+t_0)$
  2. invert the time axis: $x(-t+t_0)$: this flips the shifted function from the left to the right of the original function (again, if $t_0>0$).
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  • $\begingroup$ It is very much clear now. Thanks for the detailed description. Cheers :) $\endgroup$ Aug 13 '20 at 12:59

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