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My book give me two signals to demonstrate that the temporal translation does not alter the energy and area. It gave me

$$ x(t)=\operatorname{sinc}(t) $$

and

$$ s(t)=x(t-T)$$

and I found that energy ( with rayleigh theorem ) and area are 1 for $x(t)$. For $s(t)$ I found an area=1 but now , with Rayleigh I should demonstrate that.

$$ \int\limits_{-\infty}^{+\infty} | \operatorname{rect}(f) \,e^{- i 2 \pi f T } |^{2} \ \mathrm{d}f $$

As I made for $x(t)$,

$$ \operatorname{rect}(t) = |\operatorname{rect}(t)|^{2} $$

but solving the last integral this didn’t give me 1. Thank you so much for the help.

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    $\begingroup$ i might suggest explicitly defining your two functions $\operatorname{rect}(\cdot)$ and $\operatorname{sinc}(\cdot)$. $\endgroup$ – robert bristow-johnson Jan 28 at 15:16
  • $\begingroup$ This helps me , I wrote $$ e^{-i 2 \pi f T} $$ as $$ cos ( 2 \pi f T ) - i sen ( 2 \pi f T) $$. So in this terms $$ | e^{-i 2 \pi f T } | = 1 $$ so now I obtain an integer of 1 $\endgroup$ – Elena Martini Jan 28 at 15:52
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    $\begingroup$ i can't edit your comments (i can edit your question). try using a backslash before "sin" or "cos" or "log" or whatever function and $\LaTeX$ will show it as a function. the functions $\operatorname{rect}$ and $\operatorname{sinc}$ don't seem to be in LaTeX's list of known functions so you need to use "\operatorname{rect}". $\endgroup$ – robert bristow-johnson Jan 28 at 16:23
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    $\begingroup$ You have an issue solving the last integral, How did you go about solving it? $\endgroup$ – Engineer Jan 28 at 16:34
  • $\begingroup$ $$ \int_{-\infty}^{+\infty} 1 dt = [t]_ {-\infty}^{+\infty} $$ it’s the only thing in obtained 😓 $\endgroup$ – Elena Martini Jan 28 at 17:06

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