# Energy of a sinc signal

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.

• i might suggest explicitly defining your two functions $\operatorname{rect}(\cdot)$ and $\operatorname{sinc}(\cdot)$. – robert bristow-johnson Jan 28 '20 at 15:16
• 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 – Elena Martini Jan 28 '20 at 15:52
• 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}". – robert bristow-johnson Jan 28 '20 at 16:23
• You have an issue solving the last integral, How did you go about solving it? – Engineer Jan 28 '20 at 16:34
• $$\int_{-\infty}^{+\infty} 1 dt = [t]_ {-\infty}^{+\infty}$$ it’s the only thing in obtained 😓 – Elena Martini Jan 28 '20 at 17:06