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I have to find the Fourier transform of $$ x(t)= \frac{1}{T}e^{-\frac{t-T}{T}}u(t-T) $$

First I applied traslation property , so $$ F[x(t-T)] = X(f) e^{-i 2 \pi f T} $$

after I applied time scaling property , and I obtained $$ F[x(\frac{t}{T})] = |T| X(Tf) e^{-i 2 \pi f T} $$

Now I calculate $$ \int_{T}^{+\infty } \frac{1}{T}Te^{-\frac{t-T}{T}}u(t-T) e^{-i 2 \pi f t } $$ that’s results $$ \frac{T}{1+ 2 i\pi f t } $$

So I obtained that the final Fourier transformation should be $$ X_f = |T| \frac{T}{1 + 2 i \pi f T} e^{- 2 i \pi f T } $$

On my book the result is the same except for T|T| but I don’t know what’s wrong. Thank you so much

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2 Answers 2

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If you already know the - quite well-known :) - FT pair $$e^{-at}u(t) \longleftrightarrow \frac{1}{a+j2\pi f}$$ then by setting $a = 1/T$, you get $$e^{-\frac{t}{T}}u(t) \longleftrightarrow \frac{1}{\frac{1}{T} + j2\pi f}$$ Time delay property $$x(t-t_0) \longleftrightarrow X(f)e^{-j2\pi ft_0}$$ can be used now by setting $t_0 = T$, and then $$x(t-T) = e^{-\frac{t-T}{T}}u(t-T) \longleftrightarrow \frac{1}{\frac{1}{T} + j2\pi f}e^{-j2\pi fT}$$ Adding your $1/T$ constant in front, you get $$\frac{1}{T}e^{-\frac{t-T}{T}}u(t-T) \longleftrightarrow \frac{\frac{1}{T}}{\frac{1}{T} + j2\pi f}e^{-j2\pi fT}$$ Finally, multiplying both numerator and denominator by $T$ you get what you want: $$\frac{1}{T}e^{-\frac{t-T}{T}}u(t-T) \longleftrightarrow \frac{1}{1 + j2\pi fT}e^{-j2\pi fT}$$

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    $\begingroup$ thank you very much ! now I understood where I was wrong and I was able to solve the next exercise too. I will get this transform tattooed on my face 🙈 $\endgroup$ Commented Jan 27, 2020 at 14:46
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    $\begingroup$ What is this "notorious" book you are using? $\endgroup$
    – GKH
    Commented Jan 27, 2020 at 19:00
  • $\begingroup$ “Segnali analogici e sistemi lineari” by Armando Vannucci , my teacher $\endgroup$ Commented Jan 27, 2020 at 21:35
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Properties are great if you use them properly. In this case it was easier to just crank out the integral as it is not terrible:

$X(f)=\frac{e^1}{T}\int_T^{\infty} e^{-\big(\frac{t-T}{T}\big)}e^{-j2\pi ft}dt$

Combine the terms to get

$X(f)=\frac{e^1}{T}\int_T^{\infty}e^{-t(j2\pi f + \frac{1}{T})}dt$

Now do the integral

$X(f)=\frac{e^1}{T}\bigg[\frac{e^{-t(j2\pi f + \frac{1}{T})}}{-j2\pi f - \frac{1}{T}} \bigg]_T^{\infty}=\frac{e^1}{T}\bigg[\frac{e^{-j2\pi fT} e^{-1}}{j2\pi f + \frac{1}{T}} \bigg]=\frac{e^{-j2\pi fT}}{j2\pi fT+1}$

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