0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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: $$\frac{1}{T}e^{-\frac{t-T}{T}}u(t-T) \longleftrightarrow \frac{1}{1 + j2\pi fT}e^{-j2\pi fT}$$

|improve this answer|||||
$\endgroup$
  • 1
    $\begingroup$ 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 🙈 $\endgroup$ – Elena Martini Jan 27 at 14:46
  • 1
    $\begingroup$ What is this "notorious" book you are using? $\endgroup$ – GKH Jan 27 at 19:00
  • $\begingroup$ “Segnali analogici e sistemi lineari” by Armando Vannucci , my teacher $\endgroup$ – Elena Martini Jan 27 at 21:35
1
$\begingroup$

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

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

|improve this answer|||||
$\endgroup$

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.