0
$\begingroup$

In my previous post I asked for help for a Fourier transform of $$ t \text{rect} ( t- \frac{1}{2} ) $$ and I think I’ve understand the process. Now I’ve 2 another similar Fourier transform to do , I already solved both , but I don’t have the correct result. Can someone tell my if the 2 results I obtained are wrong ? Thank you

If $$ x(t) = t \text{rect}(t) $$ I obtained $$ X(f) = \frac {\text{sinc}(f) - \frac{1}{2} [e^{i \pi f } + e^{-i \pi f } ]}{i 2 \pi f } + \frac{1}{4} \delta(f) $$

And for $$ x(t) = \Big(t + \frac{1}{2}\Big) \text{rect}(t) $$ I obtained $$ X(f) = \frac{ \text{sinc}(f) - e^{- i \pi f } }{ i 2 \pi f } + \frac{1}{4} \delta (f) $$

$\endgroup$

1 Answer 1

0
$\begingroup$

Hint:

Derivative in frequency property of FT:

$$\frac{d}{d\omega}jF(\omega) \rightarrow tf(t)%$$

Time shift property of FT:

$$f(t-\tau) \rightarrow F(\omega)e^{-j\omega t_o}$$

FT of rect: $F(\omega) = sinc(\omega/2)$

For practice if you are learning FT I recommend deriving the properties above to confirm you can get them from the FT directly using (practice makes perfect!):

$$F(\omega) = \int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt$$

$\endgroup$
6
  • $\begingroup$ But with this I obtain that the Fourier transform of $$ t rect(t) $$ is $$ i \frac{d}{df} [sinc (f) ] $$ And that derivative of a sinc isn’t 0 ? $\endgroup$ Apr 22, 2020 at 14:35
  • $\begingroup$ What makes you get 0? $\endgroup$ Apr 22, 2020 at 14:41
  • $\begingroup$ I obtained 0 because I’m stupid :/ $$ sinc f = \frac{sen f }{f} $$ so I found that the derivative of this is $$ \frac{ [e^{ if } + e ^{ -if} ](if) - [ e^{if} - e ^{-if} ] }{2 i f^{2}} $$ now using the property should be $$ \frac{ [e^{ if } + e ^{ -if} ](if) - [ e^{if} - e ^{-if} ] }{2 f^{2}} $$ $\endgroup$ Apr 22, 2020 at 15:02
  • 1
    $\begingroup$ You're not stupid if you can figure all that out! Notice the cosine and sine equivalent in the expressions due to Euler's $\endgroup$ Apr 22, 2020 at 15:04
  • 1
    $\begingroup$ @ElenaMartini See my update $\endgroup$ Apr 22, 2020 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.