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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 $$ X(f) = \frac{ \text{sinc}(f) - e^{- i \pi f } }{ i 2 \pi f } + \frac{1}{4} \delta (f) $$

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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!):

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

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  • $\begingroup$ 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 ? $\endgroup$ – Elena Martini Apr 22 at 14:35
  • $\begingroup$ What makes you get 0? $\endgroup$ – Dan Boschen Apr 22 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 $$ \frac{ [e^{ if } + e ^{ -if} ](if) - [ e^{if} - e ^{-if} ] }{2 f^{2}} $$ $\endgroup$ – Elena Martini Apr 22 at 15:02
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    $\begingroup$ You're not stupid if you can figure all that out! Notice the cosine and sine equivalent in the expressions due to Euler's $\endgroup$ – Dan Boschen Apr 22 at 15:04
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    $\begingroup$ @ElenaMartini See my update $\endgroup$ – Dan Boschen Apr 22 at 15:09

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