# Fourier transform of t*rect(t)

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

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!):

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

• 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 ? Apr 22, 2020 at 14:35
• What makes you get 0? Apr 22, 2020 at 14:41
• 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 $$\frac{ [e^{ if } + e ^{ -if} ](if) - [ e^{if} - e ^{-if} ] }{2 f^{2}}$$ Apr 22, 2020 at 15:02
• You're not stupid if you can figure all that out! Notice the cosine and sine equivalent in the expressions due to Euler's Apr 22, 2020 at 15:04
• @ElenaMartini See my update Apr 22, 2020 at 15:09