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Find the Fourier Transform of the function t*(sent/pi*t)^2

  1. Jun 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the FT of the following signal

    The function is: [tex] f(t) = t(\frac{sen(t)}{t\pi})^2 [/tex]

    2. Relevant equations
    Fourier transform: [tex] F(\omega)= \int_{-\infty}^\infty f(t)e^{-jt\omega} [/tex]

    My attempt began with this fourier transform, and that's my goal:

    [tex] F[tf(t)]= jF'(\omega) [/tex]

    3. The attempt at a solution
    English is not my first language but i'll do my best to explain my attempt.

    With [itex] F[tf(t)]= jF'(\omega) [/itex] being [itex] f(t)=(\frac{sen(t)}{t\pi})^2 [/itex].

    It was clear that i had to find the fourier transform F(ω) of f(t) for further derivation and multiplication with "j" to finish the problem, so i said that if [itex] g(t)=\frac{sen(t)}{t\pi} [/itex], it would be [itex] f(t) = g^2(t) = g(t)g(t) [/itex] so... that F(ω) that i need is the FT of a product.

    So i concluded that the FT of f(t) it's the convolution of the FT of:
    [itex] g(t)=\frac{sen(t)}{t\pi} [/itex] with itself, in other words --> [itex] G(\omega)*G(\omega) = F(\omega) [/itex]

    So in definitive, i need the FT of [itex] g(t)=\frac{sen(t)}{t\pi} [/itex]. (if i'm not wrong)

    I didn't know how to calculate G(ω) directly but i can try to solve G'(ω) and then integrate the result, so i did this:

    [tex] \frac{dG(\omega)}{d\omega}= \int_{-\infty}^\infty \frac{sen(t)}{t\pi}\frac{de^{-jt\omega}}{d\omega}dt = \int_{-\infty}^\infty \frac{sen(t)}{t\pi}(-jt)e^{-jt\omega}dt = \frac{-j}{\pi}\int_{-\infty}^\infty sen(t)e^{-jt\omega}dt [/tex]

    [tex] \frac{-j}{\pi} \int_{-\infty}^\infty \frac{e^{jt}-e^{-jt}}{2j}e^{-jt\omega}dt = \frac{-1}{2\pi} \int_{-\infty}^\infty e^{jt-jt\omega}-e^{-jt-jt\omega}dt = \frac{-\delta(\omega + 1) + \delta(\omega - 1)}{2\pi} = G'(\omega) [/tex]

    Now i integrate G'(ω), the particularity of this is that if H(ω) is the Heaviside function, and δ(ω) is delta dirac's function i can make an equation that says that:

    [itex] H'(\omega)=\delta(\omega) \quad then \quad H(\omega) = \int \delta(\omega) [/itex]
    (I hope this is ok (?))

    So i have at last G(ω):
    [tex] G(\omega) = \frac{-H(\omega + 1) + H(\omega - 1)}{2\pi} [/tex]

    So i'll be honest, at this point, i didn't know how to do the convolution by myself (i'm learning by myself, the teacher's class was just scratching the surface of the theme so i don't understand it fully, but even so, this is part of a homework and costs points of it), so i consulted with wolfram alpha, here's the link --> http://www.wolframalpha.com/input/?...fx&f4=y&f=ConvolveCalculator.variable2\u005fy

    The result was [tex] F(\omega)= G(\omega)*G(\omega) = \frac{1 }{4\pi^2}((\omega-2)H(\omega-2)-2\omega H(\omega)+(\omega+2)H(\omega+2)) [/tex]

    Nevermind. I made a huge mistake so i edited my post, i need to derive that F(ω) to obtain F'(ω) and fulfill the TF that i defined in the beginning [itex] F[t(\frac{sen(t)}{t\pi})^2]= jF'(\omega) [/itex]. But this is ok? all that i did? i need some orientation... and there's another way to do all of this?
     
    Last edited: Jun 24, 2014
  2. jcsd
  3. Jun 24, 2014 #2

    maajdl

    User Avatar
    Gold Member

    I guess that "sen" is your symbol for the sine?
    The fast track would be to use formula 203 of this wikipedia page:

    http://en.wikipedia.org/wiki/Fourier_transform

    The derivative of a triangular function should not be a problem!
     
  4. Jun 24, 2014 #3
    Yeah, that's my sine sorry, in spanish it's "seno" so i use sen, my apologies if the notation was confusing, thank you. And... the reason why i didn't use the formula 203 is because in the remarks of the formula, they clarify that their sinc(x) is defined as
    [tex] sinc(x) = \frac{sin(x\pi )}{x\pi} [/tex] And mine would be [tex] sinc(x) = \frac{sin(x)}{x} [/tex] So i preferred to not touch that definition. But can i use that formula?
     
  5. Jun 25, 2014 #4

    maajdl

    User Avatar
    Gold Member

    Of course, you can use that formula.
    You simply need to re-arrange it a little bit.
    Make a change of variable y = Pi x and re-write formula 203 with your own variable.

    Also, you may like to derive formula 203 for your own training, which should not be too difficult.
    Doing so, you may also come close to solve your problem fully by your own means which is more fun.
     
  6. Jun 25, 2014 #5
    Hahaha yes, the feeling after you complete something that you worked on for hours straight is irreplaceable, thank you again! so in my case, f(t) being: [tex] \frac{sin(t)}{t\pi}\quad if \quad y=t\pi \quad then \quad t=\frac{y}{\pi} --> f(\frac{y}{\pi}) = (\frac{sin(\frac{y}{\pi})}{\frac{\pi}{\pi}y})^2 = (\frac{sinc(\frac{y}{\pi})}{\pi})^2 = \frac{(sinc(\frac{y}{\pi})^2)}{\pi^2} [/tex] Then, using the formula 203 (non unitary, angular frequency) my result is: [tex] F(\omega) = \frac{1}{\frac{1}{\pi}}tri(\frac{\omega}{2\pi \frac{1}{\pi}}) = \pi tri(\frac{\omega}{2}) [/tex] And it's derivative is: [tex] F'(\omega) = \pi \frac{ rect(\frac{\omega +1}{2}) - rect(\frac{\omega -1}{2})}{2} [/tex] Being "tri" the triangular function and "rect" the rectangular.

    And now i apply it to my initial transform. I suppose that that's ok isn it?
     
    Last edited: Jun 25, 2014
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