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How to calculate this inverse Fourier Transform?

  1. Apr 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Take the inverse Fourier Transform of

    [tex]5[\delta(f+100)+\delta(f-100)]\bigg(\frac{180+j2\pi f*0.0135}{1680+j2\pi f*0.0135}\bigg)[/tex]


    2. Relevant equations

    [tex]g(t)=\int_{-\infty}^{\infty} G(f)e^{j2\pi ft}dt[/tex]

    3. The attempt at a solution

    [tex]g(t)=\int_{-\infty}^{\infty} 5[\delta(f+100)+\delta(f-100)]\bigg(\frac{180+j2\pi f*0.0135}{1680+j2\pi f*0.0135}\bigg) e^{j2\pi ft}dt[/tex]

    My professor gave the hint that the dirac delta sampling property can be used, but I don't see how since I'm taking the inverse Fourier Transform and not just a regular integral.

    I've been reading about the different properties of the FT and dirac delta function, but simply can't figure out how to proceed. Any suggestions?
     
    Last edited by a moderator: Apr 6, 2014
  2. jcsd
  3. Apr 5, 2014 #2

    HallsofIvy

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    ?? The inverse Fourier Transform is a "regular integral"! And the "Dirac delta sampling property" is that [itex]\int_{-\infty}^\infty \delta(x- a)f(x)dx= f(a)[/itex]. That should make this problem easy.
     
  4. Apr 6, 2014 #3
    Ah.. I'm not sure what I was thinking, lol :-)

    OK, so having this equation:

    [tex]g(t)=\int_{-\infty}^{\infty} 5[\delta(f+100)+\delta(f-100)]\bigg(\frac{180+j2\pi f*0.0135}{1680+j2\pi f*0.0135}\bigg) e^{j2\pi ft}dt[/tex]

    When using the sifting/sampling property of the Dirac Delta function:

    [tex]\int_{-\infty}^\infty \phi (t) \delta(t - T)dt = \phi (T)[/tex]

    Should I replace [itex]f[/itex] or [itex]t[/itex] in the above equation with [itex]\pm 100[/itex]? It seems like I should replace the [itex]f[/itex]:

    [tex]g(t)=5\bigg (\frac{180+j2\pi f*0.0135}{1680+j2\pi f*0.0135}\bigg) e^{j2\pi ft} \bigg |_{f=100}+5\bigg (\frac{180+j2\pi f*0.0135}{1680+j2\pi f*0.0135}\bigg) e^{j2\pi ft} \bigg |_{f=-100}[/tex]

    [tex]g(t)=5\bigg (\frac{180+j2\pi *1.35}{1680+j2\pi *1.35}\bigg) e^{j200\pi t}+5\bigg (\frac{180-j2\pi *1.35}{1680-j2\pi *1.35}\bigg) e^{-j200\pi t}[/tex]

    But I'm not sure if that's correct?
     
    Last edited by a moderator: Apr 6, 2014
  5. Apr 6, 2014 #4
    Does anyone know if what I did is correct? Operating under that assumption, I tried converting the rectangular coordinate complex numbers to polar form:

    [tex]g(t)=5\bigg (\frac{180.2\angle 2.70^\circ}{1680.02\angle 0.29^\circ}\bigg) e^{j200\pi t}+5\bigg (\frac{180.2\angle -2.70^\circ}{1680.02\angle -0.29^\circ}\bigg) e^{-j200\pi t}[/tex]

    [tex]g(t)=5\bigg (\frac{180.2\angle 2.70^\circ}{1680.02\angle 0.29^\circ}\bigg) e^{j200\pi t}+5\bigg (\frac{180.2\angle -2.70^\circ}{1680.02\angle -0.29^\circ}\bigg) e^{-j200\pi t}[/tex]

    [tex]g(t)=5(0.11\angle 2.41^\circ) e^{j200\pi t}+5(0.11\angle -2.41^\circ) e^{-j200\pi t}[/tex]

    I think I might be doing something wrong though, because I think the answer is supposed to work out to be:

    [tex]1.1\cos(2\pi (100)t+2.41^\circ)[/tex]
     
    Last edited by a moderator: Apr 6, 2014
  6. Apr 6, 2014 #5

    vela

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    The integral should be with respect to ##f##, not ##t##.

    That's right.
     
  7. Apr 6, 2014 #6

    vela

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    You're right so far. You just have to keep going with the simplification.

    It might be simpler if you note at the start that the two terms are conjugates and use the fact that ##z+\bar{z} = 2\text{Re}(z)##.
     
  8. Apr 6, 2014 #7
    Hmm... I've been trying to simplify it and am not sure how to proceed. I tried changing the polar portion to ##\cos## and using Euler's identity to change the exponentials to ##\cos## and ##\sin##, which gives:

    ##g(t) = 5[0.11\cos(\omega t + 2.41^\circ)[\cos(200\pi t) + j\sin(200\pi t)] + 0.11\cos(\omega t - 2.41^\circ)[\cos(200\pi t) - j\sin(200\pi t)]]##

    But I feel like I'm on the wrong track here. If I multiply that out then it becomes a mess. Did I go in the wrong direction with this?

    Really appreciate your help :)
     
  9. Apr 6, 2014 #8

    vela

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    Hint: ##0.11\angle 2.41^\circ = 0.11 e^{j 2.41^\circ}##
     
  10. Apr 6, 2014 #9
    Did you by any chance mean:

    ##0.11\angle 2.41^\circ = Re(0.11 e^{j 2.41^\circ})##? Otherwise, I'm really confused :smile:

    Because:

    ##0.11\angle 2.41^\circ = 0.11\cos(\omega t + 2.41^\circ)##

    And:

    ##0.11 e^{j 2.41^\circ} = 0.11\cos(\omega t + 2.41^\circ) + j0.11\sin(\omega t + 2.41^\circ)##
     
  11. Apr 6, 2014 #10

    vela

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    What I mean is express the fractions in polar form, i.e. ##re^{j\theta}##.
     
  12. Apr 6, 2014 #11
    Like this?

    [tex]g(t)=5\bigg (\frac{180.2\angle 2.70^\circ}{1680.02\angle 0.29^\circ}\bigg) e^{j200\pi t}+5\bigg (\frac{180.2\angle -2.70^\circ}{1680.02\angle -0.29^\circ}\bigg) e^{-j200\pi t}[/tex]

    [tex]g(t)=5(0.11\angle 2.41^\circ) e^{j200\pi t}+5(0.11\angle -2.41^\circ) e^{-j200\pi t}[/tex]

    [tex]g(t)=5Re(0.11e^{j2.41^\circ})e^{j200\pi t} + 5Re(0.11e^{-j2.41^\circ})e^{-j200\pi t}[/tex]

    I can't combine the exponentials being multiplied together (because one is only the real part), can I?
     
    Last edited by a moderator: Apr 6, 2014
  13. Apr 6, 2014 #12

    vela

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    OK, let's back up one more step. Convert the complex numbers from rectangular to polar form then simplify.
     
  14. Apr 6, 2014 #13

    micromass

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    Nat3, I admire that you write all these equations in LaTeX. But one hint: if you want your formula to have its own line, then you should use tex, and not itex. I have changed this for you in your posts. The result really is a lot easier on the eyes. Check the difference between post 4 and the quoted part in post 5, for example.
     
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