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Fourier Question

  1. Aug 20, 2008 #1
    Hi there,

    I have been having problems with the question that I have attached, I have given it a go but unfortunately don't have a scanner to scan any of my work in but I will post it as soon as I can. Until then I was hoping that someone would be willing to give me a worked example of a similar question or just point me in the right direction as I am almost completely lost and the only part that I can answer confidently is part (a).


    Many thanks in advance


    Ram
     

    Attached Files:

  2. jcsd
  3. Aug 20, 2008 #2

    Redbelly98

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    What have you been able to do or try so far for part (b), "write a mathematical specification for the function"?
     
  4. Aug 21, 2008 #3
    I have gone as far as doing the fourier transform but again I'm not entirely sure that this is correct, I got:


    [tex]G(f) = [/tex][tex]\int^{\pi}_{-\pi}Sin(t)e^{-j}^{\omega}^{t}dt[/tex]

    I am not sure if this is as far as I need to go or not, I tried to complete the integration by using the integration by parts technique but got a very jumbled answer. The last part of the question (sketching the imaginary part of the transform) I haven't been able to do at all but I get the feeling it's something I could do if pointed in the right direction and if I manage to get the first part of the question right.
     
  5. Aug 21, 2008 #4

    Redbelly98

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    Looks good so far. From the question wording, it sounds like they also want an explicit expression for g(t).

    For the integral, it might be easier if you express e^(-jwt) in terms of sines and cosines.

    Are you permitted to use a table of integrals?

    edit:
    p.s Welcome to Physics Forums!
     
  6. Aug 21, 2008 #5
    Thank you very much for the welcome.

    The explisit expression I think is:

    [tex]g(t)=Sin(\omega t)[/tex]

    Would I be right in saying that [tex]e^{-j \omega t}[/tex] would be written as:

    [tex]e^{-j \omega t}=Cos(\omega t)-j Sin(\omega t)[/tex]?

    and yes we do have a limited table of integrals and derivatives.
     
  7. Aug 21, 2008 #6

    HallsofIvy

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    Yes, [itex]e^{-i\omega t}= cos(-\omega t)+ i sin(-\omega t)= cos(\omega t)- i sin(\omega t)[/itex] because cosine is an even function and sine is an odd function.

    You could also do this the other way around:
    [tex]sin(\omega t)= \frac{e^{\omega t}- e^{-\omega t}}{2i}[/tex]

    (Sorry, I just can't bring myself to write "j" instead of "i"!)
     
  8. Aug 21, 2008 #7

    Redbelly98

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    I see a problem with that expression. For example, it gives g(3pi/2) is -1, whereas the graph you provided clearly shows g(3pi/2) is zero.

    Yes, that's right. But HallsofIvy gives an even better way to evaluate the integral, using the substitution

    [tex]
    \sin (t) = \frac{e^{jt}-e^{-jt}}{2j}
    [/tex]

    Just as a general suggestion, it could help in your studies to have a full table of integrals handy. You can tell it's a full table if it has something like 700 or 800 integrals (or possibly more) included.
     
  9. Aug 25, 2008 #8
    Ok, I think I understand all that now but the next part of the question

    "sketch a labelled graph of the imaginary parts of the transform"

    is where I really start to struggle as I don't really know what it's asking for, do you think you could point me in the right direction as I haven't even been able to start this part?
     
  10. Aug 25, 2008 #9

    Redbelly98

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    Did you get an expression for the transform? You need to do that first. Then find the imaginary part of the expression.
     
  11. Aug 26, 2008 #10
    Ok I have tried to get an expression using that substitution but I still can't figure it out.
     
  12. Aug 26, 2008 #11

    Defennder

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    What have you got so far?
     
  13. Aug 26, 2008 #12

    Redbelly98

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    Yes, please write out the integral you have after making the substitution -- the one for sin(t) given in Post #7.
     
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