Fourier Transforms

Discussion in 'Homework Help' started by Kayne, Jul 28, 2011.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    Hi All,

    I am having some trouble with a question regarding Fourier Transform

    I am not asking for this to be solved just a push in the right direction as I am not sure where to start this problem.

    The attachment has the details and the question that relates to this is as follows

    Show that the Fourier Transform (spectrum) of this waveform can be written as the sum of two similar spectra centered on radian frequencies of +ωo and -ωo, respectively. Deduce the equation which describes the shape of these. Explain why the spectrum has this form.

    Thanks for your help
     
  2. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    I have found an answer which I think I am on the right track. I have attached it as Q1.

    Thanks for your time

    Kayne
     
    • Q1.pdf
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  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Are you able to do the actual integration?

    Would not the transform to the frequency domain complex equivalent be obtained from ....?

    g(\omega)=\frac{1}{2\pi}\int^{\frac{N\pi}{\omega_0}}_{-\frac{N\pi}{\omega_0}}Vsin(\omega_0 t)e^{-j\omega t}dt
     
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  4. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    Yes you are correct But I think that I have been using the incorrect formula I should have been using

    g(j\omega)=\int^{\infty}_{-\infty} f(t)e^{-j\omega_0}dt

    Below is another attempt of this question which I have found the euler form for sin and the correct formula

    g(\omega)=V\int^{\frac{NT}{2}}_{-\frac{NT}{2}}(\frac{e^{j\omega_0}-e^{-j\omega_0}}{j2}) e^{-j\omega_0}dt

    If this is correct I now have to intergrate everything which is where I get lost. I am not sure which parts I can move outside the brackets inorder to intergrat from \int^{\frac{NT}{2}}_{-\frac{NT}{2}}

    I am thinking that j2\omega_0 can be taken out so then

    g(\omega)=\frac{V}{j2\omega}\int^{\frac{NT}{2}}_{-\frac{NT}{2}}(e^{j\omega_0}-e^{-j\omega_0}) e^{-j\omega_0}dt

    Euler form is e^{j\theta}+e^{-j\theta} = (Cos\theta+jSin\theta)+(Cos\theta - jSin\theta)

    Using this do I then expand the brackets and multply and cancle out the terms or am I shooting off course again.
    E.g

    g(\omega)=\frac{V}{j2\omega}\int^{\frac{NT}{2}}_{-\frac{NT}{2}}((Cos\theta+jSin\theta)+(Cos\theta - jSin\theta))* (Cos\theta - jSin\theta)

    Then intergrate for the values?

    Thanks again for your time/help
    Kayne
     
    Last edited: Jul 28, 2011
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Be careful about the variables in the formula. Perhaps should be ....

    g(\omega)=\frac{1}{2\pi}V\int^{\frac{NT}{2}}_{-\frac{NT}{2}}(\frac{e^{j\omega_0 t}-e^{-j\omega_0 t}}{j2}) e^{-j\omega t}dt

    although the

    \frac{1}{2\pi}

    factor depends on which inverse transform version you use.

    F(t)=\int^{\infty}_{-\infty}g(\omega)e^{j\omega t} d\omega

    applies in this case. It's academic rather than particularly important in this problem.

    The integration is somewhat convoluted using the sine term and requires successive integration by parts plus some re-arranging of terms on either side of the resulting integral equation. Maybe you could use an on-line integrator to check your work if you are not confident. You can collect any common multiplier constants (including the isolated 'j' operator) outside of the integral as a combined constant multiplier.

    I didn't use the Euler form of the sine function - just did the integration as is. As stated it required successive reductions using Integration by Parts, etc.

    Post again if you can't work it out.
     
    Last edited: Jul 29, 2011
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  6. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    T.N.K I have edited my last post which I am not sure if you have seen. I will go back and have another look at the answer you provided and see if this is the same.
    I have also found an online intergrator which is helping alot

    Thanks again
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This is a good on-line integrator ....
     
  8. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    Thanks for the heads up I was using another program. So I have no had a go on with this program.It will made things a little easier.

    Can I assume that this answer is the Fourier Transform for this wave

     \frac{V*e^{j\omega x}(a cos(ax)+j\omega sin(ax))}{\omega^{2}-a^{2}}

    If so something I dont know is how to get to this from the intergration.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Using Euler's conversion (I think this ends up more complex than using original sine function)

    g(\omega)=V\int\frac{(e^{j\omega_0 t}-e^{-j\omega_0 t)}}{2j}e^{-j\omega t}dt

    g(\omega)=V\int\frac{(e^{j(\omega_0 -\omega)t}-e^{-j(\omega_0 +\omega)t)}}{2j}dt

    g(\omega)=\frac{V}{2j}\int e^{j(\omega_0 -\omega)t}-e^{-j(\omega_0 +\omega)t}dt

    g(\omega)=\frac{V}{2j}[\frac{e^{j(\omega_0 -\omega)t}}{j(\omega_0 -\omega)}-\frac{e^{-j(\omega_0 +\omega)t}}{-j(\omega_0+\omega)}]

    g(\omega)=\frac{V}{2j}e^{-j\omega t}[\frac{e^{j\omega_0 t}}{j(\omega_0  -\omega)}-\frac{e^{-j\omega_0 t}}{-j(\omega_0+\omega)}]

    g(\omega)=\frac{V}{2j}e^{-j\omega t}(\frac{1}{j})[\frac{e^{j\omega_0  t}}{(\omega_0  -\omega)}+\frac{e^{-j\omega_0  t}}{(\omega_0+\omega)}]

    g(\omega)=\frac{-V}{2}e^{-j\omega  t}[\frac{e^{j\omega_0  t}}{(\omega_0   -\omega)}+\frac{e^{-j\omega_0  t}}{(\omega_0+\omega)}]

    g(\omega)=\frac{-V}{2(\omega_0^2-\omega^2)}e^{-j\omega  t}[(\omega_0+\omega)e^{j\omega_0   t}+(\omega_0   -\omega)e^{-j\omega_0   t}]

    g(\omega)=\frac{-V}{2(\omega_0^2-\omega^2)}e^{-j\omega   t}[\omega_0( e^{j\omega_0   t}+ e^{-j\omega_0   t})+\omega( e^{j\omega_0    t}   - e^{-j\omega_0    t)}]

    g(\omega)=\frac{-V}{2(\omega_0^2-\omega^2)}e^{-j\omega    t}[2\omega_0 cos(\omega_0 t)+2j\omega sin(\omega_0t)]

    g(\omega)=\frac{V}{(\omega_^2-\omega_0^2)}e^{-j\omega    t}[\omega_0 cos(\omega_0 t)+j\omega sin(\omega_0t)]

    g(\omega)=\frac{Ve^{-j\omega    t}[\omega_0 cos(\omega_0 t)+j\omega sin(\omega_0t)]}{{(\omega_^2-\omega_0^2)}}
     
    Last edited: Jul 29, 2011
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  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You should try to do the integration without using the Euler relationship - it will improve your math no end. Particularly as you need to apply integration by parts.
     
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  11. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    I will take that on board I am sure that I will have to do more of this in the coming months.
    I can see what you mean about being messy. I will have a look at the other way of doing this and try it that way

    Thanks for your time and help.

    Hopefully I will be able to work though the rest of them
    Cheers

    kayne
     
    Last edited: Jul 29, 2011
  12. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    Sorry to re-light the fire, I have found yet another question like this one which I have been able to solve via the intergration calculator which is a great tool. What I am having touble with it calculating the intergrals upper and lower limits.

    If for instance in the orignal question the the equation was given as

    g(\omega)=V\int^{\frac{NT}{2}}_{-\frac{NT}{2}}\frac{(e^{j\omega_0 t}-e^{-j\omega_0 t)}}{2j}e^{-j\omega t}dt

    where
    \omega_o = \frac{2*pi*t}{T}

    which was solved to be

    g(\omega)=\frac{Ve^{-j\omega    t}[\omega_0 cos(\omega_0 t)+j\omega  sin(\omega_0t)]}{{(\omega_^2-\omega_0^2)}}

    The lecture told us that we can omit the negitive limit becuase we are only wanting to deal with positive time so looiking back to the first equation the limits are now

    g(\omega)=V\int^{\frac{NT}{2}}_{0}\frac{(e^{j\omega_0 t}-e^{-j\omega_0 t)}}{2j}e^{-j\omega t}dt

    I have been trying to work though the equation below to calculate the limits for this equation with not much luck. Now I have lots of theroies and scrap paper but not having any luck with solving this.

    g(\omega)=\frac{V}{2j}\int^{\frac{NT}{2}}_{0} e^{j(\omega_0 -\omega)t}-e^{-j(\omega_0 +\omega)t}dt


    If you can point me in the right direction I would appricate it. The questions that have been given are below also which might give you a better understand of what it is I have trying to be doing. Hope I have explain this well enough.


    The two questions that have been asked are as follows.

    1. The question asked the wave form can be written as the sum of two similar spectra centred on radian frquencies of \omega_o and -\omega_0 respectively. Deduce the equation which decribes the shape of these and explain why the spectrum has this form
    Seeing that this question has already been solved I thought it would be easier to use as an example.

    2. Combine the two terms which make the spectra and find a more compact expression.

    Thanks for your time
     
  13. Kayne

    Thread Starter Active Member

    Mar 19, 2009
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    I have come up with this withe the following answer in the attachement to sovle for the Limits of thie equation.

    Any feedback will be appricated.

    Thanks kayne
     
  14. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
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    I have been able to refine the answer I got the workings can be seen in the attachment. Hopefully someone can tell me if this is correct not not.

    Thanks

    Kayne
     
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