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Math Help - fourier transform of sin(3w)*cos(w)/(w^2)

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    fourier transform of sin(3w)*cos(w)/(w^2)

    Can someone tell me how to calculate the inverse fourier transform of sin(3w)*cos(w)/(w^2)
    I know that 2sin(aw)/w has an inverse fourier transform that is 1 when -a<t<a and zero otherwise and therefore the inverse fourier should be the convultion of two of these and some other thing...can you help me?
    Last edited by mr fantastic; June 29th 2011 at 02:10 PM. Reason: Title
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    Re: fourier transform of sin(3w)*cos(w)/(w^2)

    You mean the inverse of sinc is a rect (or box) function. The other function is cosinc which you'll need to find a way to invert.
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    Re: fourier transform of sin(3w)*cos(w)/(w^2)

    Quote Originally Posted by Mppl View Post
    Can someone tell me how to calculate the inverse fourier transform of sin(3w)*cos(w)/(w^2)
    I know that 2sin(aw)/w has an inverse fourier transform that is 1 when -a<t<a and zero otherwise and therefore the inverse fourier should be the convultion of two of these and some other thing...can you help me?
    This function has a non-integrable singularity at w=0, so I don't see how it can have a Fourier transform (direct or inverse).
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    Re: fourier transform of sin(3w)*cos(w)/(w^2)

    When you say that it doesn't have a FT are you considering that it can be a transform of a discrete variable signal?
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    Re: fourier transform of sin(3w)*cos(w)/(w^2)

    Where did the problem come from?
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    Re: fourier transform of sin(3w)*cos(w)/(w^2)

    My signals and systems exam
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    Re: fourier transform of sin(3w)*cos(w)/(w^2)

    OK, just wantd to be sure it wasn't a typo. I'm still thinking about how to do this and Opalg's comment. It should be pointed out that the Fourier transform of \text{sgn}(t) is \frac{2}{j\omega} which also seems to have a non-integrable singularity at w=0.
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    Re: fourier transform of sin(3w)*cos(w)/(w^2)

    Write

    \frac{\sin(3\omega)\cos \omega}{\omega^2}=\frac{1}{2}\text{sinc} (4\omega)\cdot \frac{1}{\omega}+\frac{1}{2}\text{sinc} (2\omega)\cdot \frac{1}{\omega}

    where \text{sinc } x=\frac{\sin x}{x}. Now invert using the convolution theorem.
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