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

• Jun 29th 2011, 11:50 AM
Mppl
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?
• Jun 30th 2011, 07:11 PM
ojones
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.
• Jun 30th 2011, 11:49 PM
Opalg
Re: fourier transform of sin(3w)*cos(w)/(w^2)
Quote:

Originally Posted by Mppl
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).
• Jul 1st 2011, 01:28 PM
Mppl
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?
• Jul 1st 2011, 03:10 PM
ojones
Re: fourier transform of sin(3w)*cos(w)/(w^2)
Where did the problem come from?
• Jul 1st 2011, 04:46 PM
Mppl
Re: fourier transform of sin(3w)*cos(w)/(w^2)
My signals and systems exam
• Jul 1st 2011, 05:20 PM
ojones
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 $w=0$.
• Jul 2nd 2011, 06:03 PM
ojones
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.