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?

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.

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).

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?

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

Where did the problem come from?

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

My signals and systems exam

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 $\displaystyle \text{sgn}(t)$ is $\displaystyle \frac{2}{j\omega}$ which also seems to have a non-integrable singularity at $\displaystyle w=0$.

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

Write

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

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