Hi

I am trying to find an expression for X(f), when,

x(t)=exp(-σt)sin(2∏f1t)

I have attached a screen shot of what I have worked out, but it doesn't seem to be coming together very vell, is there something I am missing in the steps? Or am I right, and how would I integrate that?

thanks.

 

The Fourier transform of

\(u(t)\epsilon^{-\sigma t}sin(\omega_{1}t)\)

per table of transforms is

\(\frac{\omega_{1}}{\omega_{1}^2+(\sigma+j\omega)^2}\)

The u(t) bit tells you the limits of integration are from 0 to plus infinity - otherwise there is no solution if the lower limit goes to minus infinity . The problem can be solved using repeated integration by parts.

Not sure if this is the same as your problem - given your equation has no reference to the u(t) term - i.e. the Heavyside Function.

 

The Fourier transform of

\(u(t)\epsilon^{-\sigma t}sin(\omega_{1}t)\)

per table of transforms is

\(\frac{\omega_{1}}{\omega_{1}^2+(\sigma+j\omega)^2}\)

The u(t) bit tells you the limits of integration are from 0 to plus infinity - otherwise there is no solution if the lower limit goes to minus infinity . The problem can be solved using repeated integration by parts.

Not sure if this is the same as your problem - given your equation has no reference to the u(t) term - i.e. the Heavyside Function.

Hi,

Thanks for your reply,

I have attached another JPEG, I have come up with an expression but not sure if it is correct, I did mean to put the limits from 0 to infinity sorry about that.
However, when i put this into matlab to plot the X(f), i just comes out wrong.

 

This is my method ...