I'm studying for an exam in a digital communications course I'm taking, and the solution to one question has me totally lost. While finding the Inverse Fourier Transform of a function, there's one step that isn't making sense to me. This is the step:

$S_{XY}(f) = \frac{1}{\alpha-i2\pi f_0}\delta(f-f_0) + \frac{1}{\alpha+i2\pi f_0}\delta(f+f_0)$

$S_{XY}(f)= \frac{2\alpha}{\alpha^2 + 4\pi^2f_0^2}[\delta(f-f_0)+\delta(f+f_0)]+\frac{i4\pi f_0}{\alpha^2+4\pi^2 f_0^2}[\delta(f-f_0)-\delta(f+f_0)]$

My algebra skills a pretty rusty, so whenever I try to carry this step out myself, I end up without the factor of two in the numerator of both the fractions in the second form. My steps are

$S_{XY}(f) = \frac{\alpha+i2\pi f_0}{\alpha^2 + 4\pi^2f_0^2}\delta(f-f_0) + \frac{\alpha-i2\pi f_0}{\alpha^2 + 4\pi^2f_0^2}\delta(f+f_0)$

$S_{XY}(f) = \frac{\alpha(\delta(f-f_0)+\delta(f+f_0)) + i2\pi f_0(\delta(f-f_0) - \delta(f+f_0)}{\alpha^2 + 4\pi^2f_0^2}$

This is the same as the solution except with the factor of two missing. For the life of me I can't figure out where the factor of two comes from. I'm expecting it to be an obvious mistake, but I've spent more time than I'd like to admit staring at this and it just doesn't make sense to me.

  • $\begingroup$ It looks like a mistake on their part. $\endgroup$ – Christopher A. Wong Feb 13 '13 at 22:20
  • $\begingroup$ Could it be that the book is multiplying your result by a factor of $2$ for the definition of the IFT (they are accounting for it)? I haven't done these in a while, but it suspiciously looks like something like that. Regards $\endgroup$ – Amzoti Feb 13 '13 at 22:30
  • $\begingroup$ @Matt: Is there any reference to the textbook where you took it from? $\endgroup$ – Kaster Feb 13 '13 at 22:43

Well, I realized the problem from the combination of the comments given and by looking at the next step in the solution. When the book actually performs the IFT, they give this

$R_{XY}(\tau)=\frac{2\alpha}{\alpha^2 + 4\pi^2f_0^2}cos(2\pi f_0 \tau)-\frac{4\pi f_0}{\alpha^2 + 4\pi^2 f_0^2}sin(2\pi f_0 \tau)$

It just seems like they randomly chose to multiply the original function by 2 a step before performing the IFT, but then ignored the $\frac{1}{2}$ term in the usual transformation pairs of cosine and sine. I can't think of any logical reason to do this and I still get the correct answer using my steps so I guess there isn't much to discuss.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.