What is the Fourier Transform of $\rect(2Bt)\cos[{\omega}_Ct + k_fm(t_k)t] $?

I got the following as the solution:

$$\DeclareMathOperator{\sinc}{sinc} \frac{1}{2} \frac{1}{2B} \sinc(\frac{\omega+{\omega}_C+k_fm(t_k) +}{4B}) + \frac{1}{2} \frac{1}{2B} \sinc(\frac{\omega-{\omega}_C-k_fm(t_k) +}{4B})$$

However, in the book it is given as:

$$ \frac{1}{2} \sinc(\frac{\omega+{\omega}_C+k_fm(t_k) +}{4B}) + \frac{1}{2} \sinc(\frac{\omega-{\omega}_C-k_fm(t_k) +}{4B})$$

Wolfram alpha shows this:


  • 1
    $\begingroup$ you should have $2B$ outside of the $\sinc$, so I'm pretty sure your book can't be right, unless $B$ is defined somewhere else. $\endgroup$ Apr 19 '20 at 17:50

If you define the $\textrm{rect}()$ function according to this definition, then your result is correct. However, there might be a problem with the definition of the $\textrm{rect}()$ function. If in your book they define that function to have unit area, then the result of the book would be correct.

  • $\begingroup$ I have used this text of B.P Lathi :academia.edu/36135973/… $\endgroup$
    – Anwesa Roy
    Apr 20 '20 at 12:28
  • $\begingroup$ Can it be possible that the book has defined the rect function to have unit area? Please refer to page no. 212. Thank you. $\endgroup$
    – Anwesa Roy
    Apr 20 '20 at 12:30
  • $\begingroup$ @AnwesaRoy: I actually don't think so, but just check if the author defines rect() somewhere in the book. But it's actually not a problem, just make sure that you understand the correctness of the result, whatever the result in the book may be. Even the best books have some errors. $\endgroup$
    – Matt L.
    Apr 20 '20 at 13:14
  • $\begingroup$ They have defined the rect(t) function just the way you have defined. Thanks for helping me out :) $\endgroup$
    – Anwesa Roy
    Apr 20 '20 at 13:34

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.