1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Generate the Fourier transform of the function

  1. Apr 14, 2016 #1
    1. The problem statement, all variables and given/known data
    a(x)=f-Nd(x) + f-(N-1)d(x) +...+ f(N-1)d(x) + fNd


    2. Relevant equations
    fd(x) = (1/a for |x-d| < a and 0 otherwise)
    Fourier transform of function g(x) is g~(p) = 1/root(2pi) ∫ dx e-ipx g(x)



    3. The attempt at a solution

    I have found the general Fourier transform for the function fd(x) and got:
    f~(p) = root(2/pi) e-ipd sin(pa)/pa

    I then tried to use the linearity of Fourier transform and added the solutions for f(x) but subbed in the various N, N-1 etc but got muddled up and a little confused.

    Is this the right thing to be doing? Is this going to come out as a nice clean answer?

    Thanks in advance.
     
  2. jcsd
  3. Apr 14, 2016 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It looks OK so far. You have an expression for the Fourier transform of ##f_d##:
    $$\hat{f_d}(p) = \sqrt{\frac{2}{\pi}} e^{-ipd} \frac{\sin(pa)}{pa}$$
    You need an expression for the Fourier transform of ##f_{kd}## where ##k## is an arbitrary integer. This will obviously be
    $$\hat{f_{kd}}(p) = \sqrt{\frac{2}{\pi}} e^{-ipkd} \frac{\sin(pa)}{pa}$$
    Now apply linearity. You want the Fourier transform of
    $$\sum_{k=-N}^{N}f_{kd}(x)$$
    which is
    $$\sum_{k=-N}^{N}\hat{f_{kd}}(p) = \sum_{k=-N}^{N} \sqrt{\frac{2}{\pi}} e^{-ipkd} \frac{\sin(pa)}{pa}$$
    Can you see how to simplify this?
     
  4. Apr 14, 2016 #3
    I don't think it's the way I'm supposed to do it as my lecturer gave a formula as a hint:
    1/(x2 + 1/x(N-1) + 1 .... + xN + x(N-1) = (x-(N+1/2) - x(N+1/2)) / (x-1/2 - x1/2)
    But I can't see how to manipulate the summation into this as there's the extra part in the exponent?
    However I had an idea; as the terms go from -N to N, there will always be a term with the negative of that exponent added to it, so can I use e-ix + eix = 2cos(x) and then add 1 due to the term when k=0?
     
  5. Apr 14, 2016 #4
    Realised this wouldn't solve the issue of the summation, just a slightly prettier equations :(
     
  6. Apr 14, 2016 #5
    I've figured it out! Thanks for all your help, much appreciated!
     
  7. Apr 14, 2016 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think the formula your lecturer meant is
    $$z^{-N} + z^{-(N-1)} + z^{-(N-2)} + \cdots + z^{-1} + 1 + z + \cdots + z^{N-1} + z^{N} = \frac{z^{-(N+1/2)} - z^{N + 1/2}} {z^{-1/2}-z^{1/2}}$$
    This can be derived as follows. We can write the left hand side more compactly as
    $$\sum_{k=-N}^{N}z^{k}$$
    Now introduce the change of variable ##m = k + N## to rewrite the sum as
    $$\begin{aligned}
    \sum_{m=0}^{2N}z^{m-N} &= z^{-N}\sum_{m=0}^{2N}z^m \\
    \end{aligned}$$
    Do you see what to do next?

    Then, assuming you have established the formula
    $$\sum_{k=-N}^{N}z^{k} = \frac{z^{-(N+1/2)} - z^{N + 1/2}} {z^{-1/2}-z^{1/2}}$$
    you can apply it to your situation by setting ##z = e^{-ipt}## and simplifying the resulting right hand side.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Generate the Fourier transform of the function
Loading...