Generate the Fourier transform of the function

  • Thread starter Poirot
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  • #1
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Homework Statement


a(x)=f-Nd(x) + f-(N-1)d(x) +...+ f(N-1)d(x) + fNd


Homework 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)



The Attempt at a Solution


[/B]
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.
 

Answers and Replies

  • #2
jbunniii
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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?
 
  • #3
94
2
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?
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?
 
  • #4
94
2
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?
Realised this wouldn't solve the issue of the summation, just a slightly prettier equations :(
 
  • #5
94
2
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?
I've figured it out! Thanks for all your help, much appreciated!
 
  • #6
jbunniii
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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.
 

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