Fourier Transforms -- Please check my solution

• warhammer
In summary, My Professor has started on the Fourier Transforms Topic in the Introductory Mathematical Physics class and gave us a small homework to try our concepts on. He assigned us to integrate two terms, one imaginary and one real, and my solution is incorrect because I mistakenly assumed that imaginary terms cancel each other out in the end.
warhammer
Homework Statement
If F(k)= A(a+k) for -a<k<0
= A(a-k) for 0<k<a
= 0 for |x| > a

then calculate ##f(x)=\frac{A}{\sqrt{2π}} \int_{-a}^0 (a+k)e^{-ikx}\,dx + \int_0^a (a-k)e^{ikx}\,dx##
Relevant Equations
##f(x)=\frac{A}{\sqrt{2π}} \int_-a^0 (a+k)e^{-ikx}\,dx + \int_0^a (a-k)e^{ikx}\,dx##
My Professor has started on the Fourier Transforms Topic in the Introductory Mathematical Physics class and gave us a small homework to try our concepts on.

I have attached a clear & legible snippet of my solution. I request someone to please have a look at it & determine if my solution is correct. (I'd also request some patience since this is my first dabble with the topic).

In the snippet, as described, I took the terms having 'i' as zero from a bit of assumed trickery, using the fact that F(k) had no terms having i as a coefficient so it must be zero in case of f(x) as well. (Not sure if this is correct way).

Then I solved for both integrals separately & added them after plugging in the desired limits.

(Edit 1- I'm fixing up on the Latex commands that I've written, please give a few moments, new to this actually.)
Mentor note: I fixed up the broken LaTeX.

Attachments

• 20220222_050453.jpg
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warhammer said:
new to this actually
Mentor note: Fixed the broken LaTeX in the previous post.
I spot a \, twice, so either you copied the stuff or you are not all that new to ##\LaTeX## ?

Anyway:
\frac{A}{√2π) should be \frac{A}{\sqrt{2π}} (curly bracket and \sqrt{..} instead of the √).

(to see ##\TeX## source, right-click on the formula):

In your " then calculate ##\displaystyle f(x)=\frac{A}{\sqrt{2π}} \left [ \int_{-a}^0 (a+k)e^{-ikx}\,dx + \int_0^a (a-k)e^{ikx}\,dx\right ]## " I don't see why the second exponent is ##+ikx## instead of ##-ikx##

Taking imaginary terms to be zero is not a good idea. Letting them cancel in the end might be good.

##\ ##

Last edited by a moderator:
warhammer
hutchphd said:
That is what you pay your professor an TA to do. Seriously.
If they are unable to answer your questions I will be very happy to help you to the fullest extent of my capabilities.
With online learning, the Professors never discuss the proceedings of the previous class and they don't entertain the students post the e-session. In my nation/university there is no concept of TAs at the UG level. They list out these problems to solve & check but never discuss the solutions. Which is why I've often had no recourse but to seek guidance on my doubts from PF Mentors.

Well then I will be glad to help. Keep up the good work, but seek avenues to self-actualize too.

warhammer
BvU said:
I spot a \, twice, so either you copied the stuff or you are not all that new to ##\LaTeX## ?
It is actually the latter, my first post using Latex. Was typing on my smartphone while viewing instructions from the Latex Help PF Webpage and now shifted to desktop as it was getting tedious.

BvU said:
In your " then calculate ##\displaystyle f(x)=\frac{A}{\sqrt{2π}} \left [ \int_{-a}^0 (a+k)e^{-ikx}\,dx + \int_0^a (a-k)e^{ikx}\,dx\right ]## " I don't see why the second exponent is ##+ikx## instead of ##-ikx##

Taking imaginary terms to be zero is not a good idea. Letting them cancel in the end might be good.

##\ ##
Oh! Thank you loads for noticing this. I think I made the error while noting down the question and continued with it throughout the solution. Will also rectify the imaginary term part so that they cancel rather than the assuming. I'm modifying the solution accordingly..

warhammer said:
It is actually the latter, my first post using Latex. Was typing on my smartphone while viewing instructions from the Latex Help PF Webpage and now shifted to desktop as it was getting tedious.Oh! Thank you loads for noticing this. I think I made the error while noting down the question and continued with it throughout the solution. Will also rectify the imaginary term part so that they cancel rather than the assuming. I'm modifying the solution accordingly..

@BvU As instructed by you, I modified my solution accordingly, continuing from the originally posted image, simultaneously correcting the second term to ##e^{-ikx}## & integrating the Imaginary Terms as well by treating 'i' as constant post expansion after Euler's Formula. Hope it is now devoid of errors..

Attachments

• 20220222_204258.jpg
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It would take me quite a while to go through that (and #1) and find out what is right and what isn't.
At this moment I don't have that time, so you'll have to check your working all by yourself.

I forgot to correct you in #1 w.r.t. the integration: to get ##f(x)## you integrate over ##k## (which you do in the handwritten stuff).

You have two integral terms $$\displaystyle f(x)=\frac{A}{\sqrt{2π}} \left [ \int_{-a}^0 (a+k)e^{-ikx}\,dk + \int_0^a (a-k)e^{-ikx}\,dk\right ]$$and if I had to do this (*), I would try to combine the two to get a real integrand of the form ##A(a-k)\left [ e^{ikx} +e^{-ikx}\right ] ##.

(*) And I had to do this and actually have done this, but unfortunately it's 50 years ago and I have a hard time reconstructing the steps .​

Unfortunately:
The reason I expect your result isn't correct comes from knowledge which you are probably not supposed to have available yet:
Your ##F(k)## is a triangle function, which is a convolution of two boxcars ( 7 here). According to 14 ibid (but the convolution in freq ##\rightarrow## product in time domain), I would expect a result looking like ##\displaystyle {\sin^2x\over x^2}##.

##\ ##

warhammer

1. What is a Fourier Transform?

A Fourier Transform is a mathematical tool used to decompose a function into its frequency components. It takes a function in the time domain and converts it into a function in the frequency domain, showing the different frequencies that make up the original function.

2. Why are Fourier Transforms important in science?

Fourier Transforms are important in science because they allow for the analysis and understanding of complex signals and data. They are used in a variety of fields, including physics, engineering, and mathematics, to analyze and interpret data in the frequency domain.

3. How is a Fourier Transform calculated?

A Fourier Transform is calculated using an integral formula, which involves taking the function in the time domain and integrating it with respect to frequency. This integral is solved using complex numbers and the result is a function in the frequency domain.

4. What is the difference between a Fourier Transform and a Fourier Series?

A Fourier Transform is used for continuous signals, while a Fourier Series is used for periodic signals. A Fourier Transform gives a continuous spectrum of frequencies, while a Fourier Series only gives discrete frequencies. Additionally, a Fourier Transform can handle non-periodic signals, while a Fourier Series cannot.

5. Can Fourier Transforms be used for image processing?

Yes, Fourier Transforms can be used for image processing. Images can be represented as functions of two variables (x and y), so a 2D Fourier Transform can be used to analyze the frequency components of an image. This allows for tasks such as filtering and compression to be performed on images.

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