Fourier Transform with inverse

  • #1
Aows

Homework Statement


Q/ in this inverse fourier problem, how did he come with the results of integration of (Sinc) function and how did he come up with those results of integration with the inverse part (as in the attached picture)
here is the problem:
https://i.imgur.com/Ir3TQIN.png

Homework Equations


all equation needed are in the relevant picture

The Attempt at a Solution


tried a lot until the integration of the sinc and results are not the same.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Q/ in this inverse fourier problem, how did he come with the results of integration of (Sinc) function and how did he come up with those results of integration with the inverse part (as in the attached picture)
here is the problem:
https://i.imgur.com/Ir3TQIN.png

Homework Equations


all equation needed are in the relevant picture

The Attempt at a Solution


tried a lot until the integration of the sinc and results are not the same.

The author goes through the calculations step-by-step. Which parts do you disagree with, and why?

Note: the author makes some rather obvious "blunders", but if you know calculus you should be able to fix those up for yourself, without help from us. The final answer is correct.
 
  • #3
Aows
The author goes through the calculations step-by-step. Which parts do you disagree with, and why?

Note: the author makes some rather obvious "blunders", but if you know calculus you should be able to fix those up for yourself, without help from us. The final answer is correct.
yes true Dr. Ray,
my question is : is it possible to get the integration of Sinc function in another way ?
 
  • #4
Ray Vickson
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yes true Dr. Ray,
my question is : is it possible to get the integration of Sinc function in another way ?

No, that was NOT your question. You asked

"Q/ in this inverse fourier problem, how did he come with the results of integration of (Sinc) function and how did he come up with those results of integration
with the inverse part (as in the attached picture)"

You were very clearly saying that you did not understand what the author was doing; you were not asking about doing it another way.
 
  • #5
Aows
No, that was NOT your question. You asked

"Q/ in this inverse fourier problem, how did he come with the results of integration of (Sinc) function and how did he come up with those results of integration
with the inverse part (as in the attached picture)"

You were very clearly saying that you did not understand what the author was doing; you were not asking about doing it another way.
Hello Dr.Ray,
i manages to solve the question later, but my last question was about finding another way of integrating the (Sinc) function if possible .

and thanks for you valuable notes and help from the beginning
 
  • #7
Ray Vickson
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Hello Dr.Ray,
i manages to solve the question later, but my last question was about finding another way of integrating the (Sinc) function if possible .

and thanks for you valuable notes and help from the beginning
Hello Dr.Ray,
i manages to solve the question later, but my last question was about finding another way of integrating the (Sinc) function if possible .

and thanks for you valuable notes and help from the beginning

I don't think he did an explicit integration; rather, he (probably) used some standard theorems about Fourier transforms and their inverses. For example, if you use a Fourier transform defined by
$$g(\theta) = ({\cal F}f)(\theta) \equiv \int_{-\infty}^{\infty} f(t) e^{-2 \pi i \theta t} \, dt, $$
then for piecewise smooth ## f ## which, together with all its derivatives, decay quickly as ##t \to \pm \infty##, we have
$$ ({\cal F}^{-1} g) (t) \equiv
\int_{-\infty}^{\infty} g(\theta) e^{+2 \pi i \theta t} \, d \theta = \frac{1}{2} \left[ f(t-) + f(t+)\right],$$
where ##f(t-)## and ##f(t+)## are the left-hand limits and right-hand limits of ##f(u)## as ##u \to t##.

In other words, at points ##t## where ##f(t)## is continuous we have ##({\cal F}^{-1} g) (t) = f(t)##, but at a point ##t## where ##f## has a "jump discontinuity" the inverse F.T. equals the average of the two ## f ## values on either side of the jump.

The ##f(t)## in your problem has jump discontinuities at ##t=-1## and ##t=1## but is continuous everywhere else (and it certainly is damped at large valus of ##|t|##), so that gives the result, up to some constants. The author uses a different definition/normalization for the Fourier transform (using ##i \omega t## instead of ##2 \pi i \theta t## in the exponent), so the inverse will have an extra multiplicative constant in it; you can work out the details for yourself.

Note: the theorem I cited above is not trivial, and may not be proved at all in an introductory course or an "engineering math" or "math for physics course". A brief proof outline is given in
https://en.wikipedia.org/wiki/Fourier_inversion_theorem
but it is very short and leaves out a lot of details for the reader to fill in if he/she can.
A longer and more detailed proof is in https://www.ima.umn.edu/~miller/fouriertransform.pdf
https://www.ima.umn.edu/~miller/fouriertransform.pdf
but it is still not easy.
 
Last edited:
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  • #9
Aows
I don't think he did an explicit integration; rather, he (probably) used some standard theorems about Fourier transforms and their inverses. For example, if you use a Fourier transform defined by
$$g(\theta) = ({\cal F}f)(\theta) \equiv \int_{-\infty}^{\infty} f(t) e^{-2 \pi i \theta t} \, dt, $$
then for piecewise smooth ## f ## which, together with all its derivatives, decay quickly as ##t \to \pm \infty##, we have
$$ ({\cal F}^{-1} g) (t) \equiv
\int_{-\infty}^{\infty} g(\theta) e^{+2 \pi i \theta t} \, d \theta = \frac{1}{2} \left[ f(t-) + f(t+)\right],$$
where ##f(t-)## and ##f(t+)## are the left-hand limits and right-hand limits of ##f(u)## as ##u \to t##.

In other words, at points ##t## where ##f(t)## is continuous we have ##({\cal F}^{-1} g) (t) = f(t)##, but at a point ##t## where ##f## has a "jump discontinuity" the inverse F.T. equals the average of the two ## f ## values on either side of the jump.

The ##f(t)## in your problem has jump discontinuities at ##t=-1## and ##t=1## but is continuous everywhere else (and it certainly is damped at large valus of ##|t|##), so that gives the result, up to some constants. The author uses a different definition/normalization for the Fourier transform (using ##i \omega t## instead of ##2 \pi i \theta t## in the exponent), so the inverse will have an extra multiplicative constant in it; you can work out the details for yourself.

Note: the theorem I cited above is not trivial, and may not be proved at all in an introductory course or an "engineering math" or "math for physics course". A brief proof outline is given in
https://en.wikipedia.org/wiki/Fourier_inversion_theorem
but it is very short and leaves out a lot of details for the reader to fill in if he/she can.
A longer and more detailed proof is in
https://www.ima.umn.edu/~miller/fouriertransform.pdf
but it is still not easy.
Thanks indeed Dr.Ray for your explanation, but can you explain more the part where the F(1) and f(-1) why they both divided by two ?
 
  • #10
Ray Vickson
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Thanks indeed Dr.Ray for your explanation, but can you explain more the part where the F(1) and f(-1) why they both divided by two ?

I just did that, in detail. Read my answer again, and compute the results for your function
$$
f(t) = \begin{cases} 0, & \;\; t < -1 \\
1 , & -1 \leq t \leq 1 \\
0 , &\;\; t > 1
\end{cases}
$$
 
Last edited:

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