Fourier transform using residues

In summary, the function g(t) = f(t).e-iwt, which is holomorphic in all C, can be solved without residues using the principal value of the integral of sin(at).
  • #1
libelec
176
0

Homework Statement


Find the Fourier transform of f(t) = e-at2, with a > 0 using the residue theorem.

The Attempt at a Solution



The problem I have is the function g(t) = f(t).e-iwt, which is holomorph in all C. Is there another way to do it without residues?
 
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  • #2
f(t)*exp(-iwt)=exp(-a*t^2-iwt). You want to complete the square and displace the integration off the x-axis using that the function is holomorphic. If you want some gory details about a very similar problem check out Count Iblis's advice in https://www.physicsforums.com/showthread.php?t=368440 You can ignore the residues, in the sense that there aren't any because there are no poles. You should still pay some attention the contours, as Count Iblis points out.
 
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  • #3
OK, thanks. I have another question though. I tried this same thing but with the sin(at)/at.

Now, I turned the integral such that [tex]\[\int\limits_{ - \infty }^\infty {\frac{{Sin(at)}}{{at}}{e^{ - iwt}}dt} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{iat}}{e^{ - iwt}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{i(a - w)t}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{} {\frac{{{e^{i(a - w)z}}}}{{az}}dz} } \right)\][/tex].

Then, evualuating the integral [tex]\[{\int\limits_\gamma {\frac{{{e^{i(a - w)z}}}}{{az}}dz} }\][/tex] I arrive to the known result that it's equal to i[tex]\[\pi \][/tex]/a. Then the Fourier transform is [tex]\[\pi \][/tex]/a.

But the Fourier transform of sinc is a rectangular function. That means that there are values of w in which the Fourier transform equals the found value, and values of w in which it equals 0. Where do I see that?
 
  • #4
Taking the imaginary part like you did there doesn't quite work here, because you have to end up with sin(at)exp(-i w t). Also, note that you would also have to take the principal value of the integral when you do that.

Instead of taking the imaginary part, you can substitute:



sin(at) = 1/(2i) [exp(i a t) - exp(-iat)]

in the integrand. Then you replace the integral by the principal value of the integral (i.e. you omit the interval from minus epsilon to epsilon and consider the limit of epsilon to zero). Then you can split up the integral in the two parts with the terms exp(i a t) and exp(- i a t). For each integral you have to consider an appropriate contour integral.
 
  • #5
So, [tex]\[\int\limits_{ - \infty }^\infty {\frac{{\sin (at).{e^{ - iwt}}}}{{at}}dt} = \int\limits_{ - \infty }^\infty {\frac{{\left( {{e^{iat}} - {e^{ - iat}}} \right).{e^{ - iwt}}}}{{2i.at}}dt} = \mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {\frac{{{e^{iat}}.{e^{ - iwt}}}}{{2i.at}}dt} - \int\limits_{ - A}^A {\frac{{{e^{ - iat}}.{e^{ - iwt}}}}{{2i.at}}dt = } \mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {\frac{{{e^{i(a}}^{ - iw)t}}}{{2i.at}}dt} - \int\limits_{ - A}^A {\frac{{{e^{ - i(a + w)t}}}}{{2i.at}}dt} \][/tex]

How is that any different from what I did before? Why should I end up with the expression sin(at)exp(-i w t), if that's the imaginary part of exp(iat).exp(-iwt)?
 
  • #6
Count Ibis said:
Taking the imaginary part like you did there doesn't quite work here, because you have to end up with sin(at)exp(-i w t).

This is what I don't understand.
 
  • #7
What is Im[exp(i a t) exp(-i w t)] ?
 
  • #8
libelec said:
OK, thanks. I have another question though. I tried this same thing but with the sin(at)/at.

Now, I turned the integral such that [tex]\[\int\limits_{ - \infty }^\infty {\frac{{Sin(at)}}{{at}}{e^{ - iwt}}dt} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{iat}}{e^{ - iwt}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{i(a - w)t}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{} {\frac{{{e^{i(a - w)z}}}}{{az}}dz} } \right)\][/tex].

Then, evualuating the integral [tex]\[{\int\limits_\gamma {\frac{{{e^{i(a - w)z}}}}{{az}}dz} }\][/tex] I arrive to the known result that it's equal to i[tex]\[\pi \][/tex]/a. Then the Fourier transform is [tex]\[\pi \][/tex]/a.

But the Fourier transform of sinc is a rectangular function. That means that there are values of w in which the Fourier transform equals the found value, and values of w in which it equals 0. Where do I see that?

You aren't going to see the step function correctly unless you are a lot more careful with defining the contours. Don't do 'Im' and consider both integrals when you expand (exp(aiz)-exp(-aiz)). Which half plane you need to close each contour in depends on the sign of (a-w) and (a+w). I would also change z into z-i*epsilon with epsilon>0 and then let epsilon -> 0 at the end. That puts the pole definitely in the upper half plane.
 
  • #9
Count Iblis said:
What is Im[exp(i a t) exp(-i w t)] ?

sin[at]*exp(-iwt).
 
  • #10
Ah, no, wait. I'm not considering the exp(-iwt) imaginary part, that's what you're saying?
 
  • #11
libelec said:
Ah, no, wait. I'm not considering the exp(-iwt) imaginary part, that's what you're saying?

That's right, if you take the imaginary part of the whole expression then you get an additional term: the imaginary part of exp(-iwt) times the real part of exp(i a t).
 

1. What is a Fourier transform using residues?

A Fourier transform using residues is a mathematical technique used to transform a function from the time domain to the frequency domain. It is based on the theory of complex analysis and uses the concept of residues, which are the residues of a complex function at its singularities, to perform the transformation.

2. How is a Fourier transform using residues different from other Fourier transforms?

A Fourier transform using residues is different from other Fourier transforms, such as the discrete Fourier transform or the fast Fourier transform, because it is based on complex analysis rather than purely numerical methods. This makes it particularly useful for analyzing functions with singularities or other complex behavior.

3. When is a Fourier transform using residues used?

A Fourier transform using residues is commonly used in signal processing, image processing, and other fields of engineering and science. It is particularly useful for analyzing signals or functions with sharp changes or singularities, which may cause other Fourier transforms to produce inaccurate results.

4. What is the process for performing a Fourier transform using residues?

The process for performing a Fourier transform using residues involves several steps, including identifying the singularities of the function, calculating the residues at each singularity, and then using a formula to transform the function into the frequency domain. This process requires knowledge of complex analysis and can be quite involved for more complex functions.

5. What are the advantages of using a Fourier transform using residues?

One of the main advantages of using a Fourier transform using residues is its ability to accurately handle functions with singularities, which may cause other Fourier transforms to produce incorrect results. It also allows for a more precise analysis of complex functions and can provide insights into their behavior in the frequency domain.

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