Help finding Fourier Transform

In summary, Dick suggests that I try integrating e^(it)/(1+t^2) over the real line first. Once I have found the poles and contour, I can use the residue theorem to evaluate the integral.
  • #1
jtleafs33
28
0

Homework Statement


Find the Fourier Transform of:

f(t)=[itex]\frac{cos(\alpha t)}{t^2+\beta^2}[/itex]

Homework Equations



F[itex](\omega)[/itex]=[itex]\frac{1}{2\pi}[/itex][itex]\int[/itex][itex]^{∞}_{-∞}[/itex][itex]\frac{cos(\alpha t)exp(i \omega t)}{t^2+\beta^2}[/itex]

The Attempt at a Solution


I start with:

[itex]cos(\alpha t)[/itex]=[itex]\frac{exp(i \alpha t)+exp(-i \alpha t)}{2}[/itex]

Substituting this in, I get:

[itex]F(\omega)[/itex]=[itex]\frac{1}{4\pi}\int^{∞}_{-∞}\frac{exp(it(\alpha+\omega))+exp(it(\omega-\alpha))}{t^2+\beta^2}[/itex]

From here I know I should be able to get this in the form of some delta functions but I can't figure out the manipulation. I'd appreciate any help!
 
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  • #2
jtleafs33 said:

Homework Statement


Find the Fourier Transform of:

f(t)=[itex]\frac{cos(\alpha t)}{t^2+\beta^2}[/itex]

Homework Equations



F[itex](\omega)[/itex]=[itex]\frac{1}{2\pi}[/itex][itex]\int[/itex][itex]^{∞}_{-∞}[/itex][itex]\frac{cos(\alpha t)exp(i \omega t)}{t^2+\beta^2}[/itex]

The Attempt at a Solution


I start with:

[itex]cos(\alpha t)[/itex]=[itex]\frac{exp(i \alpha t)+exp(-i \alpha t)}{2}[/itex]

Substituting this in, I get:

[itex]F(\omega)[/itex]=[itex]\frac{1}{4\pi}\int^{∞}_{-∞}\frac{exp(it(\alpha+\omega))+exp(it(\omega-\alpha))}{t^2+\beta^2}[/itex]

From here I know I should be able to get this in the form of some delta functions but I can't figure out the manipulation. I'd appreciate any help!

If you want to evaluate integrals like that you need to use contour integration. Find the poles, figure out contours and use the residue theorem. Have you done stuff like that?
 
  • #3
Yes I've used the residue theorem to evaluate contour integrals. We've covered so much in this class I had already forgotten that that technique would work.

Since this involves a [itex]2\pi[/itex] periodic function, I'd integrate over the unit circle in the complex plane... Use Euler's formula for the cosine substitution to get an integral in Z, and then find the poles and residues, Right?

Thanks Dick!
 
  • #4
jtleafs33 said:
Yes I've used the residue theorem to evaluate contour integrals. We've covered so much in this class I had already forgotten that that technique would work.

Since this involves a [itex]2\pi[/itex] periodic function, I'd integrate over the unit circle in the complex plane... Use Euler's formula for the cosine substitution to get an integral in Z, and then find the poles and residues, Right?

Thanks Dick!

That's the general technique alright, but the details are wrong. Your function isn't periodic and you don't integrate over the unit circle. You integrate over part of the real axis and then close the contour in the upper or lower half plane. But sure, review the technique and see how far you get.
 
  • #5
Right, this acts like a 'sinc' function, so the Jordan contour is the right selection for integration.

So the function is:

[itex]f(t)=\frac{cos(\alpha t)}{\beta^2+t^2}[/itex]

Let:
[itex]z=e^{i\alpha t}[/itex]
[itex]dz=i\alpha e^{i\alpha t}dt[/itex]
[itex]cos(\alpha t)=\frac{z+z^{-1}}{2}[/itex]

Transforming [itex] f(t)[/itex] to [itex] f(z)[/itex]

[itex]f(z)=\frac{z+z^{-1}}{2(\beta^2+t^2)} \cdot \frac{dz}{i\alpha z}[/itex]

So here with this [itex]t[/itex] in the denominator, can I replace it directly with a [itex]z[/itex], or do I need to use [itex]t=\frac{ln(z)}{i\alpha}[/itex] which I got from the second equation above?

Once I do that, I'll plug into the F.T. integral and find the value from the Residue Thm.
 
  • #6
jtleafs33 said:
Right, this acts like a 'sinc' function, so the Jordan contour is the right selection for integration.

So the function is:

[itex]f(t)=\frac{cos(\alpha t)}{\beta^2+t^2}[/itex]

Let:
[itex]z=e^{i\alpha t}[/itex]
[itex]dz=i\alpha e^{i\alpha t}dt[/itex]
[itex]cos(\alpha t)=\frac{z+z^{-1}}{2}[/itex]

Transforming [itex] f(t)[/itex] to [itex] f(z)[/itex]

[itex]f(z)=\frac{z+z^{-1}}{2(\beta^2+t^2)} \cdot \frac{dz}{i\alpha z}[/itex]

So here with this [itex]t[/itex] in the denominator, can I replace it directly with a [itex]z[/itex], or do I need to use [itex]t=\frac{ln(z)}{i\alpha}[/itex] which I got from the second equation above?

Once I do that, I'll plug into the F.T. integral and find the value from the Residue Thm.

That is WAY too fast and sloppy. You aren't going to get anywhere like this. Work out a simple example of this kind of problem first. Integrate e^(it)/(1+t^2) over the real line first. Where are the poles? What's a good contour? If you get that you are probably ready to do to think about the full problem.
 
  • #7
Okay, taking your advice.

For the example problem,

[itex]\int^{∞}_{-∞} \frac{e^{it}}{1+t^2}[/itex]
We can directly substitute t=z
[itex]\oint ^{∞}_{-∞} \frac{e^{iz}}{1+z^2} = \oint ^{\infty}_{-\infty} \frac{e^{iz}}{(1+i)(1-i)}[/itex]

Now since it's clear to see that f(z)→0 as R→∞, we can integrate over the upper half plane using a semicircular contour.

Only the pole z=i lies inside this contour.

So,

[itex] \oint ^{\infty}_{-\infty} \frac{e^{iz}}{(1+i)(1-i)}=2\pi i Res(z=i) = lim_{z→i} (z-i)\frac{e^{iz}}{(z+i)(z-i)} [/itex]
[itex] lim_{z→i}=\frac{2\pi i}{2ie} = \frac{\pi}{e}[/itex]
 
  • #8
I figured out what I was doing wrong. I realize now the substitutions I was making from the Euler formula don't work over the real line, those are more suited for integration over the unit circle. Once I chose a better way to transform into an integral in Z, this turned out to be a very simple problem.

Thanks for helping me and for scolding me for blindly pulling a formula out of the book.
 

1. What is the Fourier Transform?

The Fourier Transform is a mathematical tool used to break down a signal or function into its individual frequency components. It converts a function in the time domain to a function in the frequency domain.

2. Why is the Fourier Transform important?

The Fourier Transform is important because it allows us to analyze and understand complex signals and functions in terms of their individual frequency components. This is useful in a variety of fields, such as signal processing, image processing, and communication systems.

3. How is the Fourier Transform calculated?

The Fourier Transform is calculated using an integral over the function in the time domain. This integral is then evaluated at various frequencies to determine the amplitude and phase of each frequency component in the function.

4. What are some applications of the Fourier Transform?

The Fourier Transform has many applications, including signal filtering, spectral analysis, image reconstruction, and solving differential equations. It is also used in various fields such as physics, engineering, and mathematics.

5. Are there different types of Fourier Transforms?

Yes, there are various types of Fourier Transforms, including the Discrete Fourier Transform (DFT), Fast Fourier Transform (FFT), and Continuous Fourier Transform. Each type has its own specific applications and uses.

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