Convolution and fourier transform puzzle

In summary, the equation \vec{P}(z,t)=\int_{-\infty}^{+\infty}\epsilon_0 \tilde{\chi_e}(\tau)\vec{E}(z,t-\tau)d\tau justifies the following equality by invoking the convolution theorem: \int_{-\infty}^{+\infty} \chi(\omega)\vec{E}_0(\omega)e^{-i\omega t}d\omega=\int_{-\infty}^{+\infty} \chi(\tau)\vec{E}_0(\tau-t)d\tau. This equality can be derived from the Fourier transform identity \mathcal{F}(f*g)=\mathcal{F
  • #1
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My notes (from a physics course) justifies the following equality by invoking the convolution thm:

[tex]\int_{-\infty}^{+\infty} \chi(\omega)\vec{E}_0(\omega)e^{-i\omega t}d\omega=\int_{-\infty}^{+\infty} \chi(\tau)\vec{E}_0(\tau-t)d\tau[/tex]

From a mathematical standpoint (i.e. without reference to what the functions chi and E are), is the above some consequence of the convolution thm? The convolution thm says that [itex]\mathcal{F}(f*g)=\mathcal{F}(f)\cdot\mathcal{F}(g)[/itex] I don't see how this applies here. For one, there are two integrals mutilplied by one another on the RHS of the convolution thm. And for another, the RHS of my equation is simply the convolution of chi and E; not its Fourier transform or anything. Anyone sees?

And what if I were to add that

[tex]\vec{E}_0(\omega)=\int_{-\infty}^{+\infty} \vec{E}(T)e^{i\omega T}dT[/tex] ?

Notice the 0 subscript on the LHS but not in the integral; the difference is meant to represent two dinscting functions [of course one is just the "fourier coefficients" of the other]

Final hypothesis: perhaps my professor had another "convolution thm" in mind when he wrote that?

P.S. For clarity, I have not kept track of all the 2pi's popping up now and then in the above equations.
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  • #2
I added a 0 suscript to the RHS of the first equatiom too. Sorry about that.
  • #3
Is that also supposed to be [itex]\chi_0(\omega)[/itex]? If so, then this is a direct consequence of the Fourier transform identity you mentioned: just take the inverse Fourier transform of both sides. If not, it's obviously false: just take [itex]\chi=1[/itex]. Also, the usual notation for the Fourier transform of f is F or [itex]\tilde f[/itex]. The f0 notation doesn't make sense to me.
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  • #4
mmmh, I'm confused nowon who has a 0 and who doesnt. In my notes there are no zeroes anywhere.

The zero subscipt notation is from the fact that these E in there are electromagnetic waves. Typically, a single electromagnetic wave is noted

[tex]\vec{E}=\vec{E}_0e^{-kz-\omega t)[/tex]

so the natural extension to a wave packet is to say that we are going to superpose functions of the type

[tex]\vec{E}=\vec{E}_0(\omega)e^{-kz-\omega t)[/tex]

leading to the notation E_0 for the Fourier coefficients. (I thought you were a physics student StatusX, am I mistaken?)
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  • #5
on which chi would you put a 0 subscript?
  • #6
I am a physics student, and I have seen that notation for a plane wave. It makes sense there because then [itex]\vec E_0[/itex] is a constant vector. But I've never seen that subscript carried over to a Fourier trasnform. It's always the tilde notation (eg, griffiths).

The equation only makes sense if both functions on the LHS are Fourier transforms of the corresponding functions on the RHS.
  • #7
There's still a mistake in the sign of the exponent as far as I can tell.

Let [itex]\chi[/itex] and [itex]\vec{E}[/itex] be functions of [itex]\tau[/itex]. Then


[tex]\Rightarrow \frac{1}{\sqrt{2\pi}}\chi*\vec{E}(t)=\mathcal{F}^{-1}[\chi(\omega)\vec{E}(\omega)][/tex]

where is it implicit that when the argument is omega, we're talking about the Fourier transform of the function. Writing explicitely the last equation

[tex]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\chi(\tau)\vec{E}(t-\tau)d\tau = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\chi(\omega)\vec{E}(\omega)e^{i\omega t}d\omega[/tex]

(the 2pi's don't match those in my notes either, in my notes, I have

[tex]\frac{1}{2\pi}\int_{-\infty}^{+\infty} \chi(\tau)\vec{E}(\tau-t)d\tau=\int_{-\infty}^{+\infty} \chi(\omega)\vec{E}(\omega)e^{-i\omega t}d\omega[/tex])
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  • #8
The inverse transform is usually the one with the minus i, but this is just convention. Also, please try to use different symbols for a function and its Fourier transform. tau and omega are just dummy variables, so, eg:

[tex]\int_{-\infty}^\infty \chi(\tau) d\tau = \int_{-\infty}^\infty \chi(\omega) d\omega[/tex]
  • #9
duh :P

I want to thank you StatusX for your assistance.

Are you a grad student?
  • #10
No, I'm a senior. I actually just made a thread about applying to grad schools.
  • #11
You might not have an answer to that but...

Why are you so freakishly good at math? :biggrin:
  • #12
Thanks, I appreciate being called a freak. Maybe it's because I was born down river from a nuclear power plant and down wind from a chemical weapons testing ground.
  • #13
I have a natural talent for diplomacy. :rolleyes:
  • #14
If I take it from the top and try to recreate the result of my prof by doing every step in between...

We know that given a monochromatic plane wave of frequency [itex]\omega_0[/itex] propagating in a dielectric material in the z direction, the resulting polarization density function of the material is

[tex]\vec{P}(z,t)=\epsilon_0\chi_e(\omega_0)\vec{E}_0e^{i(k_R(\omega_0)z-\omega_0 t)}[/tex]

Now, if we shoot a wave packet

[tex]\vec{E}(z,t)=\int_{-\infty}^{+\infty}\tilde{\vec{E}}_0(\omega)e^{i(k_R(\omega)z-\omega t)}d\omega[/tex]

(happy? I put a tilde :wink: ) into the material, the resulting polarization should be

[tex]\vec{P}(z,t)=\int_{-\infty}^{+\infty}\epsilon_0 \chi_e(\omega)\tilde{\vec{E}}_0(\omega)e^{ik_R(\omega)z}e^{-i\omega t}d\omega[/tex]


[tex]\tilde{\vec{E}}_0(z,\omega)\equiv \tilde{\vec{E}}_0(\omega)e^{ik_R(\omega)z}[/tex]

Then by the convolution thm,

[tex]\vec{P}(z,t)=\int_{-\infty}^{+\infty}\epsilon_0 \tilde{\chi_e}(\tau)\vec{E}(z,t-\tau)d\tau[/tex]


[tex]\tilde{\chi_e}(\tau)=\mathcal{F}^{-1}[\chi_e(\omega)]=\int_{-\infty}^{+\infty}\chi_e(\omega) e^{-i\omega \tau}d\omega[/tex]


[tex]\vec{E}(z,\tau)=\mathcal{F}^{-1}[\vec{E}_0(z,\omega)]=\int_{-\infty}^{+\infty}\tilde{\vec{E}}_0(\omega)e^{i(k_R(\omega)z-\omega \tau)}d\omega[/tex]

is just the wave packet I sent in.

Does that look good or is the presence of the z not legal and I have to set it to 0 everywhere?
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  • #15
Yea, that looks about right, although I think there are some factors of 2pi missing. z is fine - you can just think of it as fixed throughout the calculation, then when you're done the result is true for all z.
  • #16


Related to Convolution and fourier transform puzzle

1. What is convolution in the context of Fourier transforms?

Convolution is a mathematical operation that combines two functions to create a third function that represents the amount of overlap between the original functions. In the context of Fourier transforms, convolution is used to describe the transformation of a signal from the time domain to the frequency domain.

2. How is convolution related to the Fourier transform puzzle?

The Fourier transform puzzle is a problem that involves finding the original signal from its frequency domain representation. Convolution is a key component of solving this puzzle, as it allows us to break down the signal into its individual frequency components and then reconstruct the original signal.

3. What is the difference between Fourier transform and inverse Fourier transform?

The Fourier transform is a mathematical operation that converts a signal from the time domain to the frequency domain, while the inverse Fourier transform does the opposite, converting a signal from the frequency domain back to the time domain. These two operations are essentially inverse operations of each other.

4. Can convolution and Fourier transform be applied to real-world problems?

Yes, convolution and Fourier transform are widely used in various fields such as signal processing, image processing, and audio analysis. They are powerful tools for analyzing and manipulating signals and can be applied to solve real-world problems.

5. Are there any limitations to using convolution and Fourier transform?

While convolution and Fourier transform are powerful tools, they do have some limitations. For example, convolution assumes that the signals being combined are continuous and infinite, which is not always the case in real-world applications. Additionally, Fourier transform is not suitable for analyzing signals with discontinuities or sharp changes.

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