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Subject: "fourier transform"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #683
Reading Topic #683
prathibha
Member since Jun-26-08
Jun-26-08, 06:58 AM (EST)
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"fourier transform"
 
   Fourier transform of a function f(x)is defined as
F(k) = Integrate<( f(x) Exp(-i k x ) ), {x, -Infinity,Infinity}> where the first braket is the integrand and the curly braket is the integration variable and the limits of integration. And the inverse tranform is

f(x) = Integrate<( F(k) Exp(i k x) ), {k, -Infinity,Infinity}>

where the functional form of the two functions may be different.

Now my problem is when we apply this to any physical situation, where the f(x) and F(k) will be some physical quantities eg. electric field, magnetic field, current density, charge density etc.

According to dimensional analysis, both sides of the equation should have the same dimension. So if we take a physical situation where say our function is current density. Whether the fourier transform of current density also have the dimension of current density?

Let J(r) = Integrate<( J(q) Exp( ik.r ) ), {k,-Infinity,Infinity}>

Now the exponent is dimensionless and dq has the dimension of
(L)^(-1). So should J(r)will have the dimensional of J(k) multiplied by dimension of length.

Also can somebody please say what is the dimension of dirac delta function? whether \delta(x) has the dimension of 1/x ?

Thanks


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prathibha
guest
Jun-26-08, 11:46 PM (EST)
 
1. "RE: fourier transform"
In response to message #0
 
   >Let J(r) = Integrate<( J(q) Exp( ik.r ) ), {k,-Infinity,Infinity}>

Here is a mistake it'should be "J(k)" in the RHS

>Now the exponent is dimensionless and dq has the dimension of
(L)^(-1).

Here also one mistake it should be "dk has the dimension of (L)^(-1)"

>So should J(r)will have the dimensional of J(k)
>multiplied by dimension of length.

Here also one "So should J(r)will have the dimension of J(k)
multiplied by dimension of inverse of length."

Thanks

sorry for the inconvenience


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alexbadmin
Charter Member
2239 posts
Jun-27-08, 00:20 AM (EST)
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2. "RE: fourier transform"
In response to message #1
 
   Roughly speaking, what Fourier transform does is representing a function as a sum of sinusoidal waves. (Integral is a limiting case of a sum.)

I would assume the following:

1. If any two (or more) entities can be added up they have the same dimension.

2. Since the Fourier transform adds up sines times coefficients, either sines or coefficients have the dimension of the original function.

3. However, sines appear in Fourier transforms of quantities of various dimensions, so I think it is reasonable to assume them dimensionless.

4. Thus Fourier coefficients inherit the dimension of the function.

5. This is consistent with the fact that the inverse transform gets the function back and differs from the direct transform only in a sign of the exponent.

Further, Dirac's delta is like sine. It is dimensionless.


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