Physics Help Forum Fourier transform (application in electromagnetism)

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 Jul 22nd 2017, 12:15 PM #1 Junior Member   Join Date: Jul 2017 Posts: 27 Fourier transform (application in electromagnetism) Hello buddies! I am examining a brief fragment which is dedicated to Fourier transform. My request again is addressed to the good Physicists and Mathematicians with kind heart: please, review the fragment and tell me whether it is "reliable", whether there are errors in the mathematical expressions..... These are so complicated expressions... I feel helpless to check out their correctness in terms of both Physics and Mathematics. Here it is the fragment, I did my best to translate it correctly in English. Here we go: 1. Fourier’s transform: F[f(t)] = ∫(0/∞)f(t)e^jφ(t)dt = ∫(0/∞)f(t)e^-jωtdt = F(ω) ω – angular velocity; t – time; the minus ("-") is used because of the fact that the “traversed” angle φ(t) is measured toward (in relation to) the current position. Measure of the frequency: F(sint) = F(ω) = [0; ω < 1]; [1; ω = 1]; [0; ω> 1]; ω > 0 2. Inverse Fourier’s transform: f(t)=∫(-∞/∞)F(f)e^jφ(f)df=∫(-∞/∞)F(f)e^jt2πfdf=∫(-∞/∞)F(ω/2π)e^jt2π(ω/2π)d(ω/2π)=(1/2π)∫(-∞/∞)F(ω)e^jtωdω = F{-1}[F(ω)] t - time; f = 1/T = ω/2π - frequency; T – period; the angle is “static” and because of that it is ‘unconsidered” 3. Fourier’s transform applied to sinusoidal regime (electromagnetism): f(t) = Amsin(ωot + ψ) -> F(ω) = ∫(0/∞)Amsin(ωot + ψ)e^-jωtdt = Am∫(0/∞){[e^j(ωot+ψ) - e^-j(ωo+ψ)]/2j}e^-jωtdt = (Am/2j)[e^jψ∫(0/∞)e^-j(ω-ωo)tdt - e^-jψ∫(0/∞)e^-j(ω+ωo)tdt] = (Am/2j){e^jψ[-1/j(ω-ωo)]e^-j(ω-ωo)t(∞/0) - e^-jψ[-1/j(ω+ωo)]e^-j(ω+ωo)t(∞/0)} = (Am/2j){e^jψ[1/j(ω-ωo)] - e^-jψ[1/j(ω+ωo)]} = (-Am/2){[1/(ω-ωo)]e^jψ - [1/(ω+ωo)]e^-jψ} = (Am/2){[1/(ω+ωo)]e^-jψ - [1/(ω-ωo)]e^jψ} = [Am/2(ω^2 - ωo^2)][(ω-ωo)e^-jψ - (ω+ωo)e^jψ] = Am[ωo/(ω^2 - ωo^2)][(ω-ωo)e^-jψ - (ω+ωo)e^jψ]/2ωo = Am[ωo/(ωo^2 - ω^2)][(ωo+ω)e^jψ + (ωo-ω)e^-jψ]/2ωo = F(Amsinωot)[(ωo+ω)e^jψ + (ωo-ω)e^-jψ]/2ωo; ω = ωo -> F(ω) = Ame^jψ = √2Ae^jψ -> Ae^jψ = Acosψ + jAsinψ; (Am - amplitude; A – “effective” value; ψ – “initial” stage) 4. Inverse Fourier’s transform applied to sinusoidal regime (electromagnetism): f(t) = (1/2π)∫(-∞/∞)Ame^jψe^jtωdω = (1/2π)Am∫(-∞/∞)cos(ωt+ψ)dω + j(1/2π)Am∫(-∞/∞)sin(ωt+ψ)dω = (1/2π)Am∫(-∞/∞)cosφ(ω)dω + j(1/2π)Am∫(-∞/∞)sinφ(ω)dω = Am∫(-∞/∞)cosφ(ω/2π)d(ω/2π) + jAm∫(-∞/∞)sinφ(ω/2π)d(ω/2π) = Am∫(-∞/∞)cosφ(f)df + jAm∫(-∞/∞)sinφ(f)df = Amsinφ(f)(∞/-∞) - jAmcosφ(f)(∞/-∞) = Amsinφ(f)(∞/-∞) (cosφ is an even function); φ(f) = 2πfot + ψ (f := fo) -> f(t) = Amsin(ωot+ψ)
 Jul 22nd 2017, 12:23 PM #2 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 550 sorry, I find that text a bit hard to read. A useful tool for typesetting maths is Online LaTeX Equation Editor - create, integrate and download Regarding fourier transforms I found this a useful tool. Wolfram|Alpha: Computational Knowledge Engine
 Jul 23rd 2017, 12:02 AM #3 Junior Member   Join Date: Jul 2017 Posts: 27 I edited a little bit the arrangement of the fragment. I hope I made it a little bit easier for reading and discussion. Here it is: 1. Fourier’s transform: F[f(t)] = ∫(0/∞)f(t)e^jφ(t)dt = ∫(0/∞)f(t)e^-jωtdt = F(ω) ω – angular velocity; t – time; the minus ("-") is used because of the fact that the “traversed” angle φ(t) is measured toward (in relation to) the current position. Measure of the frequency: F(sint) = F(ω) = [0; ω < 1]; [1; ω = 1]; [0; ω> 1]; ω > 0 2. Inverse Fourier’s transform: f(t)=∫(-∞/∞)F(f)e^jφ(f)df= ∫(-∞/∞)F(f)e^jt2πfdf= ∫(-∞/∞)F(ω/2π)e^jt2π(ω/2π)d(ω/2π)= (1/2π)∫(-∞/∞)F(ω)e^jtωdω = F{-1}[F(ω)] t - time; f = 1/T = ω/2π - frequency; T – period; the angle is “static” and because of that it is ‘unconsidered” 3. Fourier’s transform applied to sinusoidal regime (electromagnetism): f(t) = Amsin(ωot + ψ) -> F(ω) = ∫(0/∞)Amsin(ωot + ψ)e^-jωtdt = Am∫(0/∞){[e^j(ωot+ψ) - e^-j(ωo+ψ)]/2j}e^-jωtdt = (Am/2j)[e^jψ∫(0/∞)e^-j(ω-ωo)tdt - e^-jψ∫(0/∞)e^-j(ω+ωo)tdt] = (Am/2j){e^jψ[-1/j(ω-ωo)]e^-j(ω-ωo)t(∞/0) - e^-jψ[-1/j(ω+ωo)]e^-j(ω+ωo)t(∞/0)} = (Am/2j){e^jψ[1/j(ω-ωo)] - e^-jψ[1/j(ω+ωo)]} = (-Am/2){[1/(ω-ωo)]e^jψ - [1/(ω+ωo)]e^-jψ} = (Am/2){[1/(ω+ωo)]e^-jψ - [1/(ω-ωo)]e^jψ} = [Am/2(ω^2 - ωo^2)][(ω-ωo)e^-jψ - (ω+ωo)e^jψ] = Am[ωo/(ω^2 - ωo^2)][(ω-ωo)e^-jψ - (ω+ωo)e^jψ]/2ωo = Am[ωo/(ωo^2 - ω^2)][(ωo+ω)e^jψ + (ωo-ω)e^-jψ]/2ωo = F(Amsinωot)[(ωo+ω)e^jψ + (ωo-ω)e^-jψ]/2ωo; ω = ωo -> F(ω) = Ame^jψ = √2Ae^jψ -> Ae^jψ = Acosψ + jAsinψ; (Am - amplitude; A – “effective” value; ψ – “initial” stage) 4. Inverse Fourier’s transform applied to sinusoidal regime (electromagnetism): f(t) = (1/2π)∫(-∞/∞)Ame^jψe^jtωdω = (1/2π)Am∫(-∞/∞)cos(ωt+ψ)dω + j(1/2π)Am∫(-∞/∞)sin(ωt+ψ)dω = (1/2π)Am∫(-∞/∞)cosφ(ω)dω + j(1/2π)Am∫(-∞/∞)sinφ(ω)dω = Am∫(-∞/∞)cosφ(ω/2π)d(ω/2π) + jAm∫(-∞/∞)sinφ(ω/2π)d(ω/2π) = Am∫(-∞/∞)cosφ(f)df + jAm∫(-∞/∞)sinφ(f)df = Amsinφ(f)(∞/-∞) - jAmcosφ(f)(∞/-∞) = Amsinφ(f)(∞/-∞) (cosφ is an even function); φ(f) = 2πfot + ψ (f := fo) -> f(t) = Amsin(ωot+ψ)
Jul 23rd 2017, 12:56 AM   #4
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 Originally Posted by DesertFox ..the minus ("-") is used because of the fact that the “traversed” angle φ(t) is measured toward (in relation to) the current position.
I don't understand this comment. Why are you calling φ(t) an angle and why refer to it as the "traversed” angle? And why is it the negative sign that makes you say that?

I took an entire course on transform methods, not to mention where I used them in physics courses (E.g. quantum mechanics), and this is all news to me.

Jul 23rd 2017, 01:00 AM   #5
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 Originally Posted by Pmb I don't understand this comment. Why are you calling φ(t) an angle and why refer to it as the "traversed” angle? And why is it the negative sign that makes you say that? I took an entire course on transform methods, not to mention where I used them in physics courses (E.g. quantum mechanics), and this is all news to me.
I cannot understand his comment; that's what I'm trying to grasp.
But let's focus on the equations, on the mathematical expressions.
Are there any errors? Or they're correct?

 Jul 23rd 2017, 01:15 AM #6 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,574 Maybe tomorrow I'll go through it. But there's no physics here. So far this is just pure math. Fourier transforms have applications in other fields of science such as chemistry and computer science. Applications are: 1) Analysis of differential equations 2) Fourier transform spectroscopy 3) Quantum mechanics 4) Signal processing For details please see: See: https://en.wikipedia.org/wiki/Fourie...m#Applications Please tell me that you're not still trying to understand the text written by that idiot, are you? If you are then off to bed without your supper. Don't play with the bad boys! Last edited by Pmb; Jul 23rd 2017 at 01:37 AM.
 Jul 23rd 2017, 01:38 AM #7 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,574 You should buy a good textbook on mathematical physics. I can make a few recommendations if you'd like?
 Jul 23rd 2017, 01:58 AM #8 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 In the very first line the limits of the first integration are those for the Fourier sine transformation ( 0 to infinity), but the integrating factor is the exponential function. The Fourier exponential transformation should be negative infinity to positive infinity. I have to tell you that the standard of your mathematics is a long way from up to understanding or using Fourier transforms. This is Final undergraduate and into graduate level stuff at university. To expand on what PMB says, Fourier transforms are one of many types of integral transforms that have very wide application in applied mathematics, optics, heat, hydrodynamics, elasticity,electric circuit theory, electromagnetism, signal and communication theory and geophysics to name but a few. Integral transforms always come in pairs the transform and the inverse transform. They are useless without both parts.The most common in electrical theory is the Laplace transform. The idea of a transform is to make the solution of an equation easier to solve by transforming it into another variable. You then solve the easier equation in the other variable. But then the price for this you pay is you have to transform the solution back to the original variable, hence the need for the inverse transform. You may have done logs in school, this is a transform method for multiplying to numbers, a and b so that ab may be difficult but log(ab) = log(a) + log(b) So the log transforms a multiplication into an addition, but to get the actual product you need to take the antilog. Last comment, there is something called the Fast Fourier Transform (FFT). This is something quite different and should not be confused with the Fourier transform. The FFT is basically a numerical method, suitable for computers, of spectral or harmonic analysis of data points. Last edited by studiot; Jul 23rd 2017 at 02:37 AM.
Jul 23rd 2017, 06:22 AM   #9
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 Originally Posted by Pmb You should buy a good textbook on mathematical physics. I can make a few recommendations if you'd like?
Please, check out the notation of the Inverse Fourier transform.
t - time;
f = 1/T = ω/2π - frequency

Is it correct?
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Jul 23rd 2017, 07:05 AM   #10
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 Originally Posted by DesertFox Please, check out the notation of the Inverse Fourier transform. t - time; f = 1/T = ω/2π - frequency Is it correct?
I am unable to right not. I am ill and as a result am unable to concentrate on math.

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