July 22nd, 2017, 08:51 AM  #1 
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0  Fourier transform (application in electromagnetism)
Hello everybody again! I hope the moderator will not angry for the fact that I am starting second thread today. If something's wrong I can write my question in my first thread. I am examining the same text (actually it is compilation of very brief fragments) from the same author.... But this fragment is dedicated to different topic: Fourier transform. My request again is addressed to the good Mathematicians with kind: please, review the fragment and tell me whether it is "reliable", whether there are errors in the mathematical expressions which are used..... These are so complicated expressions... I feel helpless to check out their correctness. Here it is the fragment, I did my best to translate it correct in English. I can make additional explications of the words, if this is necessary...... 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(ωωo)tdt  e^jψ∫(0/∞)e^j(ω+ωo)tdt] = (Am/2j){e^jψ[1/j(ωωo)]e^j(ωωo)t(∞/0)  e^jψ[1/j(ω+ωo)]e^j(ω+ωo)t(∞/0)} = (Am/2j){e^jψ[1/j(ωωo)]  e^jψ[1/j(ω+ωo)]} = (Am/2){[1/(ωωo)]e^jψ  [1/(ω+ωo)]e^jψ} = (Am/2){[1/(ω+ωo)]e^jψ  [1/(ωωo)]e^jψ} = [Am/2(ω^2  ωo^2)][(ωωo)e^jψ  (ω+ωo)e^jψ] = Am[ωo/(ω^2  ωo^2)][(ωωo)e^jψ  (ω+ωo)e^jψ]/2ωo = Am[ωo/(ωo^2  ω^2)][(ωo+ω)e^jψ + (ωoω)e^jψ]/2ωo = F(Amsinωot)[(ωo+ω)e^jψ + (ωoω)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). 
July 22nd, 2017, 09:43 PM  #2  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,829 Thanks: 648 Math Focus: Yet to find out.  Quote:
What is important, and has already been mentioned, is presentation. When you try to read over this post, do you find it easy or difficult to read? I will try help with formatting first point. Quote:
 
July 22nd, 2017, 10:08 PM  #3  
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0  Quote:
I will do my best now to edit my post.... It will take time probably, because I must figure out how to rewrite them in your manner......  
July 22nd, 2017, 10:48 PM  #4  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,829 Thanks: 648 Math Focus: Yet to find out.  Quote:
 
July 22nd, 2017, 10:59 PM  #5 
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0 
I edited a little bit the arrangement; I hope I made the text at least a little bit easier for reading. 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(ωωo)tdt  e^jψ∫(0/∞)e^j(ω+ωo)tdt] = (Am/2j){e^jψ[1/j(ωωo)]e^j(ωωo)t(∞/0)  e^jψ[1/j(ω+ωo)]e^j(ω+ωo)t(∞/0)} = (Am/2j){e^jψ[1/j(ωωo)]  e^jψ[1/j(ω+ωo)]} = (Am/2){[1/(ωωo)]e^jψ  [1/(ω+ωo)]e^jψ} = (Am/2){[1/(ω+ωo)]e^jψ  [1/(ωωo)]e^jψ} = [Am/2(ω^2  ωo^2)][(ωωo)e^jψ  (ω+ωo)e^jψ] = Am[ωo/(ω^2  ωo^2)][(ωωo)e^jψ  (ω+ωo)e^jψ]/2ωo = Am[ωo/(ωo^2  ω^2)][(ωo+ω)e^jψ + (ωoω)e^jψ]/2ωo = F(Amsinωot)[(ωo+ω)e^jψ + (ωoω)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) 
July 23rd, 2017, 01:07 AM  #6 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
Did you look at the latex websites for maths I offered you in post4 of your book thread? They could save you a great deal of work. 
July 23rd, 2017, 03:52 AM  #7  
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0  Quote:
By the way, I just saw your reply on the Physics forum. Sure, I am very far from understanding Fourier's transform... I have no illusions. That's why my request is... just tell me, please, whether the quoted fragment is correct or there errors in it. That's it.... If it it too unreadable, I will soon post a better version written by latex website.....  
July 23rd, 2017, 04:07 AM  #8  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,829 Thanks: 648 Math Focus: Yet to find out.  Quote:
To avoid a possible scenario of the 'blind leading the blind', it might be best to focus on understanding these ideas separate from this one source. Then come back to it later. Also i realise you may be waiting for books to arrive, but in the meantime it may be worth browsing the web for related material.  
July 23rd, 2017, 04:28 AM  #9  
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0  Quote:
And in the meantime.. sure, I am trying to gain knowledge. It will be slow and difficult process...  
July 23rd, 2017, 04:49 AM  #10 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
Mathematics in particular is not a linear subject. By this I mean that you must learn a bit of one topic in order to understand a bit of another topic. Often knowledge of the second topic will then allow you to learn more about the first. This is a bit like the twin entwined strands of DNA, except that there are many strands or topics in Mathematics. So progress is not along a line, nor is it around a circle as then you would end up where you started. It is a sort of spiral, where you study one topic then another, then another and then back to first. Formal courses are designed to offer the correct path through this maze. Have you studied Fourier Series (or any series) yet? 

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application, electromagnetism, fourier, transform 
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