 My Math Forum Fourier transform (application in electromagnetism)
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 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(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). July 22nd, 2017, 09:43 PM   #2
Senior Member

Joined: Feb 2016
From: Australia

Posts: 1,839
Thanks: 653

Math Focus: Yet to find out.
Quote:
 Originally Posted by DesertFox Hello everybody again! I hope the moderator will not angry for the fact that I am starting second thread today.
There is no limit (that i know of) to the number of posts you can make. Only some extreme cases result in the closing of threads etc.

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:
 Originally Posted by DesertFox 1. Fourier’s transform: $\mathcal{F}[f(t)] = \int_0^\infty f(t) e^{j\phi (t)} dt = \int_0^\infty f(t)e^{-j\omega t}dt = F(\omega)$ $\omega$ – angular velocity; t – time; the minus ("-") is used because of the fact that the “traversed” angle $\phi(t)$ is measured toward (in relation to) the current position. Measure of the frequency: $F(\sin t) = F(\omega) = [0 ; \omega < 1]; [1; \omega = 1]; [0 ; \omega > 1]; \omega > 0$ July 22nd, 2017, 10:08 PM   #3
Member

Joined: Jul 2017
From: europe

Posts: 51
Thanks: 0

Quote:
 Originally Posted by Joppy There is no limit (that i know of) to the number of posts you can make. Only some extreme cases result in the closing of threads etc. 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.
I am sorry for the bad presentation of the text. I just don't have the skills (and the necessary tool) to make the equations more "readable". Actually, in the original text they are written exactly in the same way, I just copied them... And I know it is difficult for reading......
I will do my best now to edit my post.... It will take time probably, because I must figure out how to re-write them in your manner...... July 22nd, 2017, 10:48 PM   #4
Senior Member

Joined: Feb 2016
From: Australia

Posts: 1,839
Thanks: 653

Math Focus: Yet to find out.
Quote:
 Originally Posted by DesertFox I am sorry for the bad presentation of the text. I just don't have the skills (and the necessary tool) to make the equations more "readable". Actually, in the original text they are written exactly in the same way, I just copied them... And I know it is difficult for reading...... I will do my best now to edit my post.... It will take time probably, because I must figure out how to re-write them in your manner......
I forgot to mention that if you quote a post, you can view the latex commands. 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(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) 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. Thanks from DesertFox July 23rd, 2017, 03:52 AM   #7
Member

Joined: Jul 2017
From: europe

Posts: 51
Thanks: 0

Quote:
 Originally Posted by studiot 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.
I am just trying to rewrite my post by the use of latex...
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: 1,839
Thanks: 653

Math Focus: Yet to find out.
Quote:
 Originally Posted by DesertFox I am just trying to rewrite my post by the use of latex... 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.....
It seems as if you are working through a text that you aren't sure is correct or not. More than this, you haven't studied the underlying material prior to your analysis of this persons work.

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:
 Originally Posted by Joppy It seems as if you are working through a text that you aren't sure is correct or not. More than this, you haven't studied the underlying material prior to your analysis of this persons work. 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.
It is important for me to know whether he (the author) is correct or not.

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? Thanks from jks Tags application, electromagnetism, fourier, transform Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post szz Applied Math 0 December 16th, 2015 01:03 PM jsmith613 Calculus 1 October 1st, 2015 09:02 AM beckie Real Analysis 3 June 20th, 2010 12:58 PM tara Real Analysis 0 December 30th, 2009 02:10 PM aptx4869 Real Analysis 5 September 3rd, 2008 12:47 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      