Fourier Series Transform Proof Help

  • #1

Main Question or Discussion Point

Can someone fill in the blank between these two steps? I can't find fourier series proof anywhere and my professor just left it out.

(1) y(t+nT)=y(t)

(2) y(t)=[tex]A_{0}[/tex] + [tex]\Sigma^{\infty}_{n=1}[/tex][[tex]A_{n}[/tex]cos(n[tex]\omega[/tex]t) + [tex]B_{n}[/tex]sin(n[tex]\omega[/tex]t)]

(The omega is going crazy on me... it's not supposed to be superscripted, just multiplied by n and t)
 

Answers and Replies

  • #2
HallsofIvy
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Can someone fill in the blank between these two steps? I can't find fourier series proof anywhere and my professor just left it out.

(1) y(t+nT)=y(t)

(2) y(t)=[tex]A_{0}[/tex] + [tex]\Sigma^{\infty}_{n=1}[/tex][[tex]A_{n}[/tex]cos(n[tex]\omega[/tex]t) + [tex]B_{n}[/tex]sin(n[tex]\omega[/tex]t)]

(The omega is going crazy on me... it's not supposed to be superscripted, just multiplied by n and t)
What do you mean by "steps between them"? The first just says y is periodic with period T and the second is the general expression of a Fourier series of a function periodic with period [itex]2\pi/\omega[/itex]- there is no mention of "T".

As for the LaTex, I would recommend putting the entire thing in [ t e x] not just individual parts:

[tex]y(t)=A_{0}+ \Sigma^{\infty}_{n=1}[A_{n}cos(n\omega t) + B_{n}sin(n\omega t)][/tex]

It looks better and is easier to type!
 
  • #3
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I would say that a general "Fourier expansion" is actually an integral. What (1) implies is that the modes are discrete and thus the integral becomes a sum, and therefore [itex]\omega=2 \pi/T [/itex], as Halls mentioned. Maybe this is the missing step you mean?
 
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  • #4
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have u find the gap between those two statements[evotunedscc]?
 

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