Question

I have a text book which writes a function of time can be written as (V0 is the frequency)
F(t)=A02+∑m=1∞Rmcos(2πmv0t+ϕm)
and with
Am=Rmcosϕm
Bm=Rmsinϕm
then Function can be rewrite as:
F(t)=A02+∑m=1∞e2πmv0t(Am−iBm)
It seems there are some internal relations between these two functions and I have tried expand and elaborate these two functions but can't work out a direct proof can someone help me out?

Answer 1

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Answer 2

I have a text book which writes a function of time can be written as (V0 is the frequency)
\[F(t)=\frac{ A _{0} }{ 2 }+\sum_{m=1}^{\infty} R _{m} \cos(2 \pi mv _{0}t+\phi _{m})\]
and with \[A_{m}=R_{m}\cos \phi_{m} \] \[B_{m}=R_{m}\sin \phi_{m} \]
then Function can be rewrite as:
\[F(t)=\frac{ A_{0} }{2 }+\sum_{m=1}^{\infty}e ^{2\pi mv_{0}t}(A_{m}-iB_{m})\]
It seems there are some internal relations between these two functions and I have tried expand and elaborate these two functions but can work out a direct proof can someone help me out?

Answer 3

i haven't tried it on paper, but these relation should lead you to what u need,
\(\cos (A+B) = \cos A\cos B-\sin A \sin B , \quad \cos (2\pi m v_ot+\phi_o)=...?\)
\(\cos x = (e^x+e^{-x})/2 \\ \sin x =(e^x-e^{-x})/2i \)
try it and ask if you get stuck...

Answer 4

@hartnn close but those should be imaginary exponentials
@shadow_wxh This 'relationship' is merely Euler's:$$e^{i\theta}=\cos\theta+i\sin\theta\\\cos\theta=\frac12\left(e^{i\theta}+e^{-i\theta}\right)\\\sin\theta=\frac1{2i}\left(e^{i\theta}-e^{-i\theta}\right)$$

Answer 5

yes, i meant that :)

Answer 6

To see their first step,$$\cos(2\pi mv_0t+\phi_m)=\cos(2\pi mv_0 t)\cos(\phi_m)+\sin(2\pi mv_0 t)\sin\phi_m$$

Answer 7

oldrin, let shadow do the work!

Answer 8

$$e^{2\pi mv_0 t}(A-iB_m)=R_m(\cos(2\pi mv_0 t)+i\sin(2\pi mv_0 t))(\cos\phi_m-i\sin\phi_m)=\cdots?$$

Answer 9

I have done the above but missing any further lead

Answer 10

\[R_{m}\cos(2pimv_{0}t+\phi_{m})=R_{m}\cos \phi_{m}\cos(2 \pi mv_{0}t)-R_{m}\sin \phi_{m}\sin(2 \pi mv_{0}t)\]

Answer 11

\[=A_{m}\cos(2 \pi mv_{0}t)-B_{m}\sin(2 \pi mv_{0}t)\]

Answer 12

Good so far...

Answer 13

hmm gonna need a theory there

Answer 14

Any body give me a hint?

Answer 15

http://en.wikipedia.org/wiki/Fourier_series#Exponential_Fourier_series

Answer 16

Ok I try work backwards\[e ^{2 \pi m v_{0}t}(A_{m}-iB_{m})=[\cos(2 \pi mv_{0}t+i \sin(2\pi mv_{0}t)](A_{m}-iB{m})\]\[=A_{m}\cos(2\pi mv_{0}t)+iA_{m}\sin(2\pi mv_{0}t)-iB_{m}\cos(2\pi mv_{0}t)+B_{m}\sin(2\pi mv_{0}t)\]
if assume the question is true then
\[e ^{2 \pi m v_{0}t}(A_{m}-iB_{m})=A_{m}\cos(2\pi mv_{0}t)+B_{m}\sin(2\pi mv_{0}t)\]
which left
\[iA_{m}\sin(2\pi mv_{0}t)-iB_{m}\cos(2\pi mv_{0}t)=0\]
\[iR_{m} cos\phi_{m}sin(2 \pi mv_{0}t)-iR_{m}sin\phi _{m}cos(2\pi mv_{0}t)=0\]
\[iR_{m}sin(2\pi mv_{0}t-\phi_ {m})=0\]
\[2\pi mv_{0}t=\phi_{m}\] ?????Really?

Answer 17

Corrrection there is i missing in the above \[e^{2\pi imv_{0}t}\]

Answer 18

come on people try this out

Answer 19

Tried many way coudln't prove it unless m goes from minus infinity to positive infinity the closest equation is
\[F(t)=\sum_{m=-\infty}^{+\infty}C_{m}e^{2\pi imv_{0}t}\] where \[C_{m}=\frac{A_{m}-iB_{m}}{2}\] But that doesnt make them equal so i'll say there is some mistake i the text book.

Answer 20

OK anyone wants to challenge this or i'll just close it

Answer 21

huh?

Answer 22

You have any better suggestion?