Author Topic: Fourier Series (Read 2020 times) Tweet Share

QuantumJG · « **on:** July 23, 2010, 03:31:12 pm »

Find the Fourier coefficients of $|sin( \omega t )|$

now I know that since this is an even function that implies b_n=0

But what about a_n?

Now I know:

$a_{n} = \dfrac{ \omega }{ \pi } \int _{0} ^{\dfrac{2 \pi}{ \omega }} |sin( \omega t )| cos( n \omega t ) dt$

But for some reason even using the standard integrals I keep getting a_n to also be 0.

Can somebody help?

EDIT: WHOOPS!

So

$a_{n} = \dfrac{ \omega }{ \pi } \int _{0} ^{\dfrac{\pi}{ \omega }} sin( \omega t ) cos( n \omega t ) dt - \dfrac{ \omega }{ \pi } \int _{\dfrac{\pi}{ \omega }} ^{\dfrac{2 \pi}{ \omega }} sin( \omega t ) cos( n \omega t ) dt$

Thanks!

Ahmad · « **Reply #1 on:** July 23, 2010, 03:51:26 pm »

I think you missed an absolute value around sine in the integral.

/0 · « **Reply #2 on:** July 23, 2010, 05:34:10 pm »

Hmm I'm just wondering, why did you put $a_n = \frac{\omega}{\pi}\int_0^{\frac{2\pi}{\omega}}|\sin{\omega t}|\cos{n\omega t}\,dt$ instead of $a_n = \frac{1}{\pi}\int_0^{2\pi}|\sin{\omega t}|\cos{nt}\,dt$

We haven't covered Fourier Series yet, but reading ahead in my book that's the formula they give.

Given a fixed value of $\omega=1$ I tried the latter formula and got:

$|\sin{(x)}| = 4\ -\ \frac{4}{3}\cos{2x}\ -\ \frac{4}{15}\cos{4x}\ -\ \frac{4}{35}\cos{6x}\ -\ \frac{4}{63}\cos{8x}\ -\ \frac{4}{99}\cos{10x}\ -\ \frac{4}{143}\cos{12x}\ -\ ...$

Fourier analysis is cool

Cthulhu · « **Reply #3 on:** July 23, 2010, 05:52:08 pm »

My textbook uses
$a_n = \frac{1}{\pi}\int^{2\pi}_0 |\sin\omega t| \cos nt \ dt$
too.

QuantumJG · « **Reply #4 on:** July 23, 2010, 09:40:25 pm »

Quote from: /0 on July 23, 2010, 05:34:10 pm

Hmm I'm just wondering, why did you put $a_n = \frac{\omega}{\pi}\int_0^{\frac{2\pi}{\omega}}|\sin{\omega t}|\cos{n\omega t}\,dt$ instead of $a_n = \frac{1}{\pi}\int_0^{2\pi}|\sin{\omega t}|\cos{nt}\,dt$

We haven't covered Fourier Series yet, but reading ahead in my book that's the formula they give.

Given a fixed value of $\omega=1$ I tried the latter formula and got:

$|\sin{(x)}| = 4\ -\ \frac{4}{3}\cos{2x}\ -\ \frac{4}{15}\cos{4x}\ -\ \frac{4}{35}\cos{6x}\ -\ \frac{4}{63}\cos{8x}\ -\ \frac{4}{99}\cos{10x}\ -\ \frac{4}{143}\cos{12x}\ -\ ...$

Fourier analysis is cool

Definately

The only problem is that we do this last, since we do electromagnetism first then optics (I may actually come to love optics). Having said that electromagnetism looks pretty cool too.

~~With the formula I gave, both forms are equivalent its just my lecture notes use what I put up.~~

In our lecture notes:

$a_n = \frac{2}{T} \int _{0} ^{T} f(t) \cos{n\omega t}\,dt$

So I thought that if:

$T = \frac{2 \pi}{ \omega }$

Then:

$a_n = \frac{\omega}{\pi}\int_0^{\frac{2\pi}{\omega}}|\sin{\omega t}|\cos{n\omega t}\,dt$

mark_alec · « **Reply #5 on:** July 23, 2010, 10:08:59 pm »

I've never seen two Fourier theorem/integral/functions in the same notation

QuantumJG · « **Reply #6 on:** July 24, 2010, 04:46:40 pm »

Quote from: /0 on July 23, 2010, 05:34:10 pm

Hmm I'm just wondering, why did you put $a_n = \frac{\omega}{\pi}\int_0^{\frac{2\pi}{\omega}}|\sin{\omega t}|\cos{n\omega t}\,dt$ instead of $a_n = \frac{1}{\pi}\int_0^{2\pi}|\sin{\omega t}|\cos{nt}\,dt$

We haven't covered Fourier Series yet, but reading ahead in my book that's the formula they give.

Given a fixed value of $\omega=1$ I tried the latter formula and got:

$|\sin{(x)}| = \frac{2}{ \pi } -\ \frac{4}{3}\cos{2x}\ -\ \frac{4}{15}\cos{4x}\ -\ \frac{4}{35}\cos{6x}\ -\ \frac{4}{63}\cos{8x}\ -\ \frac{4}{99}\cos{10x}\ -\ \frac{4}{143}\cos{12x}\ -\ ...$

Fourier analysis is cool

This is partially incorrect.

It should be:

$|\sin{(x)}| = \frac{2}{ \pi } -\ \frac{4}{3 \pi }\cos{2x}\ -\ \frac{4}{15 \pi }\cos{4x}\ -\ \frac{4}{35 \pi }\cos{6x}\ -\ \frac{4}{63 \pi }\cos{8x}\ -\ \frac{4}{99 \pi }\cos{10x}\ -\ \frac{4}{143 \pi }\cos{12x}\ -\ ...$

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