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Fourier Series of sin (pi*t)

by anish
Tags: fourier, series
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anish
#1
Apr23-04, 12:30 AM
P: 4
I am having trouble finding the An and Bn coefficients for the fourier series f(t) = sin (pi*t) from 0<t<1, period 1

Please help! Thank you!
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ichiro_w
#2
May29-04, 11:42 PM
P: 13
Newbie wants to bump this query:

I was having trouble with this problem as well until I discovered a simple algebra error; here is how I set it up

[tex]a_0 = \frac{1}{\frac{1}{2}}\int_0^1 Sin(\pi t) \;dt = \frac{4}{\pi}[/tex]

[tex]A_N = \frac{1}{\frac{1}{2}}\int_0^1 Sin(\pi t) Cos(2 n \pi t)\; dt [/tex]

[tex]B_N = \frac{1}{\frac{1}{2}}\int_0^1 Sin(\pi t) Sin (2 n \pi t)\; dt [/tex]


A source advises the Trigonometric Identities:

[tex]2 Sin[A] Cos[B] = Sin [A+B] + Sin[A-B] [/tex]

[tex]2 Sin[A] Sin[B] = Cos[A-B] - Cos[A+B] [/tex]

let [tex] u = \pi t + 2 n\pi t = A + B \mbox{\quad and let\quad} v = \pi t - 2 n \pi t = A -B[/tex]

then

[tex]A_n = 2\; \frac{1}{2}\;\;\frac{ 1}{\pi (1+ 2 n)} \;\;\int_0^{\;\pi(1+2 n)} Sin(u)\; du[/tex]

[tex] \quad\quad\quad+ \quad 2\;\frac{1}{2}\;\;\frac{ 1}{\pi (1- 2 n)} \;\;\int_0^ {\;\pi(1-2 n)} Sin(v)\;dv [/tex]

[tex]B_n = 2\; \frac{1}{2}\;\;\frac{ 1}{\pi (1-2 n)} \;\;\int_0^{\;\pi(1-2 n)} Cos(v)\; dv[/tex]
[tex]\quad\quad + \quad 2 \;\frac{1}{2}\;\;\frac{ 1}{\pi (1+ 2 n)}\;\; \int_0^ {\;\pi(1+2 n)} Cos(u)\;du [/tex]

where the substituted limits of integration, [tex] \pi (1 \pm 2n)=\pi\pm2n\pi [/tex] for [tex] \mbox{integer}\; n\geq1[/tex] are equivalent to [tex]\pi[/tex] due to periodicity of the Sine and Cosine functions.
HallsofIvy
#3
May30-04, 12:05 PM
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Quote Quote by anish
I am having trouble finding the An and Bn coefficients for the fourier series f(t) = sin (pi*t) from 0<t<1, period 1

Please help! Thank you!
So, basically, you need to integrate sin([pi]t)sin(nt) and sin([pi]t)cos(nt).

Use the trig identities sin(a)sin(b)= (1/2)(cos(a-b)- cos(a+b)) and
sin(a)sin(b)= (1/2)(sin(a+b)+ sin(a-b)).

ichiro_w
#4
May30-04, 01:47 PM
P: 13
Fourier Series of sin (pi*t)

Yes, except for typo--second equation should read [tex]sin(a)cos(b) = 1/2 (sin[a+b] + sin[a-b])[/tex]and with the substitutions suggested these integrals tidy up rather nicely [itex]\mbox{ (hint---} B_n = 0\mbox{ for all integers,\:} n\geq 1)[/itex].
I get a good form for the actual function [itex]sin(\pi t) [/itex] being modelled by Fourier using
[tex] \frac{a_0}{2}+\sum_1^{\infty} A_n\;Cos(2 n \pi t) \quad\mbox{ \quad n is an integer}[/tex]
where
[tex]A_n =\frac{-4}{\pi (4n^2-1)}[/tex]
so
[tex]f(t)=\frac{2}{\pi}-\frac{4}{\pi}\sum_1^{\infty}\frac{Cos(2n\pi t)}{4n^2-1}=Sin(\pi t)[/tex]

[tex]f(t)\sim\frac{2}{\pi}\;-\;\frac{4}{\pi}\left(\frac{Cos(2\pi t)}{3}\;+\;\frac{Cos(4\pi t)}{15}\;+\;\frac{Cos(6\pi t)}{35}\;+\;\cdots\right ) [/tex]
ichiro_w
#5
Jun9-04, 01:57 PM
P: 13
Follows an attempt to display a plot of [itex]Sin(\pi t)[/tex] of period 1 using the initial constant term and 6 iterations of the Cos term. Click on the bitmap file.
Attached Images
File Type: bmp foo.bmp (49.8 KB, 26 views)
jdavel
#6
Jun9-04, 02:39 PM
P: 618
You guys are missing the forest for the trees!

The fourier series is a sum of sines and cosines, and it's unique. So if you can find one set of coefficients that works, you've got THE fourier series.

So if

sin(pi*t) = a0 + An*SUM[sin(n*pi*t)] + Bn*SUM[cos(n*pi*t)],

can't you just look at that and see a set of coefficients that will make the left and right hand sides of that equation the same?
ichiro_w
#7
Jun9-04, 08:20 PM
P: 13
OK, here is the same problem except make the period 2 and let the function to be modelled by Fourier series be:

[itex] f(t) =\left\{\begin{array}{cr}0&\mbox{if\;\;}-1<t<0\\sin(\pi t)&\mbox{if\quad} 0<t<1\end{array}\right [/itex]

Calculations very similar to those pictured earlier above and adjusted only slightly for the period size being doubled lead to the following series

[itex]\frac{1}{\pi} \;+\;\frac{1}{2}\;Sin(\n\pi t)\;-\frac{2}{\pi}\;\sum_{n=2}^{\infty}\;\frac{Cos(n\;\pi\;t)}{n^2-1}=f(t)[/itex]

This series produces a very nice periodic plot modelling f(t) with only 5 iterations of the Cos term .[tex]B_1=\frac{1}{2}[/tex] was calculated separately with its own integral as the general expression for [tex]B_n[/tex] is undefined at n=1 and 0 for [tex]n\geq2[/tex]. For the same reason the Cos terms are iterated starting at n=2 with [tex]A_1 = 0[/tex])

Click on the bitmap to see the plot so generated.
Attached Images
File Type: bmp foo2.bmp (49.8 KB, 26 views)


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