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Fourier Series of sin (pi*t) 
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#1
Apr2304, 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! 


#2
May2904, 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[AB] [/tex] [tex]2 Sin[A] Sin[B] = Cos[AB]  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(12 n)} Sin(v)\;dv [/tex] [tex]B_n = 2\; \frac{1}{2}\;\;\frac{ 1}{\pi (12 n)} \;\;\int_0^{\;\pi(12 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. 


#3
May3004, 12:05 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,569

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


#4
May3004, 01:47 PM

P: 13

Fourier Series of sin (pi*t)
Yes, except for typosecond equation should read [tex]sin(a)cos(b) = 1/2 (sin[a+b] + sin[ab])[/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^21)}[/tex] so [tex]f(t)=\frac{2}{\pi}\frac{4}{\pi}\sum_1^{\infty}\frac{Cos(2n\pi t)}{4n^21}=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] 


#5
Jun904, 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.



#6
Jun904, 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? 


#7
Jun904, 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^21}=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. 


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