Fourier Series of sin (pi*t)

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!

ichiro_w

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

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

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

Follows an attempt to display a plot of [latex]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

foo.bmp (49.8 KB, 24 views)

jdavel

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

OK, here is the same problem except make the period 2 and let the function to be modelled by Fourier series be:

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

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

[latex]\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)[/latex]

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

foo2.bmp (49.8 KB, 24 views)