How Do You Calculate Fourier Series Coefficients for f(t) = 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

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

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

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


A source advises the Trigonometric Identities:

2 Sin[A] Cos<b> = Sin [A+B] + Sin[A-B] </b>

2 Sin[A] Sin<b> = Cos[A-B] - Cos[A+B] </b>

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

then

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

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

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

where the substituted limits of integration, \pi (1 \pm 2n)=\pi\pm2n\pi for \mbox{integer}\; n\geq1 are equivalent to \pi due to periodicity of the Sine and Cosine functions.

 

[HallsofIvy]

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]

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

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 )

 

[ichiro_w]

Follows an attempt to display a plot of Sin(\pi t)[/tex] of period 1 using the initial constant term and 6 iterations of the Cos term. Click on the bitmap file.

 

[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 modeled by Fourier series be:

f(t) =\left\{\begin{array}{cr}0&amp;\mbox{if\;\;}-1&lt;t&lt;0\\sin(\pi t)&amp;\mbox{if\quad} 0&lt;t&lt;1\end{array}\right

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

\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)

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

Click on the bitmap to see the plot so generated.