Fourier Series of sin (pi*t)

by anish
Tags: fourier, series
 P: 4 I am having trouble finding the An and Bn coefficients for the fourier series f(t) = sin (pi*t) from 0
 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 $$a_0 = \frac{1}{\frac{1}{2}}\int_0^1 Sin(\pi t) \;dt = \frac{4}{\pi}$$ $$A_N = \frac{1}{\frac{1}{2}}\int_0^1 Sin(\pi t) Cos(2 n \pi t)\; dt$$ $$B_N = \frac{1}{\frac{1}{2}}\int_0^1 Sin(\pi t) Sin (2 n \pi t)\; dt$$ A source advises the Trigonometric Identities: $$2 Sin[A] Cos[B] = Sin [A+B] + Sin[A-B]$$ $$2 Sin[A] Sin[B] = Cos[A-B] - Cos[A+B]$$ 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.
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 Quote by anish I am having trouble finding the An and Bn coefficients for the fourier series f(t) = sin (pi*t) from 0
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)).

P: 13

Fourier Series of sin (pi*t)

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 modelled 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
$$f(t)=\frac{2}{\pi}-\frac{4}{\pi}\sum_1^{\infty}\frac{Cos(2n\pi t)}{4n^2-1}=Sin(\pi t)$$

$$f(t)\sim\frac{2}{\pi}\;-\;\frac{4}{\pi}\left(\frac{Cos(2\pi t)}{3}\;+\;\frac{Cos(4\pi t)}{15}\;+\;\frac{Cos(6\pi t)}{35}\;+\;\cdots\right )$$
P: 13
Follows an attempt to display a plot of $Sin(\pi t)[/tex] of period 1 using the initial constant term and 6 iterations of the Cos term. Click on the bitmap file. Attached Images  foo.bmp (49.8 KB, 25 views)  P: 618 You guys are missing the forest for the trees! The fourier series is a sum of sines and cosines, and it's unique. So if you can find one set of coefficients that works, you've got THE fourier series. So if sin(pi*t) = a0 + An*SUM[sin(n*pi*t)] + Bn*SUM[cos(n*pi*t)], can't you just look at that and see a set of coefficients that will make the left and right hand sides of that equation the same? P: 13 OK, here is the same problem except make the period 2 and let the function to be modelled by Fourier series be: [latex] f(t) =\left\{\begin{array}{cr}0&\mbox{if\;\;}-1<t<0\\sin(\pi t)&\mbox{if\quad} 0<t<1\end{array}\right$

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.
Attached Images
 foo2.bmp (49.8 KB, 26 views)

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