[PDF] [PDF] Fourier Sine Series Examples - MIT

[PDF] Fourier Series - Introduction

Fourier series are used in the analysis of periodic functions A periodic function From the graph (or using our calculator), we can observe that: sin(nπ) = 0

[PDF] Fourier Series - University of Miami

The idea of Fourier series is that you can write a function as an infinite series of sines and cosines Do all this by hand, no graphing calculators, though if you

[PDF] Fourier Analysis

A plot of the coefficients, {cn} versus n, is known as the spectrum of the function: it tells us how much of each frequency is present in the function The process of 

[PDF] Fourier Series

These functions are periodic with period 2 n, so their graphs trace through n periods over the interval ج ح Figure 1 shows the graphs of cos x and cos 5x, and  

[PDF] Using the Fourier Series Library (fourierlibtns)

18 oct 2018 · Select the help function in the library pane to add it to a calculator page, then execute The graph of the function and its Fourier series:

[PDF] Fourier Series - Stewart Calculus

a Fourier series EXAMPLE 1 Find the Fourier coefficients and Fourier series of the square-wave function defined by and So is periodic with period and its graph  

[PDF] Fourier Series - Stewart Calculus

and So is periodic with period and its graph is shown in Figure 1 SOLUTION Using the formulas for the Fourier coefficients in Definition 7, we have a0 1 2

[PDF] Fourier series in MATLAB

Find the Fourier coefficients using your MATLAB function: plot the Fourier coefficients vs frequency 5 The FFT Despite the fact that we presented the discrete 

[PDF] Fourier Sine Series Examples - MIT

16 nov 2007 · This turned out to be false for various badly behaved f(x), and controversy over the exact conditions for convergence of the Fourier series lasted 

[PDF] Fourier Transform - Stanford Engineering Everywhere

1 12 Appendix: Best L2 Approximation by Finite Fourier Series 3 It's clear from the geometric picture of a repeating graph that this is true To show it 

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Fourier Sine Series Examples

16th November 2007

The Fourier sine series for a functionf(x)defined onx?[0,1]writesf(x)as f(x) =∞? n=1b nsin(nπx) for some coefficientsbn. Because of orthogonality, we can compute thebnvery simply: foranygivenm, weintegratebothsidesagainstsin(mπx). Inthesummation, thisgives zero forn?=m, and?1

0sin2(mπx) = 1/2forn=m, resulting in the equation

b m= 2? 1 0 f(x) sin(mπx)dx. Fourier claimed (without proof) in 1822 thatanyfunctionf(x)can be expanded in terms of sines in this way, even discontinuous function. This turned out to be false for various badly behavedf(x), and controversy over the exact conditions for convergence of the Fourier series lasted for well over a century, until the question was finally settled by Carleson (1966) and Hunt (1968): any functionf(x)where?|f(x)|pdxis finite for somep >1has a Fourier series that convergesalmost everywheretof(x)[except at isolated points]. At points wheref(x)has a jump discontinuity, the Fourier series converges to the midpoint of the jump. So, as long as one does not care about crazy di- vergent functions or the function value exactly at points of discontinuity (which usually has no physical significance), Fourier"s remarkable claim is essentially true. To illustrate the convergence of the sine series, let"s consider a couple of examples. First, consider the functionf(x) = 1, which seems impossible to expand in sines because it is not zero at the endpoints, but nevertheless it works...if you don"t care about the valueexactlyatx= 0orx= 1. From the formula above, we obtain b m= 2? 1 0 sin(nπx)dx=-2nπ cos(nπx)????1 0

4nπ

nodd

0neven,

and thus f(x) = 1 =4π sin(πx) +43πsin(3πx) +45πsin(5πx) +···. This is plotted for 1, 2, 4, 8, 16, and 32 terms in figure. 1, showing that it does in- deed approachf(x) = 1almost everywhere. There is some oscillation at the point of discontinuity, which is known as aGibb"s phenomenon. 1

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16 termsFigure 1: Fourier sine series forf(x) = 1, truncated to a finite number of terms (from

1 to 32), showing that the series indeed converges everywhere tof(x), except exactly

at the endpoints, as the number of terms is increased. 2 Note that theneven coefficents were zero. The reason for this is simple: for even n, thesin(nπx)function isoddaround the midpointx= 1/2, whereasf(x) = 1is even around the midpoint, so the integral of their product is zero. Now, let"s try another example, one for which the endpoints are zero and there are no discontinuities, but there is a discontinuous slope:f(x) =12 - |x-12 |, which looks like a triangle when plotted. Again, this function is even around the mid-point x= 1/2, so only the odd-ncoefficients will be non-zero. For these coefficients (since the integrand is symmetric aroundx= 1/2), we only need to do the integral over half the region: b modd= 2? 1 0 f(x) sin(mπx)dx= 4? 1/2 0 xsin(mπx)dx=4(mπ)2(-1)m-12 where for the last step one must do some tedious integration by parts, and thus f(x) =4π

2sin(πx)-4(3π)2sin(3πx) +4(5π)2sin(5πx) +···.

This is plotted in figure. 2 for 1 to 8 terms-it converges faster than forf(x) = 1 because there are no discontinuities in the function to match, only discontinuities in the derivative. 3

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8 termsFigure 2: Fourier sine series (blue lines) for the triangle functionf(x) =12

- |x-12 (dashed black lines), truncated to a finite number of terms (from 1 to 32), showing that the series indeed converges everywhere tof(x). 4quotesdbs_dbs20.pdfusesText_26