# Fourier series and draw function

1. Nov 19, 2007

### karnten07

1. The problem statement, all variables and given/known data

For 0</= x </= pi, let f(x) be defined by

f(x) = { A for [x - pi/2] < alpha
{ 0 for alpha </= [x-pi/2]</= pi/2

Where A and alpha are positive constants with alpha <pi/2. I have used </ to represent less than or equal to. The square brackets are to represent single straight vertical lines.

Let f(x) be extended into an odd, periodic function of period 2pi. Sketch the graph of f(x) in the interval -3pi</= x </= 3pi, and show that the fourier series expansion of f(x) takes the form

f(x) = 4A/pi {Infinity}Sigma(m=1) ((-1)^m+1)/(2m-1) sin(2m-1)alpha sin(2m-1)x

The part in bold should be sigma with infinity above and m=1 below
2. Relevant equations

3. The attempt at a solution
I see how to draw the function, i get y = A from 0 to pi at which point it drops to zero. Because it is odd and period 2pi it will be a square wave with three horizontal maxima (effectively) from x = 2npi - (2n+1)pi.

Is this correct?

Last edited: Nov 19, 2007
2. Nov 19, 2007

### karnten07

In doing the fourier series expansion im finding it difficult to write it in terms of the sum of integrals because of the limits to use.

What im getting is sumthing like this:

2pi ao = INT(from (2n+1)pi to 2npi) + INT(from (2n+1)pi to 2npi) 0dx =

but im thinking this is wrong. Should i be integrating between -pi and pi?

In that case i would get

2pi ao = INT (-pi to 0) 0dx + INT(0 to pi) Adx = 0 + Api

Im not sure im doing this right because where does alpha come into it?

To see the question written more neatly here is the link http://www.personal.soton.ac.uk/mj2/Maths/MATH2015Problems_files/problems5.pdf [Broken]

Last edited by a moderator: May 3, 2017
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