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March 27, 2015

March 27, 2015

Posted by **Nicole** on Sunday, December 7, 2008 at 1:01pm.

I'm trying to find the Fourier Series of a function which is 1 from -pi/2 to pi/2, and zero everywhere else

inside of -pi to pi. I realize this is a square wave and I began the problem with a piecewise function

s(t) = 1 abs(t) < pi/2

0 abs(t) > pi/2

Then I had

a_k = 1/pi integral(from -pi to pi) of [cos(kt)dt]

and

b_k = 1/pi integral(from -pi to pi) of [sin(kt)dt]

but I don't know what to do from here.

Any help is greatly appreciated, thank you!

- Reiny Can't Help Here -
**Reiny**, Sunday, December 7, 2008 at 2:21pmThe last time I dealt with that topic was over 45 years ago, and frankly I don't feel confident enough to answer.

Perhaps some of the other math experts might want to take a crack at it if they are on.

- Linear Algebra -
**Damon**, Sunday, December 7, 2008 at 2:36pmThis is not a Fourier series really but a Fourier integral. The series would apply if the function were periodic.

However this function has to be zero from - oo to - pi/2

Then it jumps up one at -pi/2

Then it jumps down one at +pi/2

That is a unit step up at -pi/2

minus

A unit step up at + pi/2

- Linear Algebra -
**Damon**, Sunday, December 7, 2008 at 2:50pm

- Linear Algebra -
- Linear Algebra -
**Damon**, Sunday, December 7, 2008 at 6:19pmI think you were doing fine now that I look at what you did.

a_k = 1/pi integral(from -pi to pi) of [cos(kt)dt]

= 1/(k pi) [sin k pi - sin(-k pi)

= (2/k pi)sin k pi

which is the answer

and

b_k = 1/pi integral(from -pi to pi) of [sin(kt)dt]

= 1/(k pi) [-cos k pi + cos -k pi}

= 0 because cosine is even

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