# Linear Algebra

posted by on .

Hello,

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

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 - ,

The 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 - ,

This 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 - ,

http://en.wikipedia.org/wiki/Rectangular_function

• Linear Algebra - ,

I 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

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