Linear Algebra
posted by Nicole .
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
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
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 
Damon
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  Damon

Linear Algebra 
Damon
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
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
Respond to this Question
Similar Questions

Linear Algebra
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) … 
Math (College Level Mathematics)
Fourier sin series for f(x) = 1, 0 < x < Pie is given by 1 = 4/n E 1/ (2n1) times sin (2n1) x, (0 < x < n). Using this, find the Fourier sinc series for f(x)= 1, on 0 < x < c where c > 0. Then find the Fourier … 
Fourier Series
A periodic function f(t), with period 2π is defined as,f(t) = c for 0 < t < πf(t) = c for π < t < 0where c = 1.4, Taking π = 3.142, calculate the Fourier sine series approximation up to the 5th harmonics … 
math
Anyone can help me on this qns? The Fourier series expansion for the periodic function,f(t) = sin tis defined in its fundamental interval. Taking π = 3.142, calculate the Fourier cosine series approximation of f(t), up to the 
math
The Fourier series expansion for the periodic function,f(t) = sin tis defined in its fundamental interval. Taking π = 3.142, calculate the Fourier cosine series approximation of f(t), up to the 6th harmonics when t = 1.09. Give … 
math
The Fourier series expansion for the periodic function,f(t) = sin tis defined in its fundamental interval. Taking π = 3.142, calculate the Fourier cosine series approximation of f(t), up to the 6th harmonics when t = 1.09. Give … 
Math, Fourier Series
For Fourier Series of f(x)=sinx which is an even function, bn should be 0. However, I solved that b1=1 while the rest of the terms =0, meaning bn=0. Is there a mistake? 
Math Fourier series
Evaluate the formula for cn in Fourier :integral of e^kx dx = e^kx /k :unless k=0: Type your formula for c0 and cn (n>0) into the indicated spaces. Then rewrite the Fourier series in terms of sines and cosines. Simplify as far as … 
Math Fourier series
Evaluate the formula for cn in Fourier :integral of e^kx dx = e^kx /k :unless k=0: Type your formula for c0 and cn (n>0) into the indicated spaces. Then rewrite the Fourier series in terms of sines and cosines. Simplify as far as … 
Math Fourier series
Evaluate the formula for cn in Fourier :integral of e^kx dx = e^kx /k :unless k=0: Type your formula for c0 and cn (n>0) into the indicated spaces. Then rewrite the Fourier series in terms of sines and cosines. Simplify as far as …