Complex Fourier Series for cos(t/2)

Aows

Homework Statement


Q:/ Find the complex form of Fourier series for the following periodic function whose definition in one period is given below then convert to real trigonometry also find f(0).
f(t)=cos(t/2), notes: (T=2*pi) (L=pi)


Homework Equations


1) f(t)=sum from -inf to +inf (Cn exp(j*n*(pi/L)*t)
2) Cn=(1/2pi) *integration from -L to +L (f(t) exp (-j * n (pi/L)* t) *dt


The Attempt at a Solution


i failed at finding the solution to the Cn coefficient
 
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Aows said:

Homework Statement


Q:/ Find the complex form of Fourier series for the following periodic function whose definition in one period is given below then convert to real trigonometry also find f(0).
f(t)=cos(t/2), notes: (T=2*pi) (L=pi)

Homework Equations


1) f(t)=sum from -inf to +inf (Cn exp(j*n*(pi/L)*t)
2) Cn=(1/2pi) *integration from -L to +L (f(t) exp (-j * n (pi/L)* t) *dt

The Attempt at a Solution


i failed at finding the solution to the Cn coefficient

Well, post what the equations look like when you substitute cos(t/2) in for f(t). Also, for evaluating the integral, it might help to convert it to exponentials, using:

cos(x) = \frac{1}{2} (e^{i x} + e^{-ix})
 
stevendaryl said:
Well, post what the equations look like when you substitute cos(t/2) in for f(t). Also, for evaluating the integral, it might help to convert it to exponentials, using:

cos(x) = \frac{1}{2} (e^{i x} + e^{-ix})
Hello,
what do you want me to post ?
 
here is part of the question
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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