Complex Fourier Series for cos(t/2)

In summary, the complex form of the Fourier series for the periodic function f(t)=cos(t/2) is given by the equation f(t)=sum from -inf to +inf (Cn exp(j*n*(pi/L)*t). To find the coefficient Cn, we use the equation Cn=(1/2pi) *integration from -L to +L (f(t) exp (-j * n (pi/L)* t) *dt, where L=pi in this case. By substituting cos(t/2) in for f(t), we can simplify the equation to Cn = 1/2pi * integration from -pi to pi [(cos(t/2)) * (e^(-j*n
  • #1
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
 
Physics news on Phys.org
  • #2
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 [itex]cos(t/2)[/itex] in for [itex]f(t)[/itex]. Also, for evaluating the integral, it might help to convert it to exponentials, using:

[itex]cos(x) = \frac{1}{2} (e^{i x} + e^{-ix})[/itex]
 
  • #3
stevendaryl said:
Well, post what the equations look like when you substitute [itex]cos(t/2)[/itex] in for [itex]f(t)[/itex]. Also, for evaluating the integral, it might help to convert it to exponentials, using:

[itex]cos(x) = \frac{1}{2} (e^{i x} + e^{-ix})[/itex]
Hello,
what do you want me to post ?
 
  • #4
here is part of the question
7SYajnJ.jpg
 

1. What is a Complex Fourier Series?

A Complex Fourier Series is a mathematical representation of a periodic function as a sum of complex exponential functions. It is used to decompose a signal into its individual frequency components.

2. How is a Complex Fourier Series different from a regular Fourier Series?

A regular Fourier Series uses only real exponential functions, while a Complex Fourier Series uses both real and imaginary exponential functions. This allows for a more accurate representation of periodic functions with complex values.

3. What is the purpose of using a Complex Fourier Series?

The purpose of using a Complex Fourier Series is to analyze and manipulate signals in the frequency domain. It allows for the identification of specific frequency components within a signal, which can then be modified or filtered as needed.

4. How is a Complex Fourier Series calculated?

A Complex Fourier Series is calculated using the Fourier coefficients, which are obtained by integrating the function over one full period. These coefficients are then used to construct the series, which is an infinite sum of complex exponential functions.

5. What are some applications of Complex Fourier Series?

Complex Fourier Series have many applications in engineering, physics, and mathematics. They are used in signal processing, image and audio compression, solving differential equations, and analyzing periodic phenomena such as sound waves and electromagnetic waves.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
298
  • Calculus and Beyond Homework Help
Replies
1
Views
228
  • Calculus and Beyond Homework Help
Replies
1
Views
542
  • Calculus and Beyond Homework Help
Replies
1
Views
355
  • Calculus and Beyond Homework Help
Replies
2
Views
379
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
422
  • Calculus and Beyond Homework Help
Replies
3
Views
379
  • Calculus and Beyond Homework Help
Replies
16
Views
571
  • Calculus and Beyond Homework Help
Replies
6
Views
917
Back
Top