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Complex fourier series

  • Thread starter reece
  • Start date
9
0
1. Homework Statement
Solve [tex]\frac{1}{1}[/tex] [tex]\int^{0}_{-1}[/tex] -t e[tex]^{-j2\pi*nt}[/tex]dt


2. Homework Equations
So I use integration by parts
u = -t and dv = e[tex]^{-j2\pi*nt}[/tex] , du= -1 and v = [tex]\frac{1}{-j2\pi*n}[/tex]e[tex]^{-j2\pi*nt}[/tex]


3. The Attempt at a Solution
after integration by parts I get:
[tex]\frac{e^-j2\pi*nt}{j2pi*n}[/tex] + [tex]\frac{1}{(j2pi*n)^{2}}[/tex] [1 - e[tex]^{-j2\pi*nt}[/tex]]

So basically im suppose to end up with [tex]\frac{1}{j2pi*n}[/tex] but it doesnt work out like it should.
Any help would be good thanks.
 
Last edited:

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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893
First, you certainly do not get anything involving "t" when you have evaluated at t= 0 and -1! Did you mean
[tex]\frac{e^{2\pi i n}}{2\pi i n}+ \frac{1- e^{2\pi i n}}{(2\pi i n)^2}[/tex]

What is [itex]e^{-2\pi i nt}[/itex]?
(Sorry, but I just can't force my self to use "j" instead of "i"!)
 
9
0
woops my bad, those t's arent spose to be in there. But the rest is right.

And yes it is what you typed out.

Im leading to think I shouldnt get the same answer as what I got using the Trig-Complex Relationship equation. Which is why im getting something different.

The exponential part is what I integrate with to evaluate the complex fourier series. See the equation. -t is the function.
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
Once again, what is
[tex]e^{-2\pi i n}[/tex]?
 
9
0
In trig FS, you evaluate each component a0, an, bn. Well the exponential replaces that I think.

im guessing you use Euler's Rules (i think) to evaluate it which says

sin theta = 1 / 2i [ e ^ i*theta - e ^ -i*theta]
cos theta = 1 /2 [ e ^ i*theta + e ^ -i*theta]

Hopefully that answered that.
 

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