• Support PF! Buy your school textbooks, materials and every day products Here!

Fourier Transform

  • Thread starter roldy
  • Start date
  • #1
232
1

Homework Statement


I need to determine the first three terms in the Fourier series pictured in the attachment.
Did I define the peace-wise functions correctly?

I'm re-posting this with the tex code instead of the attached document.

Homework Equations


[tex]
a_o=\frac{1}{2L}\int_{-L}^Lf(t)dt
[/tex]

[tex]
a_n=\frac{1}{L}\int_{-L}^Lf(t)cos\left(\frac{n\pi{t}}{L}\right)dt
[/tex]


The Attempt at a Solution


The plot of the series is symmetric, so therefore I am only going to find what [tex]a_o[/tex] and [tex]a_n[/tex].

[tex]
a_o=\frac{2}{3T} \left[\int_\frac{-3T}{4}^\frac{-T}{4}(-1)dt+\int_\frac{-T}{4}^\frac{T}{4}(1)dt+\int_\frac{T}{4}^\frac{3T}{4}(-1)dt\right]
[/tex]
[tex]
=\frac{2}{3T}\left[-\left(-T/4-(-3T/4)\right)+\left(T/4-(-T/4)\right)+\left(3T/4-T/4\right)\right]

=\frac{2}{3T}\left(T/4-3T/4+T/4+T/4+3T/4-T/4\right)=1/3
[/tex]

[tex]
a_n=\frac{4}{3T}\left[\int_\frac{-3T}{4}^\frac{-T}{4}(-1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt+\int_\frac{-T}{4}^\frac{T}{4}(1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt+\int_\frac{T}{4}^\frac{3T}{4}(-1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt\right]
[/tex]

[tex]
=\frac{4}{3T}\left[\left[-sin\left(\frac{n\pi{\left(-T/4\right)}}{3T/4}\right)+sin\left(\frac{n\pi{\left(-3T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}+\left[sin\left(\frac{n\pi{\left(T/4\right)}}{3T/4}\right)-sin\left(\frac{n\pi{\left(-T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}
[/tex]
[tex]
-\left[sin\left(\frac{n\pi{\left(3T/4\right)}}{3T/4}\right)-sin\left(\frac{n\pi{\left(T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}
[/tex]


[tex]
\frac{1}{n\pi}\left[\left[-sin\left(\frac{-n\pi}{3}\right)+sin\left(-n\pi\right)\right]+\left[sin\left(\frac{n\pi}{3}\right)-sin\left(\frac{-n\pi}{3}\right)\right]-\left[sin\left(n\pi\right)-sin\left(\frac{n\pi}{3}\right)\right]\right]
[/tex]

Distributing the signs through and simplifying:

[tex]
\frac{1}{n\pi}\left[4sin\left(\frac{n\pi}{3}\right)-2sin\left(n\pi\right)\right]
[/tex]

So for n=1,2,3 I get

[tex]
a_1=\frac{2\sqrt{3}}{2\pi}
[/tex]

[tex]
a_2=\frac{\sqrt{3}}{2\pi}
[/tex]

[tex]
a_3=0
[/tex]
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
What is f?
 
  • #3
232
1
What is f?
It's not f. It's f(t), meaning f of t. If you look at the attached picture I came up with f(t) on three intervals. I'm not sure if they are correct. They look correct to me.
 
  • #4
lanedance
Homework Helper
3,304
2
yeah i get its f(t) but can't see any attached pic? pretty important part of the problem
 
  • #5
232
1
Sorry about that. For some reason it didn't attach. Here's another try at it.
 

Attachments

  • #6
lanedance
Homework Helper
3,304
2
ok, so why integrate over more that a period?

3T/4 - (-3T/4) = 3T/2
 
  • #7
232
1
I think that's where I might of made the mistake. I think I should of done it from -T/4 to 3T/4.
 
  • #8
lanedance
Homework Helper
3,304
2
That sounds better... Note that the integral over sin(tnpi/T) would be non zero over that interval as well, you could try your new interval on a sin as a check
 

Related Threads on Fourier Transform

Replies
7
Views
1K
  • Last Post
Replies
8
Views
950
  • Last Post
Replies
2
Views
538
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
843
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
377
Top