Calculating Fourier Coefficients for an Odd, Periodic Function

In summary, the given function f(t) is an odd, periodic function with period 1. Using the half range formula for Fourier coefficients, the solution for the first part of the question is (22*cos(n*pi) - 11*n^2*pi^2 + 22*n*pi*sin(n*pi) - 22)/(2*(n^3 * pi^3)). To find the Fourier coefficient for b5, this solution can be substituted with n = 5.
  • #1
TW Cantor
54
1

Homework Statement


f(t) is an odd, periodic function with period 1 and:

f(t) = -5.5 + 22*t2 for -0.5 ≤ t < 0

i) find the Fourier coefficient bn
ii) find the Fourier coefficient b5


Homework Equations


bn = (2/T) * ∫ f(t) *sin((2*n*∏*t)/T) dt between T/2 and -T/2

sin(n*∏) = 0 for all values of n
cos(n*∏) = 1 for even values of n
cos(n*∏) = -1 for odd values of n


The Attempt at a Solution



by integrating f(t)*sin((2*n*∏*t)/T) with respect to t i get:
-(11*sin(n*∏) - 11*n*∏*cos(n*∏))/(n^3 * ∏^3)

replacing the sin(n*∏) with 0 i get:
11*cos(n*∏))/(n^2 * ∏^2)

so this should be the solution to the first part of the question but when i put n into the equation as 5, i get the wrong answer. can anyone see where I've gone wrong?
 
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  • #2
Can you show your intermediate steps when performing the integral?

In particular, I want to see what the bounds of the integral were and what expression you used for f(t).

Note that your function is only partly defined by this equation:

[itex]f(t) = -5.5 + 22t^2[/itex] for [itex]-0.5 \leq t < 0[/itex]

This defines it on an interval of length 0.5. But the period is 1. So what is an equation for the function for the interval [itex]0 < t < 0.5[/itex]? (You need to use the fact that the function is odd.)

P.S. It won't make any difference when integrating, but what does the oddness of the function imply that f(0) must be?
 
  • #3
well I am not really sure what to use as the bounds for the integral... i got confused when it said the period was 1 but the function is only true between -0.5 < t < 0. i know since its odd its asymettric about the vertical axis and since its periodic it might have something to do with that? i could use the equation f(-t) = -f(t) maybe?
 
  • #4
TW Cantor said:
well I am not really sure what to use as the bounds for the integral... i got confused when it said the period was 1 but the function is only true between -0.5 < t < 0. i know since its odd its asymettric about the vertical axis and since its periodic it might have something to do with that? i could use the equation f(-t) = -f(t) maybe?

Yes, f(-t) = -f(t) is the defining characteristic of an odd function. Try using that to define what f(t) must be for 0 < t < 0.5.
 
  • #5
so for 0 < t < 0.5, f(t) = 5.5 - 22*t^2 ?

then do i add the integral of:
f(t)*cos((2*pi*n*t)/T) dt between -0.5 < t < 0
and -f(t)*cos((2*pi*n*t)/T) dt between 0 < t < 0.5

because won't that just equal zero?
 
  • #6
how would you integrate for this problem?
 
  • #7
TW Cantor said:
so for 0 < t < 0.5, f(t) = 5.5 - 22*t^2 ?

then do i add the integral of:
f(t)*cos((2*pi*n*t)/T) dt between -0.5 < t < 0
and -f(t)*cos((2*pi*n*t)/T) dt between 0 < t < 0.5

because won't that just equal zero?

TW Cantor said:
how would you integrate for this problem?

You use the half range formulas for the coefficients. Since f(t) is odd you know an = 0 for all n. The half range formula for ##b_n## for a function of period ##2p##, which should be in your text, is$$
b_n =\frac 2 p\int_0^p f(t)\sin(\frac{n\pi t}{p})\,dt$$
 
  • #8
TW Cantor said:
so for 0 < t < 0.5, f(t) = 5.5 - 22*t^2 ?

then do i add the integral of:
f(t)*cos((2*pi*n*t)/T) dt between -0.5 < t < 0
and -f(t)*cos((2*pi*n*t)/T) dt between 0 < t < 0.5

because won't that just equal zero?

Yes, that's right. The cosine terms are all even, and their coefficients will all be zero for an odd function.

What about the sine terms?
 
  • #9
ahh! I've been using sin instead of cos! i can't believe i did that.

so I've integrated
2/T ∫ f(t)*sin(n*π*t/T) dt between 0.5 and 0

and i get:
(22*cos(pi*n) - 11*n^2 *pi^2 + 22*n*pi*sin(n*pi) -22)/(2*(n^3 * pi^3))

since I've integrated for only half the period and between 0 < t < 0.5 i have to multiply by -2 to get it into the full period and between -0.5 < t < 0 right? and then i can replace sin(n*pi) and cos(n*pi)?
 
  • #10
TW Cantor said:
ahh! I've been using sin instead of cos! i can't believe i did that.

so I've integrated
2/T ∫ f(t)*sin(n*π*t/T) dt between 0.5 and 0

and i get:
(22*cos(pi*n) - 11*n^2 *pi^2 + 22*n*pi*sin(n*pi) -22)/(2*(n^3 * pi^3))

since I've integrated for only half the period and between 0 < t < 0.5 i have to multiply by -2 to get it into the full period and between -0.5 < t < 0 right? and then i can replace sin(n*pi) and cos(n*pi)?

I didn't check whether you calculated the integral correctly, but your method looks correct now.
 
  • #11
yeah I am getting the right answer now :-) thanks for your help!
 

1. What is a Fourier Series Problem?

A Fourier Series Problem is a mathematical problem that involves representing a periodic, or repeating, function as a sum of sine and cosine functions. This allows for the analysis and manipulation of complex functions in a simpler manner.

2. Why are Fourier Series important in science?

Fourier Series are important in science because they provide a way to approximate and analyze complex, periodic phenomena such as sound waves, electromagnetic waves, and other physical processes. They also have applications in signal processing, image analysis, and data compression.

3. How is a Fourier Series calculated?

A Fourier Series can be calculated using a mathematical formula known as the Fourier series formula, which involves integrating the function over one period and then using trigonometric identities to simplify the resulting expression. Alternatively, there are also computer algorithms that can calculate Fourier Series numerically.

4. What is the difference between a Fourier Series and a Fourier Transform?

A Fourier Series represents a periodic function as a sum of sine and cosine functions, while a Fourier Transform represents a non-periodic function as a sum of sine and cosine functions with different frequencies. In other words, a Fourier Series is used for periodic functions, while a Fourier Transform is used for non-periodic functions.

5. What are some real-world applications of Fourier Series?

Fourier Series have numerous real-world applications, including in signal processing for filtering and noise reduction, in image and audio compression, and in solving differential equations in physics and engineering. They are also used in the analysis of musical tones, ocean waves, and many other physical phenomena.

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