# Homework Help: Fourier Series Problem

1. Mar 2, 2012

### TW Cantor

1. The problem statement, all variables and given/known data
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

2. Relevant 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

3. 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 ive gone wrong?

2. Mar 2, 2012

### jbunniii

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:

$f(t) = -5.5 + 22t^2$ for $-0.5 \leq t < 0$

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 $0 < t < 0.5$? (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. Mar 2, 2012

### TW Cantor

well im 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. Mar 2, 2012

### jbunniii

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. Mar 2, 2012

### TW Cantor

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 wont that just equal zero?

6. Mar 2, 2012

### TW Cantor

how would you integrate for this problem?

7. Mar 2, 2012

### LCKurtz

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. Mar 2, 2012

### jbunniii

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

9. Mar 2, 2012

### TW Cantor

ahh! ive been using sin instead of cos! i cant believe i did that.

so ive 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 ive 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. Mar 2, 2012

### jbunniii

I didn't check whether you calculated the integral correctly, but your method looks correct now.

11. Mar 2, 2012

### TW Cantor

yeah im getting the right answer now :-) thanks for your help!