# Thread: Problem with Fourier Series

1. ## Problem with Fourier Series

I have a problem with the below fourier series

-2, -2 < t < -1
-1, -1 < t < 0
f(t) = 1, 0 < t < 1
2, 1 < t < 2

f(t + 4) = f(t)

I want to intergrate by parts to go on to solve bn but dont how what U will be.

2. Originally Posted by hunterage2000
I have a problem with the below fourier series

-2, -2 < t < -1
-1, -1 < t < 0
f(t) = 1, 0 < t < 1
2, 1 < t < 2

f(t + 4) = f(t)

I want to intergrate by parts to go on to solve bn but dont how what U will be.
You want to find the Fourier series of

$f(t)=\begin{cases} -2, -2 < t < -1 \\ -1, -1 < t < 0 \\ 0, 0 < t < 1 \\ 1, 1 < t < 2 \\
\end{cases}$

You should not need integration by part. The function is constant on each interval.

Here is a PDF with the series.

3. its $f(t)=\begin{cases} -2, -2 < t < -1 \\ -1, -1 < t < 0 \\ 1, 0 < t < 1 \\ 2, 1 < t < 2 \\
\end{cases}$

4. That make the problem easier. $f(x)$ is an odd function so all of the $a_n$ must be zero.

$\displaystyle b_n=\frac{1}{2}\left[ \int_{-2}^{-1}-2\sin\left(\frac{n\pi x}{2} \right)dx + \int_{-1}^{0}-\sin\left(\frac{n\pi x}{2} \right)dx+\int_{0}^{1}\sin\left(\frac{n\pi x}{2} \right)dx+\int_{1}^{2}2\sin\left(\frac{n\pi x}{2} \right)dx \right]$

$\displaystyle b_n=-\frac{2}{\pi n}\left[ 2\cos(n\pi)-\cos\left( \frac{n \pi}{2}\right)-1\right]$

5. so what is the value of U? is it 2 + 1 or 3?

6. Originally Posted by hunterage2000
so what is the value of U? is it 2 + 1 or 3?
I don't understand what you mean. What is U?? and 2+1=3 so I am really confused.

7. U as in the intergration by parts U.

U = ?
dv/dt = cos (n*pi*t/2)
du/dt = ?
v = -2/n*pi sin(n*pi*t/2)

8. As I said in my first post you do not need integration by parts.

$\displaystyle \int \sin(ax)dx=-\frac{1}{a}\cos(ax)$

$\displaystyle \int \sin\left(\frac{\pi n}{L}x \right)dx=-\frac{L}{\pi n}\cos\left(\frac{\pi n}{L}x \right)$