- #1
wahoo2000
- 37
- 0
Find the Fourier series of the 2Pi-periodic function
f(x)=
{0 , Abs(x) <= Pi/2
{Abs(x)-Pi/2, Pi/2 < Abs(x) <= Pi
My attempt at a solution
I have sketched the function... It equals zero between -Pi/2 and Pi/2 and it equals Pi/2 at -Pi and Pi. Then the pattern is repeated with peaks at +/- Pi, 3Pi, 5Pi...
To find the Fourier series I have to identify a_0, a_n and b_n.
Here, b_n = 0 because the function is even, right?
a_0= (1/(2Pi))*AUC = Pi/8 (AUC=area under curve from -Pi to Pi (i.e. one period))
then a_n = (1/Pi)*integral_-Pi_Pi f(x)*cos(nx) dx= (2/Pi)*integral_0_Pi f(x)*cos(nx) dx (because even function)
= (2/Pi)integral_0_Pi (Abs(x)-Pi/2))*cos(nx) dx = (integration by parts) = (2/(nPi))[(Abs(x)-Pi/2)*sin(nx)] (evaluate for x=Pi minus for x=Pi/2, because the function is zero anyway between 0 and Pi/2 --> integral is zero) - (2/(nPi))*integral_Pi/2_Pi 1*sin(nx) dx
[(Abs(x)-Pi/2)*sin(nx)] (evaluate for x=Pi minus for x=Pi/2) I think is zero so just the last expression left, but I just can't figure it out.
Sorry for the format of this post, I can hardly figure out what I wrote myself so it's a miracle if somebody can help me out of this :) I now saw the "insert image" feature, which will be my choice for the maths next time I post :)
f(x)=
{0 , Abs(x) <= Pi/2
{Abs(x)-Pi/2, Pi/2 < Abs(x) <= Pi
My attempt at a solution
I have sketched the function... It equals zero between -Pi/2 and Pi/2 and it equals Pi/2 at -Pi and Pi. Then the pattern is repeated with peaks at +/- Pi, 3Pi, 5Pi...
To find the Fourier series I have to identify a_0, a_n and b_n.
Here, b_n = 0 because the function is even, right?
a_0= (1/(2Pi))*AUC = Pi/8 (AUC=area under curve from -Pi to Pi (i.e. one period))
then a_n = (1/Pi)*integral_-Pi_Pi f(x)*cos(nx) dx= (2/Pi)*integral_0_Pi f(x)*cos(nx) dx (because even function)
= (2/Pi)integral_0_Pi (Abs(x)-Pi/2))*cos(nx) dx = (integration by parts) = (2/(nPi))[(Abs(x)-Pi/2)*sin(nx)] (evaluate for x=Pi minus for x=Pi/2, because the function is zero anyway between 0 and Pi/2 --> integral is zero) - (2/(nPi))*integral_Pi/2_Pi 1*sin(nx) dx
[(Abs(x)-Pi/2)*sin(nx)] (evaluate for x=Pi minus for x=Pi/2) I think is zero so just the last expression left, but I just can't figure it out.
Sorry for the format of this post, I can hardly figure out what I wrote myself so it's a miracle if somebody can help me out of this :) I now saw the "insert image" feature, which will be my choice for the maths next time I post :)