Understanding Fourier Series: Solving a Problem with Sinusoidal Functions

[ognik]

Hi, appreciate some help with this FS problem - $f(t)= 0$ on $[-\pi, 0]$ and $f(t)=sin\omega t$ on $[0,\pi]$

I get $a_0=\frac{2}{\pi}$ and $b_1 = \frac{1}{2}$, which agree with the book; all other $b_n = 0$ because Sin(mx)Sin(nx) orthogonal for $m \ne n$

But $a_n =\frac{1}{\pi}\int_{0}^{\pi}Sin(\omega t) Cos (n \omega t) \,d\omega t $ - so this should also be 0 because the terms are orthogonal, but the book's answer has Cos terms in even powers of n?

 

[Dethrone]

Hi ognik,

I could be wrong, but I don't think the orthogonality property applies here because intuitively speaking, you have an odd function that is evaluated at half the period, not the full period.

i.e Consider $\int_0^{\pi}\sin\left({x}\right)\cos\left({nx}\right) \,dx$. For $n=1$, it indeed evaluates to 0, but not for any other $n$.

I suggest that you integrate it directly, which is of the form $\int \sin\left({Ax}\right)\cos\left({Bx}\right) \,dx$.

 

[ognik]

Thanks - so does orthogonality only apply for a period of $2\pi$? Both $[0,2\pi]$ and $[-\pi, \pi]$ ?

 

[Dethrone]

I believe that it has to be symmetric about the x-axis, so any period such that $-L<x<L$. That means that $[0,2\pi]$ won't work, but $[-\pi, \pi]$ will work.

There are however, cases where we can work with half the period. Scroll down to the last section of the page:
Pauls Online Notes : Differential Equations - Periodic Functions & Orthogonal Functions

 

[ognik]

I believe that it has to be symmetric about the x-axis, so any period such that $-L<x<L$. That means that $[0,2\pi]$ won't work, but $[-\pi, \pi]$ will work.

Hmm...my book uses $[0,2\pi]$ when talking about this orthogonality?

 

[Dethrone]

Hmm...my book uses $[0,2\pi]$ when talking about this orthogonality?

Hmm...I was talking in generality, but it does seem like it would work for $[0, 2\pi]$ in this case, my bad.
This is because $\sin Ax\cdot\cos Bx$ can be split into two individual sine terms, each of which evaluate to 0 if you integrate from $[0, \text{period}]$. (A complete sine oscillation has an equal positive and negative area which cancel)

 

[ognik]

For myself, thinking about it further, it seems that a full period from any start point would also work (so $[x_0, x_0+2\pi]$ - because the same length of curve(s) would be included in the interval regardless of where in the curve we start from.

 

[Dethrone]

Yep, that's also correct, as in any phase shift of the function.