Proving the Fourier Series for a Real, Odd Function

[robgb]

The Fourier Series!

Hi guys, I'm having a bit a trouble helping my daughter with this question on the Fourier series approximation:

The Fourier series for a real, odd function, f(t) can be written as:
f(t) = [SUM to infinity, n=1, of]: b[subscipt n] sin(nwt)
where f(t=T)=f(t) and w=(2[pie])/T

Prove that b[subscipt n] = 2/T [integral between T/2 & -T/2 of]: f(t)sin(nwt)dt

Sorry about not knowing how to do all the symbols etc, but if you write it out I'm pretty sure it makes sence.

Anyway, if anyone could give me a hand with how to go about proving this, it would be greatly appreciated.

Many thanks, Robert.

 

[chroot]

Hey Robert,

You have a function of t represented by the series

$$f(t) = \sum_{n=1}^\infty b_n \sin{n \omega t}$$

And you'd like to know what the $$\inline{b_n}$$ are. You're also given the condition $$\inline{f(t + T) = f(t)}$$, which just indicates the function's period is $$\inline{T}$$.

You're not even required to figure out what the $$\inline{b_n}$$ are; you're just asked to demonstrate that they are, in fact

$$b_n = \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt$$.

To do this, first note the following fact:

$$\begin{equation*}<br /> \begin{split}<br /> \int_{-T/2}^{T/2} \sin^2{\left( \frac{2 \pi n t}{T} \right) }\ dt &= \frac{T}{2}<br /> \end{split}<br /> \end{equation*}$$

for all n, and use it to perform the following steps:

$$\begin{equation*}<br /> \begin{split}<br /> \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt &= \int_{-T/2}^{T/2} \left[ \sum_{n=1}^\infty b_n \sin{n \omega t} \right] \sin{n \omega t}\ dt\\<br /> &= \sum_{n=1}^\infty \int_{-T/2}^{T/2} b_n \sin^2{n \omega t}\ dt\\<br /> &= \frac{T}{2} b_n<br /> \end{split}<br /> \end{equation*}$$

Does this make sense?

- Warren

 

[robgb]

Thanks mate, yep i essentially understand, just 1 question though:

How did you get from the second line to the third line in the following:

$$\begin{equation*}<br /> \begin{split}<br /> \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt &= \int_{-T/2}^{T/2} \left[ \sum_{n=1}^\infty b_n \sin{n \omega t} \right] \sin{n \omega t}\ dt\\<br /> &= \sum_{n=1}^\infty \int_{-T/2}^{T/2} b_n \sin^2{n \omega t}\ dt\\<br /> &= \frac{T}{2} b_n<br /> \end{split}<br /> \end{equation*}$$

Could you work me though that bit in a little more detail mate?

Thanks a lot,
Robert.

 

[lethe]

Originally posted by robgb
Thanks mate, yep i essentially understand, just 1 question though:

How did you get from the second line to the third line in the following:

Could you work me though that bit in a little more detail mate?

Thanks a lot,
Robert.

i think chroot pulled a fast one on you. there is a bit more to the trick that you have to realize to make the infinite summation turn into a single term, with one integration.

to see this, first let me remind you of some trigonometric identities:

$$\cos(a\pm b)=\cos a\cos b \mp\sin a\sin b$$
so
$$\begin{align}<br /> &\frac{1}{2}(\cos(a-b)-\cos(a+b))\\<br /> &=-\frac{1}{2}\cos a\cos b+\frac{1}{2}\sin a\sin b\\<br /> & -\frac{1}{2}\cos a\cos b+\frac{1}{2}\sin a\sin b\\<br /> &= \sin a\sin b<br /> \end{align}$$

we need to use this identity to evaluate some integrals:

$$\begin{align}<br /> &\int_{-T/2}^{T/2}\sin(\frac{2\pi mt}{T})\sin(\frac{2\pi nt}{T})\ dt=\\<br /> &\frac{1}{2}\int_{-T/2}^{T/2}\cos\frac{2\pi(n-m)t}{T}\ dt-\frac{1}{2}\int_{-T/2}^{T/2}\cos\frac{2\pi(n+m)t}{T}\ dt=\\<br /> &\left\frac{T}{4\pi(n-m)}\sin\frac{2\pi(n-m)t}{T}\right]^{T/2}_{-T/2}\\<br /> &-\left\frac{T}{4\pi(n+m)}\sin\frac{2\pi(n+m)t}{T}\right]^{T/2}_{-T/2}\\<br /> &=0<br /> \end{align}$$
note that this integration is not valid when n=m, since it involves a division by zero. in the n=m case, i have:
$$\begin{align}<br /> \int_{-T/2}^{T/2}\sin^2\frac{2\pi nt}{T}\! dt\\<br /> =\frac{1}{2}\int_{-T/2}^{T/2}dt-\frac{1}{2}\int_{-T/2}^{T/2}\cos\frac{4\pi nt}{T}dt\\<br /> =T/2<br /> \end{align}$$

so, now, armed with these integrals, we can do the problem.

i am going to rename the dummy index n, in the infinite summation to m. you will see why.
$$f(t) = \sum_m^\infty b_m \sin\frac{2\pi mt}{T}$$

then


$$\begin{equation*}<br /> \begin{split}<br /> \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt &= \int_{-T/2}^{T/2} \left[ \sum_m^\infty b_m \sin\frac{2\pi mt}{T}\right] \sin\frac{2\pi nt}{T}dt\\<br /> &= \sum_m^\infty b_m\int_{-T/2}^{T/2} \sin\frac{2\pi nt}{T}\sin\frac{2\pi mt}{T}\ dt<br /> \end{split}<br /> \end{equation*}$$

this summation contains an infinite number of terms, one for every value of m, but only when m=n is the term nonzero, so i will drop the summation, and change m to n, so i am only considering one integral, and the value of that integral is T/2. this is the only nonzero term of the infinite summation.

$$\int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt =\frac{T}{2}b_n$$
or
$$b_n=\frac{2}{T}\int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt$$

and that s it. any questions?