help on fourier series

could someone explain how i can develop a finite fourier series for a function such as cos(t)sin(5t) with a set period of 2pi - lecturer was not very helpful at all in explaining

could someone explain how i can develop a finite fourier series for a function such as cos(t)sin(5t) with a set period of 2pi - lecturer was not very helpful at all in explaining

Since you have $f(t)=\cos(t)\sin(5t)$ with $f(t+2\pi)=f(t)$ then since it is an odd function, you it will have the form:

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

where

$\displaystyle b_n = \frac{2}{\pi} \int_{0}^{\pi} f(t)\sin(nt).dt$

Since you have $f(t)=\cos(t)\sin(5t)$ with $f(t+2\pi)=f(t)$ then since it is an odd function, you it will have the form:

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

where

$\displaystyle b_n = \frac{2}{\pi} \int_{0}^{\pi} f(t)\sin(nt).dt$

honestly, im struggling with this question so reckon i'm just going to give up

Rewrite it as a sum of 2 sins using standard trig identities and you immediately get the Fourier series, no integration required.

$\cos(t)\sin(5t) = \frac{1}{2}( \sin(4t)+ \sin(6t))$ using the sum of two sines identity, so $\displaystyle b_n= \frac{1}{\pi} \int_{0}^{\pi} \sin(4t)\sin(nt)+ \sin(6t)\sin(nt) .dt$

Then using the product of two sines identity you get $\sin(4t)\sin(nt)=\frac{1}{2}[\cos([4-n]t)+\cos([4+n]t)]$

So your integral just simply goes to:

$\displaystyle b_n=\frac{1}{2\pi} \int_{0}^{\pi} \cos[(4-n)t]+\cos[(6-n)t]+\cos[(4+n)t]+\cos[(6+n)t].dt$

which is straightforward to integrate and then simplify.



Sorry, I don't see it

I've not mastered Fourier series yet (nor the use of these sum identities) but I'd like to see how this can be done in such a short time, as you seem to say

$\frac{1}{2}( \sin(4t)+ \sin(6t))$ is a Fourier series, and since Fourier series are unique, it is also the Fourier series...

I thought F.s. needs to be in terms of infinite sums of sines/cosines?

$\cos(t)\sin(5t) = \frac{1}{2}( \sin(4t)+ \sin(6t))$ using the sum of two sines identity, so $\displaystyle b_n= \frac{1}{\pi} \int_{0}^{\pi} \sin(4t)\sin(nt)+ \sin(6t)\sin(nt) .dt$

Then using the product of two sines identity you get $\sin(4t)\sin(nt)=\frac{1}{2}[\cos([4-n]t)+\cos([4+n]t)]$

So your integral just simply goes to:

$\displaystyle b_n=\frac{1}{2\pi} \int_{0}^{\pi} \cos[(4-n)t]+\cos[(6-n)t]+\cos[(4+n)t]+\cos[(6+n)t].dt$

which is straightforward to integrate and then simplify.



Sorry, I don't see it

I've not mastered Fourier series yet (nor the use of these sum identities) but I'd like to see how this can be done in such a short time, as you seem to say

$\frac{1}{2}( \sin(4t)+ \sin(6t))$ is a Fourier series, and since Fourier series are unique, it is also the Fourier series...

ok ive managed to work myself to just after the integral - couple of questions now though:

1) why is the limit between 0 and pi if the period of the function is 2pi?

2) my integral is all in terms of sin with n and t as constants - i know the limit value applies to t but then im stuck with a huge answer in terms of sin and n

I'd always recommend DFrank's way in this scenario, but if you insist...

$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin nt \, \mathrm{d}t = \frac{2}{\pi} \int_0^{\pi} f(t) \sin nt$ since both f(t) and sin nt are odd their product is even.




Your limits are pi. So you get sin (npi) and other terms of similar form. You known sin(npi) = 0, sin(pi/2 + npi) = 1, etc...(assuming what you have already is correct, I think RDK's post has some mistakes)