How to Expand a Function in a Half-Range Fourier Series

[hurcw]

Half wave Fourier series help !

I am trying desperately to figure out a Fourier series question i have been given and have hit a big mental wall can someone please help and maybe point me in the right direction.
The problems i have been given are:-
a) f(t)=e[^2t] 0<t<1 (cosine series)
b) f(t)=e[^2t] 0<t<∏ ( sine series)

working on the sine series i have that the a[n] terms will =0 and there will be no a[0][/2] term
What i have so far is:-
b[n]=[2][/p] ∫p][/2]&-[p][/2] F(t)sin ([2n∏t][/p]).dt
=[2][/2] ∫1 & -1 [e][^2t]sin (2n∏t).dt
how does this look so far?
Any help is appreciated, thanks

 

[chiro]

Hey hurcw and welcome to the forums.

I'm a little confused at your question.

The first thing is that if you want to do a series decomposition for a function, then for both the sine and the cosine parts, your function will be the same and the interval will be the same. You have two intervals: one is 0 < t < 1 and the other is 0 < t < ∏. Unless you get some kind of specific cancellation as a result of the integral, you can't just do this.

Also I a little skeptical that your function will only give you b[n] terms to be non-zero since e^(2t) is neither even nor odd.

It might help if you showed all your calculations for the integral with respect to the cosine bases and also with respect to the sine bases.

If you can't use latex (there is a guide in the help section), then you could scan your work from paper and add it as an attachment (this is a common way to do things).

 

[hurcw]

Thanks for the reply, what i ahve is two different questions, but one is a cosine series and the other a sine series, or at least that's how i understand it.
I have the answers but i have to prove my method and i just keep getting lost.
I sounds like i have messed up fro the beginning by thinking it is an odd series.
We have done a previous example of t² and had to define it so that the resulting function only had a) cosine terms, b) sine terms, so maybe this is what i ahve to do on this example, if that make any sense.

 

[chiro]

Thanks for the reply, what i ahve is two different questions, but one is a cosine series and the other a sine series, or at least that's how i understand it.
I have the answers but i have to prove my method and i just keep getting lost.
I sounds like i have messed up fro the beginning by thinking it is an odd series.
We have done a previous example of t² and had to define it so that the resulting function only had a) cosine terms, b) sine terms, so maybe this is what i ahve to do on this example, if that make any sense.

I can understand if you have to find only the sine or cosine terms but trying to force the function to give only sine or cosine terms won't work if f(t) = e^(2t).

For a function to have only sine terms your function must be odd which means f(t) = -f(-t) and for even the function is f(x) = f(-x). The t² is an even function while t³ is an odd function.

If you have to just find the sine component and the cosine component of the different series then you just calculate the a[n] and b[n] components using the Fourier series projection formula.

 

[hurcw]

What i need to prove is that
a) f(t) = e^2t 0<t<1
= (e²-1)/2+Ʃ((4/(4+n²∏²))(e²COS(n∏)COSn∏t)
b)f(t) = e^2t 0<t<∏
= Ʃ(2n∏/(4+n²∏²))(1-e²COS(n∏))SIN n∏t

If this make sense.

 

[chiro]

What i need to prove is that
a) f(t) = e^2t 0<t<1
= (e²-1)/2+Ʃ((4/(4+n²∏²))(e²COS(n∏)COSn∏t)
b)f(t) = e^2t 0<t<∏
= Ʃ(2n∏/(4+n²∏²))(1-e²COS(n∏))SIN n∏t

If this make sense.

For these you will have to take the Fourier transform involving both the sines and cosines and then simplify or create substitutions that match up with the results.

So what you should probably do is calculating the a[n] and b[n] terms first for a) and then showed what you have gotten. Then from this you can try fiddling around until you get the result.

Basically do the same for b as well.

The only thing though is that you need to make sure you have the right window size for each function because the interval is different. If you don't understand what I mean take a look this:

http://en.wikipedia.org/wiki/Fourie...general_interval_.5Ba.2C.C2.A0a_.2B_.CF.84.5D

 

[hurcw]

Not really getting what you mean, i will have to look after work and see how i get on just to get me started i think.
I appreciate the help

 

[chiro]

Not really getting what you mean, i will have to look after work and see how i get on just to get me started i think.
I appreciate the help

What I mean for the interval is that you instead of dividing by pi, you have to divide by something else. In the link I provided look at the value of tau - \tau.

Once you calculate the a[n] and b[n] for this interval, then you can get your expression in terms of a[n]xsin(2pinx) + b[n]xsin(2pinx] for whatever a[n] and b[n] they are and then you will be able to do simplifications which will give you the formula. This is the idea for both exercises.

 

[LCKurtz]

What i need to prove is that
a) f(t) = e^2t 0<t<1
= (e²-1)/2+Ʃ((4/(4+n²∏²))(e²COS(n∏)COSn∏t)
b)f(t) = e^2t 0<t<∏
= Ʃ(2n∏/(4+n²∏²))(1-e²COS(n∏))SIN n∏t

If this make sense.

It makes perfect sense to expand a function in a half-range series. Just use the half-range formulas. For a half range sine series$$
b_n=\frac 2 p\int_0^p f(t)\sin(\frac{n\pi}{p}t)\, dt,\ a_n=0$$In your example, $p=\pi$. For the half-range cosine the formulas are$$
b_n=0,\ a_n=\frac 2 p\int_0^p f(t)\cos(\frac{n\pi}{p}t)\, dt,\ a_0=\frac 1 p \int_0^p f(t)\ dt$$In your example $p=1$ for the cosine series.