How to Determine the Correct Fourier Series for a Given Waveform?

[Gremlin]

Sorry i see what you're saying now. I shall try n=3 in what i have now and hope for 1/pi

 

[Gremlin]

No it doesnt.

 

[Gremlin]

I'm still trying, and failing, to nail this.

Can someone tell me where the attached goes off course please? There's not a lot to it, but already if n=3 i don't get 1/pi.

 

[gneill]

You're having problems with the integration. The arguments of the sine functions are not simple x's, so dx is not the differential element for them to be integrated as just a sine function. In other words,

$\int sin(u) du = -cos(u) + C$ which is fine since du is the differential element of u.

but if $u = (n+1) x$, then $du = (n + 1) dx$

You need to modify the integral so that the correct differential element is accounted for.

 

[Gremlin]

Ok thanks, i shall have another look.

 

[Birchyuk]

Good evening,

I am also struggling on this question and have been for a solid week.

I am currently trying to calculate a_n. I so far have:

I have assumed for the time being x=wt

\int sinxcosnx.dx
\frac{1}{2} \int sin(n+1)x+sin(n-1)x.dx
\frac {1}{2} (\frac {-cos(n+1)x} {n+1} + (\frac {-cos(n-1)x}{n-1})

I have input the limits of x between pi and \frac {pi}{2} and subtracted there. Furthermore i have done the same to 2pi and \frac {3pi}{2} and added these but my result continually comes out at 0. I am expecting only odd numbers of n will produce a result. I have tried a few methods but none of them seem to be working.My hunch is that I am not integrating correctly or I am misunderstanding the relevance of (wt)?

Any assistance would be appreciate greatly.

 

[bizuputyi]

Hi Birchyuk,

Try this:

a_n = \frac{1}{\pi} \int sinxcosx.dx + \frac{1}{\pi} \int sinxcosx.dx

Insert your limits and put the equation in wolfram or mathlab. What do you get?

'V' won't make any difference as long as it's 1 but to be accurate 'V' should also be in the equation.

 

[Birchyuk]

Hi bizuputyi,

Thanks for the reply

Inputting this i get the following:

https://www4c.wolframalpha.com/Calculate/MSP/MSP103820ai5432f8fdg93e0000303fdbg66ci2bbf8?MSPStoreType=image/gif&s=4 and https://www5b.wolframalpha.com/Calculate/MSP/MSP8661cb3a0cb5a74g28300003gh8hbag7c679080?MSPStoreType=image/gif&s=33

I have produced a table in wolfram which is as below: (i have to admit i am not an expert with wolfram)

http://www.wolframalpha.com/input/?...n*3*pi/2))+(cos(n*2*pi))))/(n^2-1)),{n,2,10}]

This again shows i have both even and odd harmonics which i believe is wrong

 

[bizuputyi]

I know this is a shortcut but it is perfectly acceptable to produce this:

http://www.wolframalpha.com/input/?...2*pi}}]*cos(n*x)+from+x+=+0+to+2*pi,{n,1,10}]

Then put a1, a3, a5... in your F.S final result.

 

[Birchyuk]

Hi Bizuputyi,

Thanks for this. For my own piece of mind what is going wrong as this does not make sense to me why this is not working the way i think it should?

Thanks.

 

[macca67]

Hi Guys,

I am really struggling with this same question, here is what I have done so far, but there is a mistake (a few probably) in there and I can't find it!

fx=Vsin (wt)
x=wt

So a₀= 1/π∫f(x)dx

= V/π∫sin (x) dx [lim π/2 to π] + V/π∫sin (x) dx [lim 3π/2 to 2π]

So next I integrated the above to give
=V/π[-cosx] [lim π/2 to π] + V/π[-cosx] [lim 3π/2 to 2π]
=V/π[0]
=0

so I proved a₀=0

Next on to a_n

a_n= V/π∫sin (x).cos (nx) dx [lim π/2 to π] + V/π∫sin (x).cos (nx) dx [lim 3π/2 to 2π]

using the trig rule 2sin A.cos B = [sin (A+B) + sin (A-B)]/2 I changed the above to

a_n= V/2π∫sin (x+nx)+sin (x-nx) dx [lim π/2 to π] + V/2π∫sin (x+nx)+sin (x-nx) dx [lim 3π/2 to 2π]

this was then integrated to give

= V/2π[(-cos(x+nx)/n+1)-(cos(x-nx)/n-1)]+V/2π[(-cos(x+nx)/n+1)-(cos(x-nx)/n-1)]

When I put n=1 it all cancels out to 0, but by what's written in this thread I should have -2(V/2π) to give (-V/π)

For b_n I end up with

b_n=V/2π∫cos (x-nx)-cos (x+nx) dx [lim π/2 to π] + V/2π∫cos (x-nx)-cos (x+nx) dx [lim 3π/2 to 2π]

which integrated to

V/2π[(sin(x).cos(x)-(x-sin (x))] [lim π/2 to π] +V/2π[(x-1/4sin(4x))-(x-sin1/3(3x))] [lim 3π/2 to 2π]

b_n=1/2 when n=1 (I put the n=1 in before integration as suggested by Gneil to get this, I don't understand why it should make a difference though?)

where have I gone wrong?

I think b_n is correct, but my a_n isn't

any help gratefully received!

EDIT- Just noticed the red bit above is wrong, not doing too well here!

 

[macca67]

One theory I have is when I use the trig rule, I change the sin(x) of the b_n function to a cos then when I integrate it goes back to a sin, after integration should b_n= (something) cos(something)?

 

[macca67]

ok,

b_n now sorted using integration by parts.

Still having difficulty with a_n