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    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
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    (Original post by SassyPete)
    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
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    (Original post by SassyPete)
    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
    (Original post by RDKGames)
    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
    Rewrite it as a sum of 2 sins using standard trig identities and you immediately get the Fourier series, no integration required.
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    (Original post by RDKGames)
    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
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    (Original post by SassyPete)
    honestly, im struggling with this question so reckon i'm just going to give up
    \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.

    (Original post by DFranklin)
    Rewrite it as a sum of 2 sins using standard trig identities and you immediately get the Fourier series, no integration required.
    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
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    (Original post by RDKGames)
    \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...
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    (Original post by DFranklin)
    \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?
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    (Original post by RDKGames)
    I thought F.s. needs to be in terms of infinite sums of sines/cosines?
    No (or at least, it can be an infinite series where all but a finite number of terms are 0). If you go through the maths you've done (and assuming there's not anything odd going on with periods that I've missed), you'll find all the other terms are zero.
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    (Original post by RDKGames)
    \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

    could you not use eulers equation prior?
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    (Original post by DFranklin)
    \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
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    (Original post by SassyPete)
    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?
    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.


    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
    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)
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    (Original post by Zacken)
    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)
    ahh i tried to follow RDKs post. so should i be using a different method to what he suggested?
 
 
 
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