Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    5
    ReputationRep:
    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
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (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
    Offline

    17
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (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
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (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
    Offline

    17
    ReputationRep:
    (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...
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (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?
    Offline

    17
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (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?
    • Thread Starter
    Offline

    5
    ReputationRep:
    (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
    Offline

    22
    ReputationRep:
    (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)
    • Thread Starter
    Offline

    5
    ReputationRep:
    (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?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Will you be richer or poorer than your parents?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.