Fourier Transform of x(t)= 8 sin(8(pi)t + pi/8)

Discussion in 'Homework Help' started by hitmen, Mar 19, 2009.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    Using the formula sin (A+B) = sinAcosB+cosAsinB I am able to reduce the equation to

    (8 cos (pi/8) ) sin( 8(pi)t ) + (8 sin (pi)/8 )cos( 8(pi)t )

    Now I do not know how to do the fourier transform or what equation to use. pls help
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
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    Hi hitmen,

    Sorry if the page didn't load up with your function entered.

    The result is complex but presumably explainable in physical terms.

    Since the time domain function transforms {maps?} as single frequency value into the frequency {fourier} domain one just gets a single fourier frequency value - mathematically expressed as a dirac or delta based function....?

    I've attached a printout of the "valid" entry for your function on the WIMS site

    (I used s as the default variable - maybe ω would be better so as not to confuse with Laplace)

    View attachment WIMS Home.pdf

    How you derive this from a knowledge of the transform I'll to leave to you to discover - the website has some information which may be useful ...?

    :)
     
  4. GirishC

    Active Member

    Jan 23, 2009
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    Now 8Cos(pi/8) and 8sin(pi/8) is a constant term added to your time varying signal...its a DC..so you do not need to worry about it much....

    Now you have two terms as

    7.39*sin(8*PI*t) + 3.06*cos(8*PI*t) = 7.39*sin(4t) + 3.06*cos(4t)

    i.e. you have two components at 4t
     
  5. mik3

    Senior Member

    Feb 4, 2008
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