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Fourier series expansion. Find value of a term in expansion

  1. Jan 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Fourier series expansion of a signal f(t) is given as
    f(t) = summation (n = -inf to n = +inf) [3/(4+(3n pi)2) ) * e j pi n t

    A term in expansion is A0cos(6 pi )
    find the value of A0

    Repeat above question for A0 sin (6 pi t)

    2. Relevant equations
    Fourier expansion is summation n = -inf to +inf Cn ejwnt
    where Cn is Integration over T0 x(t) e-jwnt dt

    3. The attempt at a solution
    upload_2017-1-28_15-14-27.png
    In book they've said Cn is right.
    But then they say an is 2 times real part of Cn
    and have done an = 6/[4 + (18 pi)2

    For third part they've put A0sin (6 pi t) = bn sin (n w t)
    and for n = 6, this is zero.

    I didn't understand this part. Why did they multiply it by 2 first and then how did it become zero in the second.

    I understand that sin ( 6 pi) is zero but how can sin ( 6 pi t) be zero? Wouldn't this vary as 't' varies?
     
  2. jcsd
  3. Jan 28, 2017 #2

    pasmith

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    Homework Helper

    The coefficients of [itex]\cos n\pi t[/itex] and [itex]\sin n\pi t[/itex] come not just from [itex]c_n[/itex], but also [itex]c_{-n}[/itex]: [tex]
    c_ne^{i n\pi t} + c_{-n}e^{-in\pi t} = (c_n + c_{-n})\cos n\pi t + i(c_n - c_{-n})\sin n\pi t[/tex]
     
  4. Jan 28, 2017 #3
    Oh yeah. Thanks. I didn't think about C-n.
     
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