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Thread: Fourier expansion of f(t)=cos(t), what? How?

  1. #1
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    Default Fourier expansion of f(t)=cos(t), what? How?

    How do you make a fourier expansion of
    f(t)=cos(t); with a period of 2pi and with -pi < t< pi

    Using the Fourier formula:

    f(t)=Ao + SUM (An*cos(2Pi*n*t/T)+Bn*sin(2Pi*n*t/T)

    where t=Pi or -Pi
    where n=1 (number of terms)

    and

    Ao=1/T Integral of f(t)*d(t)
    where f(t) is the cos(t)



    I KEEP GETTING ZERO for all Ao, An, and Bn. Guys in my class have the same thing and I dont think it is correct. There is something more to this.

    Does anyone have experience doing this?

  2. #2
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    Default

    Figured it out, "trick" question...the cos function is already in its fourier series so nothing more is needed.

    But now that the whole class was stumped on the assignment he just gave us completion/attempt marks. Waste of my time

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