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Thread: Fourier trigonometric series - amplitude and phase spectrum

  1. #1
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    Fourier trigonometric series - amplitude and phase spectrum

    Hello,

    I don't understand why is amplitude spectrum of Fourier series defined as:

    c_0 = |a_0|
    c_n = \sqrt{a_n^{2} + b_n^{2}}

    and why is phase spectrum:

    Arg(a_n - b_n)

    Can someone give me an understandable easy explanation.
    Thanks.
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  2. #2
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    Try writing
    a_n \cos{ (nt)} + b_n \sin{ (nt) }= c_n \cos {(nt+\alpha)} and solving for c_n and \alpha.

    Hint: use \cos{ (A+B) }= \cos{ (A)}\cos{(B)}-\sin{(A)}\sin{(B)} on the right hand side.
    Last edited by Archie; Jul 13th 2017 at 07:30 AM.
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  3. #3
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    Ok, I got this:

    c_n = \frac{a_n(cos(nt)+b_n(nt)}{cos(nt)cos(\alpha)-sin(nt)(sin(\alpha)}

    How to get from this to Pythagorean theorem?
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  4. #4
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    No, we equate coefficients:
    \begin{align*} c_n \cos{(nt-\alpha_n)} &= c_n \big( \cos{(\alpha_n)}\cos{(nt)} + \sin{(\alpha_n)}\sin{(nt)} \big) \\ &= c_n \cos{(\alpha_n)}\cos{(nt)} + c_n \sin{(\alpha_n)}\sin{(nt)} = a_n \cos{(nt)} + b_n \sin{(nt)} \end{align*}

    And so we can say that
    a_n = c_n \cos{(\alpha_n)} \quad b_n = c_n \sin{(\alpha_n)}

    Thus,
    a_n^2 + b_n^2 = c_n^2 \quad \tan{(\alpha_n)} = \frac{b_n}{a_n}
    Last edited by Archie; Jul 14th 2017 at 08:48 AM.
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  5. #5
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    Hey thank you.

    I have another question. In this equation the left hand side is the definition of Fourier trigonometric series (Infinite sum of cos and sin terms). How did you get to the right hand side?

    a_n \cos{ (nt)} + b_n \sin{ (nt) }= c_n \cos {(nt+\alpha)}

    Well, I have another question, but let's stick with one at the time.
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  6. #6
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    It's a trigonometric identity.
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  7. #7
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    Which trigonometric identity? I didn't find it on wikipedia or any other site.

    And another question: Why are there two these series? (Exponential Fourier series and Trigonometric)
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  8. #8
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    No one knows which trigonometric identity is this? What about other question?
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  9. #9
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    Re: Fourier trigonometric series - amplitude and phase spectrum

    The "trigonometric identity" in question is cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)

    Taking A= nt, B= \alpha, cos(nt+\alpha)= cos(nt)cos(\alpha)- sin(nt)sin(\alpha)

    So a_ncos(nt)+ b_nsin(nt)= c_ncos(nt+ \alpha) with a_n= c_ncos(\alpha) and b_n= c_nsin(\alpha)
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