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Integration of Mutual Inductance

  1. Dec 18, 2011 #1
    Hello. I am working through the book "Analysis of Electric Machinery and Drive Systems" by Krause. On p.51, the mutual inductance between two windings (a and b) of a 3-phase machine (generator) is calculated. I am struggling to follow the integration performed on eqn 1.5-15 to arrive at eqn. 1.5-16

    I have given the root of the problem: a double integral involving the product of trig. functions

    1. The problem statement, all variables and given/known data

    ∫sin ρ ∫cos(ζ - 2∏/3)(A - Bcos(2(ζ - θ)) dζ dρ

    first integral is done from ∏→2∏
    second integral is done from ρ → ρ + ∏

    result given in textbook is:

    ∏/2 * (A + Bcos(2(θ - ∏/3)))
    2. Relevant equations



    3. The attempt at a solution

    I tried expanding out the cosine and sine terms but the problem quickly blows up... :(

    EDIT: I did perform the integral using MATLAB Symbolic Toolbox. The result agrees with the given answer. I'm just at a loss of how to do this by hand. It seems very challenging so I'm wondering if I'm missing a trick.

    MATLAB Code

    %2011-12-18
    %Derivation of main integral in eqn. 1.5-15 --> 1.5-16 on p.51
    syms xi A B phi theta
    first_int = int(cos(xi - 2*pi/3)*(A - B*cos(2*xi - 2*theta)), xi, phi, phi+pi);
    second_int = int(first_int*sin(phi), phi, pi, 2*pi)
     
    Last edited: Dec 18, 2011
  2. jcsd
  3. Dec 18, 2011 #2

    Dick

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    Homework Helper

    I'd try expanding cos(ζ - 2∏/3)(A - Bcos(2(ζ - θ))) to Acos(ζ - 2∏/3)-Bcos(ζ - 2∏/3)cos(2(ζ - θ)). The term with one cos should be no problem. For the second one use cos(X)cos(Y)=(1/2)(cos(X+Y)-cos(X-Y)).
     
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