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Math Help - [SOLVED] Matlab symbolic "solve" gives unknown parameter "z"

  1. #1
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    Joined
    Apr 2010
    Posts
    7

    [SOLVED] Matlab symbolic "solve" gives unknown parameter "z"

    Hello

    I am trying to symbolically solve for 'ws' in the following equations:

    Code:
    syms N ep es ws
    t1=ep*es; t1_=sqrt(1-t1^2); t2=1/ws; t2_=sqrt(1-t2^2);
    e1=0.5*(1-t1_^0.5)/(1+t1_^0.5); e2=0.5*(1-t2_^0.5)/(1+t2_^0.5);
    q1=e1+2*e1^5; q2=e2+2*e2^5;
    K1=pi/2*(1+2*q1+2*q1^4)^2; K2=pi/2*(1+2*q2+2*q2^4)^2;
    K1_=K1/pi*log(1/q1); K2_=K2/pi*log(1/q2);
    k=N*K1/K1_;
    But when try to solve, I get this:

    Code:
    solve(k-K2/K2_,ws)
     
    ans =
     
      1/((z^2 + 1)^(1/2)*(1 - z)^(1/2)*(z + 1)^(1/2))
     -1/((z^2 + 1)^(1/2)*(1 - z)^(1/2)*(z + 1)^(1/2))
    What is that "z" parameter? "z <enter>" says "??? Undefined function or variable 'z'."
    Also, I tried simplifying the equations, willingly accepting less precision in favor of a clear answer:

    Code:
    q1=e1; q2=e2;
    K1=pi/2*(1+2*q1)^2; K2=pi/2*(1+2*q2)^2;
    K1_=K1/pi*log(1/q1); K2_=K2/pi*log(1/q2);
    k=N*K1/K1_;
    ...but I get the same answer!(?) Can anyone shed some light in this? I googled for some answers and the closest I could find was that "z" replaces some really complicated formulae, all looking the same, that would have only filled the display, so Matlab simplified them with "z", but it said that revealing it wasn't easy. That's it, not very helpful.


    Regards,
    Vlad.
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  2. #2
    Newbie
    Joined
    Apr 2010
    Posts
    7
    I found a solution to it:
    Elliptic( Cauer, Zolotarev ) approximation method for obtaining transfer functions of linear filters

    There are some simplifications which approximate the nome q=q1^(1/N) and then k can be calculated through a series.

    I still don't know what Matlab's mysterious "z" means so, if anyone has answers, don't hesitate.


    Regards,
    Vlad.
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