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Solve an equation with angle of 2 transfer functions to get the capacitor?

Posted 1 month ago
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Hello, I have a little Mathematica notebook that equalizes the angle of 2 transfer functions to calculate a capacitor with Solve[]. In an other Computer Mathematics System I get a solution without problem, but Mathematica gives an error back. Can anybody help me, who is better in Mathematica syntax? Thanks, Regards Leo

Za = I*w*L + 1/(I*w*C + 2/Z0);
Zb = 1/(I*w*C) + 1/(1/(I*w*L) + 2/Z0);
Zl = Re[1/(1/Za + 1/Zb)] - I*Im[1/(1/Za + 1/Zb)];
Zal = 1/(1/Zl + 1/Za);
Zbl = 1/(1/Zl + 1/Zb);
WU2U0a = Arg[Zbl/2/(1/(1/(Zbl + I*w*L) + I*w*C) + Z0/2)];
WU2U0b = Arg[-Zal/
     2/(Z0/2 + 1/(1/(1/(Zal + 1/(I*w*C))) + (1/(I*w*L))))];
Z0 = 44 - I*15; w = 2*Pi*868*^6; L = 1.8*^-9;

Solve[WU2U0a == WU2U0b && C > 0 && C < 50*^-12, C]
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9 Replies

Do not use C as a variable, since it is a built-in symbol in the system. We put X =WU2U0a - WU2U0b . In this problem we have a special case when X goes through 0 abruptly - see fig. Thus, all solvers fail on this problem. Fortunately, we can approximately determine the root of the equation as the minimum point X. It is possible that this solution makes sense for technical applications in trigger generators.

p = 9; Za = I*w*L + 1/(I*w*C1 + 2/Z0);
Zb = 1/(I*w*C1) + 1/(1/(I*w*L) + 2/Z0);
Zl = Re[1/(1/Za + 1/Zb)] - I*Im[1/(1/Za + 1/Zb)];
Zal = 1/(1/Zl + 1/Za);
Zbl = 1/(1/Zl + 1/Zb);
WU2U0a = Arg[Zbl/2/(1/(1/(Zbl + I*w*L) + I*w*C1) + Z0/2)];
WU2U0b = Arg[-Zal/
     2/(Z0/2 + 1/(1/(1/(Zal + 1/(I*w*C1))) + (1/(I*w*L))))];
Z0 = 44 - I*15; w = Rationalize[2*Pi*868*^6, 10^-p]; L = 18*^-10;
X = WU2U0a - WU2U0b;
Block[{$MinPrecision = p, $MaxPrecision = p}, 
 NMinimize[{X, 0 <= C1 < 5*10^-11}, C1, WorkingPrecision -> p]]
{-2.96652668, {C1 -> 2.77937850*10^-11}}
{Plot[X, {C1, 2.77`30*10^-11, 2.79`30*10^-11}], 
 Plot[{WU2U0a, WU2U0b}, {C1, 2.77`30*10^-11, 2.79`30*10^-11}, 
  PlotLegends -> "Expressions"]}

fig1

In this problem, you compare impedances, not phases. This is another task, it has another solution in the complex plane. And in the original problem, we looked for real roots that physically correspond to the capacitance C1.

Many solutions in the complex plane and only one real root, which I have indicated. But if you change the frequency w, as suggested by the German mathematician, the roots will be infinitely many.

Posted 1 month ago

Hello Alexander, the circuit is a monofrequent balun. With another Computer Mathematics System (MuPad) it is easy to solve the problem. The problem is solved now. It was my mistake.

Thanks a lot.

Regards.

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It is possible that some of the solutions found with MuPad are erroneous. It is easy to check, for example, that C1 = 1.606139208*10^-11 is not a root of the equation X = 0, indeed

In[3]:= X /. C1 -> 1.606139208*10^-11
Out[3]= 2.15442

Or have you found another solution? Does it coincide with mine C1 -> 2.77937850*10^-11?

Posted 1 month ago

At the moment I have the problem, that I get different solutions with different calculations with Mathematica. Sometimers I get 2 real solutions, sometimes I get only complex solutions.- I will check it tomorrow. I am confused.

My results are in the attachment.

I will check it tomorrow, it is late here. I will message.

Regards

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Posted 1 month ago

It is a fact that there are many solutions for C1 that angle of transfer function 1 is equal to angle of transfer function 2. A German mathematician told me yesterday the problem is a quotient with 2 polynomials in w with the order 48.

It depends upon the methode to find some solutions. Different methods deliver different solutions. MuPad delivers 13 real solutions for C1. Mathematica's Solve and NSolve have different methodes, so it sometimes delivers different solutions depending on whether methode was internal used. -

Maybe I will check if there are wrong solutions when I have time.

I guess all solutions are right and we should accept that.

Regards Leo

Posted 1 month ago

It is a fact that there are many solutions for C1 that angle of transfer function 1 is equal to angle of transfer function 2. A German mathematician told me yesterday the problem is a quotient with 2 polynomials in w with the order 48.

It depends upon the methode to find some solutions. Different methods deliver different solutions. MuPad delivers 13 real solutions for C1. Mathematica's Solve and NSolve have different methodes, so it sometimes delivers different solutions depending on whether methode was internal used. -

Maybe I will check if there are wrong solutions when I have time.

I guess all solutions are right and we should accept that.

MuPad: C1 -> 1.606139208*10^-11 In the attachment You can bottom see that the 2 voltages have the same phase in the load impedance. So C1 -> 1.606139208*10^-11 is a correct solution.

This C1 solution is the best, because with it Zgenerator will be Zinput*, so it is close to matching.

Regards Leo

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Posted 1 month ago

MuPad.pdf at the bottom in the graph shows that C1 is a correct solution to the problem. The two voltages are in phase in the load impedance, which we wanted. Please see also Zin and compare with Z0 at the top. There is adaptation with C1. I think Mathematica NSolve fails here because it does not provide all the real solutions. -

I also send You the basics to that problem with the circuit and the equitations.

Have fun - regards Leo

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