solving a nonlinear equation with complex numbers

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Mimo T
Mimo T am 7 Nov. 2024 um 19:37
Bearbeitet: Paul am 7 Nov. 2024 um 22:15
I want to get the values of resistance and reactance of a parallal impedance that gives me specific value for magnitude and phase of the reflection coefficient. Resistance must be positive only while X must be a real number. This is my code
syms R
syms X
Z=(R*1i*X/(R+1i*X));
r=(Z-Zo)/(Z+Zo);
assume(R,'positive')
assume(X,'real')
eqn=[abs(r)==0.5, angle(r)==55*pi/180];
[R1,X1]=vpasolve(eqn,[R,X])
I can not add these constains like solve function and sometimes R becomes negative. (solve function can't be used here because it gives me a warning that vpasolve is used).
Trying to add R>0 in cases vpasolve gives positive R, leads to empty solution.
  2 Kommentare
Walter Roberson
Walter Roberson am 7 Nov. 2024 um 20:15
Unfortunately your code does not define Zo.
Mimo T
Mimo T am 7 Nov. 2024 um 21:47
Yes, I missed it when I copied the code.
Zo=50;

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Akzeptierte Antwort

Walter Roberson
Walter Roberson am 7 Nov. 2024 um 22:09
Bearbeitet: Walter Roberson am 7 Nov. 2024 um 22:10
Zo = 50;
syms R
syms X
Z=(R*1i*X/(R+1i*X));
r=(Z-Zo)/(Z+Zo);
assume(R,'positive')
assume(X,'real')
eqn=[abs(r)==0.5, angle(r)==55*pi/180];
[R1,X1]=vpasolve(eqn,[R,X], [0 inf; -inf inf]);
disp(char(R1))
121.57176242340307307386879418841
disp(char(X1))
111.30878870807649324383788998856

Weitere Antworten (2)

Paul
Paul am 7 Nov. 2024 um 22:14
Bearbeitet: Paul am 7 Nov. 2024 um 22:15
Does this solution work?
syms R
syms X
Zo = sym(50);
Z=(R*1i*X/(R+1i*X));
r=(Z-Zo)/(Z+Zo);
assume(R,'positive')
assume(X,'real')
eqn=[abs(r)==0.5, angle(r)==55*pi/180];
[R1,X1]=vpasolve(eqn,[R,X],[0,inf;-inf,inf]) % specify initial search bounds
double([R1,X1])
ans = 1×2
121.5718 111.3088
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%sol = solve(eqn,[R,X],'ReturnConditions',true)

John D'Errico
John D'Errico am 7 Nov. 2024 um 21:00
Bearbeitet: John D'Errico am 7 Nov. 2024 um 21:03
syms R positive % positive implicitly implies real also
syms X real
syms Zo
Z=(R*1i*X/(R+1i*X));
r=(Z-Zo)/(Z+Zo)
r = simplify(r)
Suppose we knew the value of Zo? I'll pick 0.5 as an example.
r_Zo = subs(r,Zo,0.5)
eqn = [abs(r_Zo)==0.5, angle(r_Zo)==55*pi/180]
Now, can we solve this?
RX = solve(eqn,[R,X],returnconditions = true)
RX = struct with fields:
R: z1 X: z2 parameters: [z1 z2] conditions: abs((z1*1i)/2 - z2/2 + z1*z2)/abs(z2/2 - (z1*1i)/2 + z1*z2) - 1/2 == 0 & - (11*pi)/36 + angle(((z1*1i)/2 - z2/2 + z1*z2)/(z2/2 - (z1*1i)/2 + z1*z2)) == 0 & 0 < z1
Essentially, solve is telling us it cannot find an algebraic form for the solution. However, as long as Zo is known, we can use a numerical tool like vpasolve.
RX = vpasolve(eqn,[R,X])
RX = struct with fields:
R: 1.2157176242340307307386879418841 X: 1.1130878870807649324383788998856
The problem is, if Zo is left unknown, then there will be no analytical solution for the problem. You cannot use solve, as I just showed, and vpasolve cannot be employed to solve problems with unknown symbolic dependencies.
(Sadly, Answers is bugged, and it is not showing the symbolic output. One year they will fix this, what was working nicely only a month or so ago, and what does OCCASIONALLY work to this day. SIGH.)
  2 Kommentare
Mimo T
Mimo T am 7 Nov. 2024 um 22:01
but vpasolve returns R with a negative value. This is my question that vpasolve doesn't take into account the conditions
Mimo T
Mimo T am 7 Nov. 2024 um 22:02
I am also sure that there is an admittance (which is Z) on smith chart with positive value of R that is the solution

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