Problem with the product of complex numbers

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clement
clement 2013-5-24
Hello,
I calculated the equivalent impedance of an RLC circuit, and I would like this one to be completely resistive (complex part equals to 0). So I declared my variables as 'syms' and I used the function 'solve' to obtain the equivalent impedance litterally like:
% syms R X Y Z
% Zeq=solve('(R+i*X)*(-i*Y)/(R+i*X-i*Y)=Z',Z)
The problem is that Matlab gives me a solution like this:
%Zeq =
% -(Y*(R + X*i)*i)/(R + X*i - Y*i)
But I would like something like: Zeq = A + i*B.
Could anyone help?
Thanks

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采纳的回答

Jonathan Epperl
Jonathan Epperl 2013-5-24
Probably simplify(Zeq) will do that.
  2 个评论
clement
clement 2013-5-24
Thanks it's working! But now I've got another problem... When I multiply tne numerator of my fraction by the complex conjugate of the denominator, Matlab gives me this:
%sol = -Y*(R + X*i)*(X - Y + R*i)
instead of A + i*B. And this time 'simplify', or 'factor' don't work.
Walter Roberson
Walter Roberson 2013-5-24
expandsol = expand(sol);
A = real(expandsol);
B = imag(expandsol);

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更多回答（1 个）

Walter Roberson
Walter Roberson 2013-5-24
You cannot do that unless you add the assumption that the variables are real-valued
syms R X Y Z real
Zeq = simplify(solve((R+i*X)*(-i*Y)/(R+i*X-i*Y)-(Z),Z));
A = simplify(real(Zeq));
B = simplify(imag(Zeq));
A + B*i
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Walter Roberson
Walter Roberson 2013-5-25
Before R2011b, "==" was processed as a logical relationship to be evaluated and the result of the logical evaluation to be passed into solve(). But those versions also did not know how to compare a symbol (with any content) against a number, so the expression would generate an error... unless, of course, A was a number instead of a symbol.
Jonathan Epperl
Jonathan Epperl 2013-5-25
I see, I didn't know that -- so A-50 is the more robust syntax...

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