plotting equation which is function of 2 variable
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I have one equation in terms of 2 variable A,B, I want to plot magnitude vs phase for different value of A and B .
i don't know how to do this . i was trying but it won't work
for A = -20:1:20
for B = -20:1:20
g=(1+i*B)/(1+i*A);
M=20*log10(abs(g)); %Magnitude matrix of Compensator
M = round(M,2);
P =rad2deg(angle(g)); %Phase Matrix of compensator
P = round(P,2);
plot(P,M);
hold on
end
end
trying to plot this figure
then
Find value of A and B for particualr value of phase and magnitude.
采纳的回答
Your code is close. I would take advantage of meshgrid instead of the for loops. Here's what I get making that change. Also, there is no need to round.
[A,B] = meshgrid(-20:1:20);
g=(1+i*B)./(1+i*A);
M=20*log10(abs(g)); %Magnitude matrix of Compensator
P=rad2deg(angle(g)); %Phase Matrix of compensator
plot(P,M,'k')
8 个评论
Looking at your plot, you don't want even spacing in you A & B values. Hard code the values displayed on the plot.
[A,B] = meshgrid([-80 -33 -14 -6 -2.7 -1.2 -0.5 0 .5 1.2 2.7 6 14 33 80]);
g=(1+i*B)./(1+i*A);
M=20*log10(abs(g)); %Magnitude matrix of Compensator
P =rad2deg(angle(g)); %Phase Matrix of compensator
plot(P,M,'k',P,-M,'k')
RJS
2021-10-6
Cris LaPierre
2021-10-6
The trick for me was to realize that the lines of B are the negative of the magnitudes of A, so I just plotted +M and -M against phase.
plot(P,M,'k',P,-M,'k')
% (^,^^) ->2nd plot using negative magnitude.
You could compute it if you want using g=(1+i*A)./(1+i*B);
Cris LaPierre
2021-10-6
Perhaps try run your equations in reverse?
Perhaps try adding an assumption that a and b are real.
syms a b real
g=(1+i*b)/(1+i*a);
E = [20*log10(abs(g)) == 10, rad2deg(angle(g)) == 12];
S = vpasolve(E,a,b)
Now test the result to see if it results in the expected M and P values.
A=double(S.a);
B=double(S.b);
g=(1+i*B)./(1+i*A);
M=20*log10(abs(g)) % Should be 10
P =rad2deg(angle(g)) % Should be 12
RJS
2021-10-9
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