Why are condition2 condition3 and condition4 not logical?
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zhenyu zeng
le 22 Juil 2019
Commenté : Image Analyst
le 23 Juil 2019
a=0.336;
Ta=9.476;
Te=1.208;
Tw=1.498;
eqh=[0.661;0.619;0.568];
ex=[-1.24;-1.346;-1.441];
en=-ex;
ey=[0.376;0.705;0.968];
eqx=[-0.309;-0.357;-0.392];
eh=[1.594;1.583;1.545];
eyqh=[0.642;0.78;0.897];
a0=a*Ta*Te^2;
r = -100:1:100;
[bt,Td]=meshgrid(r);
i=1;
figure('Name','Stable');
Kp=1./bt;
exqh(i)=en(i).*eqh(i)+eqx(i).*eh(i);
Ki=1./(bt.*Td);
a1=Ta.*Tw.*eqh(i)+a.*Te^2.*en(i)+a.*Te^2.*ey(i).*Kp;
a2=a.*Te^2.*Ki*ey(i)+Ta+Tw.*exqh(i)-eyqh(i).*Kp.*Tw;
a3=en(i)+ey(i).*Kp-eyqh(i).*Ki.*Tw;
a4=ey(i).*Ki;
condition1 =a1>0; output = ones(length(r));condition2=zeros(size(condition1));condition3=zeros(size(condition1));condition4=zeros(size(condition1));
for j=1:numel(a1)
condition2(j)=det([a1(j) a0;a3(j) a2(j)])>0;
condition3(j) =det([a1(j) a0 0;a3(j) a2(j) a1(j);0 a4(j) a3(j)])>0;
condition4(j)=det([a1(j) a0 0 0;a3(j) a2(j) a1(j) a0;0 a4(j) a3(j) a2(j);0 0 0 a4(j)])>0;
end
output(~(condition1 & condition2 & condition3 & condition4)) = 0;
imshow(output, 'xdata', r, 'ydata', r);
axis on;
Why are condition2 condition3 and condition4 not logical?
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Image Analyst
le 22 Juil 2019
You defined them to be doubles by instantiating them with the zeros() function. Try using false() instead of zeros():
condition2 = false(size(condition1));
condition3 = false(size(condition1));
condition4 = false(size(condition1));
2 commentaires
Image Analyst
le 23 Juil 2019
Please, can you then "Accept this answer" by clicking on that link on this page? Thanks in advance.
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