Why are condition2 condition3 and condition4 not logical?
1 view (last 30 days)
Show older comments
zhenyu zeng
on 22 Jul 2019
Commented: Image Analyst
on 23 Jul 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?
0 Comments
Accepted Answer
Image Analyst
on 22 Jul 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 Comments
Image Analyst
on 23 Jul 2019
Please, can you then "Accept this answer" by clicking on that link on this page? Thanks in advance.
More Answers (0)
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!