Why don't i get a vector for phi1?

1 次查看(过去 30 天)
t = pi/2:0.01:2*pi+(pi/2);
phi1 = (a^2+a^2+d^2+2*a*d*cos(pi-t)-d^2)/(2*b*sqrt(a^2+d^2+2*a*d*cos(pi-t)));

采纳的回答

Mischa Kim
Mischa Kim 2018-5-24
编辑:Mischa Kim 2018-5-24
All constants need to be defined. Then use:
t = pi/2:0.01:2*pi+(pi/2);
phi1 = (a^2+a^2+d^2+2*a*d*cos(pi-t)-d^2)./(2*b*sqrt(a^2+d^2+2*a*d*cos(pi-t)));
plot(t, phi1)
t is a vector, therefore, you need to use the ./ operator.

更多回答(0 个)

标签

产品


版本

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!