why am I getting matrix dimension error..?
显示 更早的评论
when i tried to run this code , its displaying the "matrix dimension must agree". I want the plot for phase angle at various frequencies. and when i used for loop i got the result only for last iteration and plot shows only last iterated point. can i get a help on this.
clc;
clear;
close all;
d=0.23;
theta=40;
f=1:18;
lamda(f)=3*0.1/f;
phaseangle(lamda)=(2*pi/(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
%%%%%
clc;
clear;
close all;
d=0.23;
theta=40;
for f=1:1:18;
lamda=3*0.1/f;
phaseangle=(2*pi/(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
end
采纳的回答
Yongjian Feng
2021-7-29
There are several errors typos.
clc;
clear;
close all;
d=0.23;
theta=40;
f=1:18;
lambda =3*0.1./f;
phaseangle=(2*pi./(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
%%%%%
clc;
clear;
close all;
d=0.23;
theta=40;
for f=1:1:18
lambda=3*0.1/f;
phaseangle=(2*pi./(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
end
3 个评论
Sai Monika Ananthoju
2021-7-29
Yongjian Feng
2021-7-29
You actually want this, right?
clc;
clear;
close all;
d=0.23;
theta=40;
f=1:18;
lambda=3*0.1./f;
phaseangle=(2*pi./(lambda))*d*sin(theta);
plot(f,phaseangle,'-o')
xlabel('frequency(Ghz)')
ylabel('phase angle')
Sai Monika Ananthoju
2021-7-29
编辑：Sai Monika Ananthoju
2021-7-30
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