Vectors must be the same lengths; plotting

Question

alexr 2011-5-2

2
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/6670-vectors-must-be-the-same-lengths-plotting

评论： Komal Kumawat 2020-10-2

I have googled this to death, the common error was people looping arrays. I just have 2 simple sets and cant see where the error is.

simulated=[0.0000000 4.3830000 4.9470000 2.0180000 0.1761000 0.0162800 4.3650000 4.9540000 2.0170000 0.1760000 0.0162700 ]
actual=   [0.0750000 4.0750000 4.7630000 3.5130000 0.5125000 0.1375000 2.6380000 4.5750000 4.8880000 1.5130000 0.3250000 ]
x = 0:0.0002:0.01;
plot(x, simulated, x, actual);
legend('Simulated', 'Actual')
ylabel('Current (mA)')
xlabel('Time ( mS )')
title('Current across R2 as a function of VS','FontSize',12)

4 个评论
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alexr 2011-5-3

Great thanks, I thought only the number of elements in the simulated length had to match the actual. It works now

waqas muhammad 2018-1-16

can you please tell me how did you solve the problem of number of elements and how you made it equal?

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Answer 1

Smith 2016-10-30

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/6670-vectors-must-be-the-same-lengths-plotting#answer_241327

在 MATLAB Online 中打开

You just need to modify variable x to match the size of variable simulated or variable actual.

simulated=[0.0000000 4.3830000 4.9470000 2.0180000 0.1761000 0.0162800 4.3650000 4.9540000 2.0170000 0.1760000 0.0162700 ]
actual=   [0.0750000 4.0750000 4.7630000 3.5130000 0.5125000 0.1375000 2.6380000 4.5750000 4.8880000 1.5130000 0.3250000 ]
x = linspace(0,0.01,length(actual));
plot(x, simulated, x, actual);
legend('Simulated', 'Actual')
ylabel('Current (mA)')
xlabel('Time ( mS )')
title('Current across R2 as a function of VS','FontSize',12)

3 个评论
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Jan 2016-11-14

@assiya malik: Please do not post a new question as a comment to an answer, but open a new thread. Note that the problem is hidding inside "calculated by Runge Kutta 4 method" and not shown here. It does not matter how you define the variables before you redefine them by any computations.

Komal Kumawat 2020-10-2

在 MATLAB Online 中打开

%*********************************************************% %Defining the frequency vector and the mass matrix, %damping matrix, the stiffness matrix and the amplitude of %the excitation force. %*********************************************************% f=linspace(0,0.7,50); m=[1 0;0 2]; k=[2 -1;-1 1]; c=[0.002 -0.001;-0.001 0.001]; fi=[1;1]; %*********************************************************% %Calculating the amplitude and the phase for each frequency %defined by the frequency vector. %*********************************************************% for i = 1:50 omega(i)=2*pi*f(i); %omega in terms of frequency omega2(i)=omega(i)*omega(i); % squaring omega a11=-omega2(i)*m+k; % representing the left hand… a12=omega(i)*c; % matrix of the single matrix… a21=-omega(i)*c; % equation a22=-omega2(i)*m+k; a=[a11 a12;a21 a22]; b=inv(a); c1=[0;0;fi]; d(1,i)=b(1,:)*c1; d(2,i)=b(2,:)*c1; d(3,i)=b(3,:)*c1; d(4,i)=b(4,:)*c1; x(1,i)=sqrt(abs(d(1,i))^2+abs(d(3,i))^2); x(2,i)=sqrt(abs(d(2,i))^2+abs(d(4,i))^2); p(1,i)=atan(d(1,i)/d(3,i))*180/pi; if p(1,i)<0 % to check whether the angle is negative or not. p(1,i)=180+p(1,i); else

 p(1,i)=p(1,i); 
 end 
 p(2,i)=atan(d(2,i)/d(4,i))*180/pi; 
 if p(2,i)<0 
 if d(4,i)<0 
 p(2,i) = -180 + p(2,i) 
 else
 p(2,i)=180+p(2,i); 
 end 
 else 
 p(2,i)=p(2,i); 
 end
end
figure(1) 
plot(f,x(1,:));grid 
xlabel(‘Frequency’) 
ylabel(‘Amplitude of Mass 1’) 
figure(2) 
plot(f,x(2,:));grid 
xlabel(‘Frequency’) 
ylabel(‘Amplitude of Mass 2’) 
figure(3) 
plot(f,p(1,:));grid 
xlabel(‘Frequency’) 
ylabel(‘Phase of Mass 1’) 
figure(4) 
plot(f,p(2,:));grid 
xlabel(‘Frequency’) 
ylabel(‘Phase of Mass 2’)