I just need this one TINY code to execute :( Any idea what the problem with this simple equation is?

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KLETECH MOTORSPORTS 2020-11-8

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编辑： Walter Roberson 2020-11-15

I need to put this Either in another function or as a seperate function. Or the script. It's accepting L, M, and b values, but i am not getting an output value for y.

suggestions?

prompt1 = 'Enter value of L';
prompt2 = 'Enter value of M';
prompt3 = 'Enter value of b';
L = input(prompt1)
M = input(prompt2)
b = input(prompt3)
y = (12*9.81*L)/(4*L^.2)+(b.^2)-(12*L*M)+(12*M.^2);

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Answer 1

Ameer Hamza 2020-11-8

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编辑：Ameer Hamza 2020-11-15

在 MATLAB Online 中打开

You need to print the output. For example using disp()

prompt1 = 'Enter value of L';
prompt2 = 'Enter value of M';
prompt3 = 'Enter value of b';
L = input(prompt1)
M = input(prompt2)
b = input(prompt3)
y = (12*9.81*L)/(4*L^.2)+(b.^2)-(12*L*M)+(12*M.^2);
disp(y)

Or remove the semicolon from the last line

y = (12*9.81*L)/(4*L^.2)+(b.^2)-(12*L*M)+(12*M.^2)
                                                 %^ removed semicolon from here

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KLETECH MOTORSPORTS 2020-11-8

在 MATLAB Online 中打开

I tried that and it worked!!

I have to incorporate it into the rest of my code, and i’ve tried to do it as follows but there seems to be a problem.

Any idea what i can do to correct it?

function xdot = pend_cl(t1,x)     %for linear
 
function z = wsq
 
prompt1 = 'Enter value of L';
prompt2 = 'Enter value of M';
prompt3 = 'Enter value of b';
L = input(prompt1)
M = input(prompt2)
b = input(prompt3)
disp(z) = (12*9.81*L)/((4*L^.2)+(b.^2)-(12*L*M)+(12*M.^2)); 
end 
 
xdot = [x(2); wsq*x(1)];
end
 
function ydot = pend_cn(t2,y)    %for non linear
 
function z = wsq 
 
prompt1 = 'Enter value of L';
prompt2 = 'Enter value of M';
prompt3 = 'Enter value of b';
L = input(prompt1)
M = input(prompt2)
b = input(prompt3)
disp(z) = (12*9.81*L)/((4*L^.2)+(b.^2)-(12*L*M)+(12*M.^2)); 
end
 
ydot = [y(2); wsq*sin(y(1))];
 
end
 
 
 clear all;
clc;
clf;
tic;
 
tspan = 0:0.01:4.5;
a=pi/2;
b=0;
 
 
x0 = [a; b];                %a is initial disp, 
                            %b is initial velocity
[t1,x] = ode45(@pend_cl,tspan,x0);
 
X1 = x(:,1);       %X1 is angular disp for linear
X2 = x(:,2);       %X2 is angular accln for linear
 
y0 = [a ; b];
 
[t2,y] = ode45(@pend_cn,tspan,y0);
 
Y1 = y(:,1);       %Y1 is angular disp for non linear                
Y2 = y(:,2);       %Y2 is angular accln for non linear   
 
figure(1);
subplot(2,2,1);
plot(t1,X1); 
xlabel('Time (s)');
ylabel('Displacement (rad)');
hold on;
grid on;
plot(t2,Y1);
% legend('Linear','Non Linear');
subplot(2,2,2);
% figure(2);
plot(t1,X2);
xlabel('Time (s)');
ylabel('Velocity (rad/s)');
 
hold on;
grid on;
plot(t2,Y2);
subplot(2,2,3);
plot(X1,X2);
hold on;
plot(Y1,Y2);
xlabel('Displacement (rad)');
ylabel('Velocity (rad/s)');
 
grid on;
 
 
 
toc;

Walter Roberson 2020-11-8

编辑：Walter Roberson 2020-11-15

A scalar is any array that holds exactly one piece of information -- not empty, and not two or more values.

The operators * and ^ and / and \ are defined as being matrix operations:

A*B is algebraic matrix multiplication, inner product of A and B . For A*B to be valid, size(A,2) must be the same as size(B,1) and they cannot have 3 or more dimensions
A^B is matrix power; when B is a positive integer it is like A*A*A*A...*A for a total of B times, where * here is algebraic matrix multiplication (inner product); the same size restrictions exist as for *
A/B is like A * pinv(B) but not exactly the same . It is a least-squared operation suitable for solving linear equations. A\B is like pinv(A)*B

These differ from the .* and .^ and ./ and the (rare) .\ operators, all of which are element-by-element operations. C = A.*B is C(J,K) = A(J,K) .* B(J,K) -- multiplication of corresponding locations; likewise for the other operations mentioned.

The matrix operator * acts like .* if one of the operands is scalar instead of matrice. 3*A is A with each element multiplied by 3. The ^ and / operators act like .^ and ./ if both of the operands are scalar.

You have written (12*9.81*L)/((4*L^.2)+(b.^2)-(12*L*M)+(12*M.^2)) . If L happens to be something that is not a scalar, then almost everything you wrote there will work -- the L.^2 would work. For example [2 5]^2 would fail because it would be [2 5]*[2 5] and size() of the second dimension of the first operand is 2 but size() of the first dimension of the second operand is 1 which does not match. [2 5]^2 is not [4 25]. But [2 5].^2 is [4 25] . You wrote your expression nearly all with the dot operators, so nearly all of it will work if L or M happen to be matrices. But you used / and / between two things that are not scalar is going to give you results that you do not expect. You need ./ in that expression if there is a chance that L is not a scalar. If L and M are both something that might not be scalar then L*M should be L.*M to get your code to work the way you expect.

KLETECH MOTORSPORTS 2020-11-15