Different results with same code??

Please consider 2 scripts below, this supposed to be output the same results of D but it doesn't

script 1:

m_s = 2000;                        %(kg)   Sprung mass 
m_u = 200;                         %(kg)   Un-sprung mass
k_s = 200000;                      %(N/m)  Suspension stiffness
k_u = 870000;                      %(N/m)  Tire stiffness
c_s = 6000;                        %(Ns/m) Damping coefficient

M = [m_u 0;0 m_s];
K = [k_s+k_u -k_s; -k_s k_s];
A = M\K;

[V,D]  = eig(A); 
D

script 2:

m_s = 2000;                        %(kg)   Sprung mass 
m_u = 200;                         %(kg)   Un-sprung mass
k_u = 870000;                      %(N/m)  Tire stiffness
c_s = 6000;                        %(Ns/m) Damping coefficient
syms  k_s

M = [m_u 0;0 m_s];
K = [k_s+k_u -k_s; -k_s k_s];
A = M\K;
[V,D]   = eig(A);

k_s = 200000;
D=double(subs(D))

THANK YOU!

[ali hassan]

are your inputs complete??

i dont think so

[KSSV]

Give the complete code which you have tried.....the inputs which lead to the given output is not shown. Without this how you expect us to help you?

[Tran Gia Minh]

Sorry for that, INPUT is another file which contains input variables

I have just updated the scripts, if you have time please take a look.

[ali hassan]

try this. hope it will work

https://www.mathworks.com/matlabcentral/answers/664433-different-results-with-same-code#comment_1164073

%%script 1:
m_s = 2000;                        %(kg)   Sprung mass 
m_u = 200;                         %(kg)   Un-sprung mass
k_s = 200000;                      %(N/m)  Suspension stiffness
k_u = 870000;                      %(N/m)  Tire stiffness
c_s = 6000;                        %(Ns/m) Damping coefficient

M = [m_u 0;0 m_s];
K = [k_s+k_u -k_s; -k_s k_s];
A = M\K;

[V,D]  = eig(A); 
D



%% %%script 2:
m_s = 2000;                        %(kg)   Sprung mass 
m_u = 200;                         %(kg)   Un-sprung mass
k_u = 870000;                      %(N/m)  Tire stiffness
c_s = 6000;                        %(Ns/m) Damping coefficient
syms  k_s

M = [m_u 0;0 m_s];
K = [k_s+k_u -k_s; -k_s k_s];
A = M\K;
[V,D]   = eig(A);

k_s = 200000;
% D=double(subs(D))
double(subs(D))

[ali hassan]

plzzz reply if it worked??

i am sure it worked.

you were actually making a mistake using sub.firstly, try to keep code simple. u dont need to use symbolic variables when you are going to give the value later coz there is no reason to add lines.

hope answers the question

BEST REGARDS

[Tran Gia Minh]

Unfortunately, they still give different result,

Thank you for your patient Ali Hassan!

I try symbolic variables for a specific purpose of my own.

[Tran Gia Minh]

in script 1 you provided, result of D is:

D =

1.0e+03 *

5.3690 0

0 0.0810

and script2:

D =

1.0e+03 *

0.0810 0

0 5.3690

[ali hassan]

https://www.mathworks.com/matlabcentral/answers/664433-different-results-with-same-code#comment_1164123

brother these are eigen values and you have got two answers and the position does'nt matter in my reckoning.

did u get it or i need to explain more??

if u want i can

[ali hassan]

actually i hope you know but just to remind you.

actually D is a scalor value and when it is multipled with the eigen vector it satisfies the equation

w’A = Dw’

here D is a scalor value and you have got two different values which means when you will multiply your two values one by one with the eigen vector, you must get the left side of equation(w’A) and it is true in your case. so dont get puzzled with the sequence.

Answer 1

[VBBV]

D=double(subs(D,k_s))

Use the variable k_s as argument, I get the same results now

 m_s = 2000;                        %(kg)   Sprung mass 
m_u = 200;                         %(kg)   Un-sprung mass
k_s = 200000;                      %(N/m)  Suspension stiffness
k_u = 870000;                      %(N/m)  Tire stiffness
c_s = 6000;                        %(Ns/m) Damping coefficient

M = [m_u 0;0 m_s];
K = [k_s+k_u  -k_s; -k_s  k_s];
A = M\K;

[V,D]  = eig(A); 
D
D =
   1.0e+03 *
    5.3690         0
         0    0.0810      

clear
m_s = 2000;                        %(kg)   Sprung mass 
m_u = 200;                         %(kg)   Un-sprung mass
k_u = 870000;                      %(N/m)  Tire stiffness
c_s = 6000;                        %(Ns/m) Damping coefficient
syms  k_s

M = [m_u 0;0 m_s];
K = [k_s+k_u  -k_s; -k_s  k_s];
A = M\K;
[V,D]   = eig(A);

k_s = 200000;
D=double(subs(D,k_s))
D =
   1.0e+03 *
    5.3690         0 
      0    0.0810

[VBBV]

Use clear statment in the begnning of 2nd script

Answer 2

[VBBV]

It depends on values for input variables inside the M and K matrices If you use different values for each of script the resulting D will not be same

[ali hassan]

***if we use same values for each of script the resulting D will not be same.

right???

Answer 3

[Steven Lord]

m_s = 2000;                        %(kg)   Sprung mass 
m_u = 200;                         %(kg)   Un-sprung mass
k_s = 200000;                      %(N/m)  Suspension stiffness
k_u = 870000;                      %(N/m)  Tire stiffness
c_s = 6000;                        %(Ns/m) Damping coefficient

M = [m_u 0;0 m_s];
K = [k_s+k_u -k_s; -k_s k_s];
A = M\K;

[Vdbl,Ddbl]  = eig(A); 
Ddbl;

syms  k_s

M = [m_u 0;0 m_s];
K = [k_s+k_u -k_s; -k_s k_s];
A = M\K;
[Vsym,Dsym]   = eig(A);

k_s = 200000;
Dsym2=double(subs(Dsym));

isequal(sort(diag(Ddbl)), sort(diag(Dsym2)))

ans = logical
   1

Ddbl and Dsym2 do contain the same diagonal elements, but the order of those elements along the diagonal is not the same. If you swap the columns of Vsym accordingly you'll see that each eigenvector in Vdbl is a non-zero scalar multiple of the corresponding eigenvector in Vsym. If v is an eigenvector of A with eigenvalue d, k*d is also an eigenvector of A with eigenvalue d for any non-zero scalar k.

format longg
Vdbl./fliplr(double(subs(Vsym)))

ans = 2×2

-0.0189755917710498 0.982462249755453 -0.0189755917710498 0.982462249755453

●

So same answers, just in a different order in the output.