Unable to isolate variable from expression using 'solve'.
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Pedro Ricardo Garcia de Oliveira
on 19 Jun 2015
Commented: Pedro Ricardo Garcia de Oliveira
on 19 Jun 2015
Hello, I am trying to reproduce all steps to create a transfer function from the beginning.
As seen on the code bellow, I managed to get to the equation F(s), then I isolated Xo using solve.
syms xi(t) xo(t) t B M K s Xo Xi;
xo2 = diff(xo(t),2);
xo1 = diff(xo(t),1);
xi2 = diff(xi(t),2);
xi1 = diff(xi(t),1);
f = xo2 + (B/M)*xo1 + (K/M)*xo -((B/M)*xi1 + (K/M)*xi);
F = laplace(f,t,s);
F = subs(F,{'xo(0)','D(xo)(0)','xi(0)','laplace(xo(t),t,s)','laplace(xi(t),t,s)'},{0,0,0,Xo,Xi})==0;
FXo = solve(F,Xo)==Xo;
pretty(FXo)
Which results in:
K Xi + B Xi s
-------------- == Xo
2
M s + B s + K
In order to create a transfer function I need Xo/Xi , so I used solve again, but this time I used:
solve(FXo,Xo/Xi)
This code results in:
ans =
Xi: [0x1 sym]
s: [0x1 sym]
Then I modified the 2 last lines of the code to:
FXo = solve(F,Xo)==1
pretty(FXo)
solve(FXo,1/Xi)
But it resulted the same. I also tried to use only one solve with Xo/Xi as parameter but it didn't work. Thanks in advance!
2 Comments
Walter Roberson
on 19 Jun 2015
In your
f = xo2 + (B/M)*xo1 + (K/M)*xo -((B/M)*xi1 + (K/M)*xi);
it is better to use
f = xo2 + (B/M)*xo1 + (K/M)*xo(t) -((B/M)*xi1 + (K/M)*xi(t));
Saves problems when trying to cross-verify with other packages.
Accepted Answer
Walter Roberson
on 19 Jun 2015
You appear to be attempting to solve for an expression Xo/Xi instead of solving for a variable. When you using the symbolic toolbox solve() function, then any expressions you give that are not pure variables are taken as expressions that must be solved for equality with 0.
In order to get Xo/Xi what you should be doing is dividing both sides of FXo by Xi. Just divide, not solve().
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