Chancing variable in laplace transformation

% mass of person and parachute
Mj = 70; % kg
Mp = 25; % kg

% friction cof.
Bj = 13; % N/(m/s)
Bp = 140; % N/(m/s)
syms Xp(t) Xj(t) t s XP(s) XJ(s)

% elastic cons.
k = 400; % N/m

% gravity
g = 9.81; % m/s^2
D1XP=diff(Xp,t);
D2XP=diff(D1XP);
D1XJ=diff(Xj,t);
D2XJ=diff(D1XJ);

eq1=Mp*diff(Xp,t,2)-Mp*g-k*(Xj-Xp)+Bp*diff(Xp,t)==0;
eq2=Mj*diff(Xj,t,2)-Mj*g+k*(Xj-Xp)+Bj*diff(Xj,t)==0;
f1=laplace(eq1,s);
f1=subs(f1,{laplace(Xp(t),t,s),Xp(0),D1XP(0)},{Xp(s),0,0})
f2=laplace(eq2,s);
f2=subs(f2,{laplace(Xj(t),t,s),Xj(0),D1XJ(0)},{Xj(s),0,0});

ı am traying to change laplace(Xj(t), t, s) in f1 as a some funciton of Xp ı can derive an expression for Xj in terms of Xp from eq2 but how can ı ımplement in f1 ?

Answer 1

[Vinay]

Hii @YUNUS EMRE,

The laplace(Xj(t), t, s) value from f1 can be expressed in terms of Xp(s) by isolating the value using the 'isolate' command as showf1_sub = subs(f1, XJ(s), rhs(XJ_expr));

disp(f1_sub)n below

XJ_expr1 = isolate(f1, laplace(Xj(t), t, s));
disp(XJ_expr1)

The expression for the Xj(s) in terms of laplace(Xp(t), t, s) can be solved using the 'isolate' in the equation f2 as

XJ_expr = isolate(f2, XJ(s));
disp(XJ_expr)

The value of XJ(s) from f2 is substituted in place of laplace(Xj(t), t, s) in f1 using the 'subs' command as

f1_sub = subs(f1, laplace(Xj(t), t, s), rhs(XJ_expr));
disp(f1_sub)

The code below shows the implementation of changing the variable

% Define constants
Mj = 70; % kg
Mp = 25; % kg
Bj = 13; % N/(m/s)
Bp = 140; % N/(m/s)
k = 400; % N/m
g = 9.81; % m/s^2

% Define symbolic variables
syms Xp(t) Xj(t) t s XP(s) XJ(s)

% Define differential equations
eq1 = Mp*diff(Xp, t, 2) - Mp*g - k*(Xj - Xp) + Bp*diff(Xp, t) == 0;
eq2 = Mj*diff(Xj, t, 2) - Mj*g + k*(Xj - Xp) + Bj*diff(Xj, t) == 0;

% Take Laplace transforms
f1 = laplace(eq1, t, s);
f1 = subs(f1, {laplace(Xp(t), t, s), Xp(0), subs(diff(Xp, t), t, 0)}, {XP(s), 0, 0});
f2 = laplace(eq2, t, s);
f2 = subs(f2, {laplace(Xj(t), t, s), Xj(0), subs(diff(Xj, t), t, 0)}, {XJ(s), 0, 0});

% Isolate XJ(s) in terms of XP(s) from f2
XJ_expr = isolate(f2, XJ(s));
disp(XJ_expr)


% Substitute the isolated XJ(s) from f2 in place of laplace(Xj(t), t, s)
f1_sub = subs(f1, laplace(Xj(t), t, s), rhs(XJ_expr));
disp(f1_sub)
% Simplify the resulting equation
f1_simplified = simplify(f1_sub);

% Display the simplified equation
disp('The simplified equation in terms of XP(s) is:');
disp(f1_simplified);

Kindly refer to the below documentations for more details:

'Isolate' : https://in.mathworks.com/help/symbolic/sym.isolate.html

'subs' : https://in.mathworks.com/help/symbolic/sym.subs.html

I hope this helps!