Problem with syms and symfun

Hi guys,

I had my first code run which gives me the equations of motion.

eqx1 = M1*y1dd - F + K*(s0 + y1 - y2 - r*sin(ph))

eqx2 = F + M2*y2dd - K*(s0 + y1 - y2 - r*sin(ph))

eqx3 = I*phdd + F*r*cos(ph) - K*r*cos(ph)*(s0 + y1 - y2 - r*sin(ph))

where all the variables were defined as symm

syms M1 M2 I K;
syms t y1 y1d y1dd y2 y2d y2dd ph phd phdd;

Now, I tried replacing the y1d, y1dd, y2d, y2dd etc with diff(y1,t), diff(y1,t,2) respectively.

fx1=subs(eqx1,{y1,y1d,y1dd,y2,y2d,y2dd,ph,phd,phdd},{y1(t),diff(y1,t),diff(y1,t,2),y2(t),diff(y2,t),diff(y2,t,2),ph(t),diff(ph,t),diff(ph,t,2)})==0

Output:

 fx1=M1*D(D(y1))(t) - F + K*(s0 + y1 - y2 - r*sin(ph)) == 0

now when I run

diff(fx1), i get 

>> diff(fx1)
ans =
K == 0

>> diff(fx1,t)

ans =

M1*diff(D(D(y1))(t), t) == 0

Now, I tried directly defining f (same as fx1)

>> f = M1*diff(y1,t,2) - F + K*(s0 + y1(t) - y2(t) - r*sin(ph(t))) == 0

f(t) =

M1*D(D(y1))(t) - F + K*(s0 + y1(t) - y2(t) - r*sin(ph(t))) == 0

>> diff(f)

ans(t) =

M1*D(D(D(y1)))(t) - K*(D(y2)(t) - D(y1)(t) + r*D(ph)(t)*cos(ph(t))) == 0

how do I get the above for fx1.

How do I come from fx1 to f

Answer 1

[Walter Roberson]

syms fx1(t) 
fx1(t)  = subs(....)

[yashwant kolluru]

i already tried that, unfortunately it didn't work!

The below is the output I got

fx1(t)=subs(eqx1,{y1,y1d,y1dd,y2,y2d,y2dd,ph,phd,phdd},{y1(t),diff(y1,t),diff(y1,t,2),y2(t),diff(y2,t),diff(y2,t,2),ph(t),diff(ph,t),diff(ph,t,2)})==0

fx1(t) =

M1*D(D(y1))(t) - F + K*(s0 + y1 - y2 - r*sin(ph)) == 0

>> diff(fx1)

ans(t) =

M1*D(D(D(y1)))(t) == 0

>> diff(fx1,t)

ans(t) =

M1*D(D(D(y1)))(t) == 0

[Walter Roberson]

Your posted code is not functional in R2016b

syms M1 M2 I K;
syms t y1 y1d y1dd y2 y2d y2dd ph phd phdd;

%you did not declare these
syms F s0 r

%you cannot have y1(t) and ph(t) on the values to substitute in unless they have been declared as functions
syms y1(t) y2(t) ph(t)

eqx1 = M1*y1dd - F + K*(s0 + y1 - y2 - r*sin(ph));
eqx2 = F + M2*y2dd - K*(s0 + y1 - y2 - r*sin(ph));
eqx3 = I*phdd + F*r*cos(ph) - K*r*cos(ph)*(s0 + y1 - y2 - r*sin(ph));

%for convenience, they make the code easier to debug
source = {y1(t), diff(y1,t), diff(y1,t,2), y2(t), diff(y2,t), diff(y2,t,2), ph(t), diff(ph,t), diff(ph,t,2)};
dest = {y1, y1d, y1dd, y2, y2d, y2dd, ph, phd, phdd};

fx1 = subs(eqx1,dest,source)==0;

Be sure to clear your workspace before executing these; existing definitions of syms y1 y2 ph can mess up what happens later.

[yashwant kolluru]

I did replacing y1 to y1(t) and others also.But that also doesn't work.

The below is the output.

eqx1 =

M1*y1dd - F + K*(s0 + y1 - y2 - r*sin(ph))

eqx2 =

F + M2*y2dd - K*(s0 + y1 - y2 - r*sin(ph))

eqx3 =

I*phdd + F*r*cos(ph) - K*r*cos(ph)*(s0 + y1 - y2 - r*sin(ph))

>> syms y1(t) y2(t) ph(t)
>> source = {y1(t), diff(y1,t), diff(y1,t,2), y2(t), diff(y2,t), diff(y2,t,2), ph(t), diff(ph,t), diff(ph,t,2)};
dest = {y1, y1d, y1dd, y2, y2d, y2dd, ph, phd, phdd};
>> fx1 = subs(eqx1,dest,source)==0

fx1 =

M1*D(D(y1))(t) - F + K*(s0 + y1 - y2 - r*sin(ph)) == 0

>> diff(fx1)

ans =

K == 0

>>

Note : I am using matlab 2012b

[Walter Roberson]

We need your exact code, all the syms declarations, the actual assignments to eqx1, everything.

[Walter Roberson]

I just installed R2012b, and copied and pasted the code I posted above, and it worked fine.

>> fx1 = subs(eqx1,dest,source)==0

fx1(t) =

M1*D(D(y1))(t) - F + K*(s0 + y1(t) - y2(t) - r*sin(ph(t))) == 0