Find the transfer function of a differential equation symbolically

As an exercise, I wanted to verify the transfer function for the general solution of a second-order dynamic system with an input and initial conditions—symbolically.

I found a way to get the Laplace domain representation of the differential equation including initial conditions but it's a bit convoluted and maybe there is an easier way:

syms u(t) y(t) dydt(t) t s Y(s) U(s) y0 dydt0 u0 omega_n z K
% Standard form of second-order system
eqn_t = ((1/omega_n^2)*diff(y(t), t, 2) + (2*z/omega_n)*diff(y(t), t) + y) / K == u(t);
% In Laplace domain
eqn_s = subs(laplace(eqn_t), [laplace(y(t), t, s), laplace(u(t), t, s), diff(y(t), t)], [Y(s), U(s), dydt(t)])
% Set initial conditions to zero to get transfer function
eqn_s0 = subs(eqn_s, [y(0), dydt(0)], [0, 0])

This produces:

eqn_s =

-(dydt(0) + s*y(0) - omega_n^2*Y(s) - s^2*Y(s) + 2*omega_n*z*(y(0) - s*Y(s)))/(K*omega_n^2) == U(s)

eqn_s0 =

(omega_n^2*Y(s) + s^2*Y(s) + 2*omega_n*s*z*Y(s))/(K*omega_n^2) == U(s)

Which I think is the correct expression.

Next I want to rearrange this to get an expression for the transfer function (G(s)=Y(s)/U(s)). How do I do that? I tried this:

G(s) = Y(s)/U(s);
solve(eqn_s0,G(s))

But this produces:

ans =

struct with fields:

s: [0×1 sym]

z: [0×1 sym]

What does this mean? Both s and z are 'Empty sym: 0-by-1'.

What I want is:

G(s) =

(K*omega_n^2)/(omega_n^2 + 2*z*omega_n*s + s^2)

I'm sure there is a much easier way of doing all this...

Answer 1

[Bill Tubbs]

@madhan ravi was right. Using the simplify command works in this case:

rhs(eqn_s0) / U(s) / (simplify(lhs(eqn_s0)) / Y(s))

ans =

(K*omega_n^2)/(omega_n^2 + 2*z*omega_n*s + s^2)

Answer 2

[Bill Tubbs]

Maybe the problem is something to do with simplifying the left hand side when Y(s) is a function.

If you substitute Y(s) for a new symbolic variable and dividing by U(s) after solving it seems to work:

syms Ytemp
solve(subs(eqn_s0,Y(s),Ytemp),Ytemp)/U(s)

This produces:

ans =

(K*omega_n^2)/(omega_n^2 + 2*z*omega_n*s + s^2)

ADDITIONAL COMMENT:

Maybe this boils down to a more fundamental question. If you take the following expression, MATLAB doesn't simplify it:

syms a b
(a*b + b)/b

ans =

(b + a*b)/b

(2*b + b)/b

ans =

3

Maybe there a reason for that. Anyone know?

[madhan ravi]

Well see Stars answer above.

[madhan ravi]

You need to expand/simplify that expression , replying to your additional comment.