Find the transfer function of a differential equation symbolically
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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...
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Bill Tubbs
2020-6-6
编辑:Bill Tubbs
2020-6-6
2 个评论
madhan ravi
2020-6-6
编辑:madhan ravi
2020-6-6
You need to expand/simplify that expression , replying to your additional comment.
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