plotting transfer function / time domain version

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Miller
Miller 2023-4-17
回答： Sam Chak 2023-4-17
Hi guys,
I am trying to plot a Laplace domain function I have, however when I use fplot or step() the graph shows nothing at all. How do I fix this?
Code Below:
clear
clc
P = [8.925*10^-20 11.925*10^-15 17.995*10^-10 6.625 10^5];
poles = roots(P);
poles = poles';
Q = [2.25*10^-20 5.25*10^-15 1.5*10^-5 3*10^-5];
zeros = roots(Q);
zeros = zeros';
FT = tf(zeros,poles);
syms s t z
snum = poly2sym(zeros,s);
sden = poly2sym(poles,s);
step(FT);
FT_time_domain = ilaplace(snum/sden);
FT_time_domain = simplify(FT_time_domain,'Steps',10);
FT_time_domain = collect(FT_time_domain, exp(-t))

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回答（2 个）

Star Strider
Star Strider 2023-4-17
编辑：Star Strider 2023-4-17
Ran in:
I’m assuming here that ‘Q’ is the numerator polynomial and ‘P’ is the denominator poplynomial. (If that is not correct, it is easy to switch them.)
Try something like this —
P = [8.925*10^-20 11.925*10^-15 17.995*10^-10 6.625 10^5];
poles = roots(P);
poles = poles';
Q = [2.25*10^-20 5.25*10^-15 1.5*10^-5 3*10^-5];
zeros = roots(Q);
zeros = zeros';
FT = tf(Q,P)
FT = 2.25e-20 s^3 + 5.25e-15 s^2 + 1.5e-05 s + 3e-05 -------------------------------------------------------------- 8.925e-20 s^4 + 1.193e-14 s^3 + 1.8e-09 s^2 + 6.625 s + 100000 Continuous-time transfer function.
syms s t z
snum = poly2sym(Q,s);
sden = poly2sym(P,s);
FTsym = vpa(snum / sden, 5)
FTsym = 
step(FT);
FT_time_domain = ilaplace(snum/(sden*s)); % Step Response Requires: snum/sden*(1/s)
FT_time_domain = simplify(FT_time_domain,'Steps',10);
FT_time_domain = collect(FT_time_domain, exp(-t))
FT_time_domain = 
figure
fplot(FT_time_domain, [0 1.5E-5])
xlabel('Time (s)')
ylabel('Amplitude (units)')
The system appears to be unstable.
EDIT — (17 Apr 2023 at 13:02)
Initially forgot the multiplication to produce a step response in the symbolic code. Now added, with comment.
.
Sam Chak
Sam Chak 2023-4-17
Ran in:
Hi @Miller
If you want to use the step(sys) function, then the sys must be a SISO or MIMO Linear System Model.
In your case, it can descibed by a transfer function model in the s-domain.
Q = [2.250e-20 5.250e-15 1.500e-5 3e-5];
P = [8.925e-20 11.925e-15 17.995e-10 6.625 10^5];
G = tf(Q, P)
G = 2.25e-20 s^3 + 5.25e-15 s^2 + 1.5e-05 s + 3e-05 -------------------------------------------------------------- 8.925e-20 s^4 + 1.193e-14 s^3 + 1.8e-09 s^2 + 6.625 s + 100000 Continuous-time transfer function.
step(G)
p = pole(G) % two of the poles have positive real parts, thus the system is unstable.
p =
1.0e+06 * -4.2411 + 0.0000i 2.0613 + 3.6406i 2.0613 - 3.6406i -0.0151 + 0.0000i

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