multiplication to produce a step response in the symbolic code. Now added, with comment. plotting transfer function / time domain version
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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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Réponses (2)
Star Strider
le 17 Avr 2023
Modifié(e) : Star Strider
le 17 Avr 2023
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)
syms s t z
snum = poly2sym(Q,s);
sden = poly2sym(P,s);
FTsym = vpa(snum / sden, 5)
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))
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.
multiplication to produce a step response in the symbolic code. Now added, with comment. .
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Sam Chak
le 17 Avr 2023
Hi @Miller
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)
step(G)
p = pole(G) % two of the poles have positive real parts, thus the system is unstable.
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