Stability of Transfer Function
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Hello,
I can't understand how to define the stability of a Transfer Function (Stable, Unstable or Marginally Stable)
- f(t) = 0, as t(s) = inf, then the system is Stable.
- f(t) = inf, as t(s) = inf, then the system is UnStable.
- f(t) ~= 0 or inf , as t(s) = inf, then the system is Marginally Stable.
I don't want to use the "isstable", because it only tells you if it is stable or not.
It doesn't inform you in case if it is Marginally Stable.
%%
clc;
clear all;
close all;
%% 3
%% 3.1
numA = [-7 2];
denA = [4 4];
GA = tf(numA,denA);
%% 3.2
numB = [4 -7];
denB = [2 2];
GB = tf(numB,denB);
%% 3.3
numC = [1 1];
denC = [4 -7 2];
GC = tf(numC,denC);
%% 3.4
numD = [1 1];
denD = [4 7 -2];
GD = tf(numD,denD);
%% 5
%% 5.1
numA = [1];
denA = [1 3 2 0];
GA = tf(numA,denA);
%% 5.2
numB = [1];
denB = [1 1 -2 0];
GB = tf(numB,denB);
%% 5.3
numC = [1];
denC = [1 2 0 0 0];
GC = tf(numC,denC);
%% 5.4
numD = [1];
denD = [1 4 3];
GD = tf(numD,denD);
%% 6
%% 6.1
numA = [1];
denA = [1 4 0 0 0];
GA = tf(numA,denA);
%% 6.2
numB = [1 2];
denB = [1 2 8];
GB = tf(numB,denB);
%% 6.3
numC = [1 -1 -2];
denC = [1 7 13];
GC = tf(numC,denC);
%% 6.4
numD = [1];
denD = [1 -1 -2 0];
GD = tf(numD,denD);
Thanks
6 commentaires
David Goodmanson
le 18 Mar 2021
I assumed that in the question, G(s) is the laplace transform of f(t). And I assumed that the question meant: based on your knowledge of G(s), what happens to f(t) as t --> inf? I bypassed the notation t(s) because I did not know what that means.
Paul
le 18 Mar 2021
Modifié(e) : Paul
le 18 Mar 2021
Ah, now I see our disconnect. I thought the OP wants to know "how to define the stability of a Transfer Function." The question then goes on to state that stability of G(s) is defined by the properties of f(t) (which I too assume is the inverse Laplace transform of G(s), which by all rights should be called g(t)).
So I assumed the question is to determine (not define) the external stability of the system represented by the transfer function G(s) from the properties of G(s) s.t. the properties of G(s) are consistent with the stability definitions as given by the three criteria on f(t) (which aren't quite right either). In this light, I don't believe the FVT applies, because the FVT assumes certain conditions on G(s), and verifyhing whether or not those assumptions hold is basically (with one exception) the answer to the question (as I interpreted it).
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