Б. О. Джолдошев а из Института автоматики и информационных технологий нан кр, г. Бишкек; «Cинтез кибернетических автоматических систем с использованием эталонной модели»
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| By comparing the stability conditions given by (5) and (6) we find that the signs of the expressions in the second inequalities are opposite. Also we can see that the signs of expressions in the first inequalities can be opposite due to squares of the parameters k1 and k3 if we properly set their values.
Let us suppose that parameter T1 can be perturbed but remains positive. If we set k2 and k3 both negative and  then the value of parameter T2 is irrelevant. It can assume any values both positive and negative (except zero), and the system given by (2) remains stable. If T2 is positive then the system converges to the equilibrium point (3) (becomes stable). Likewise, if T2 is negative then the system converges to the equilibrium point (4) which appears (becomes stable). At this moment the equilibrium point (3) becomes unstable (disappears).
Let us suppose that T2 is positive, or can be perturbed staying positive. So if we can set the k2 and k3 both negative and  then it does not matter what value (negative or positive) the parameter T1 would be (except zero), in any case the system (2) will be stable. If T1 is positive then equilibrium point (3) appears (becomes stable) and equilibrium point (4) becomes unstable (disappears) and vice versa, if T1 is negative then equilibrium point (4) appears (become stable) and equilibrium point (3) becomes unstable (disappears).
Results of MatLab simulation for the first and second cases are presented in Fig. 2 and 3 respectively. In both cases we see how phase trajectories converge to equilibrium points  and .
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