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86 CHAPTER 7 Reactivity feedbacks
7.10 Explanation of stability using state-space
representation
Express the three differential Eqs. (7.7)–(7.9) in the state-space form as.
dX
(7.10)
dt ¼ AX + Bf
X is the vector of state variables and f is a forcing term. Matrices A (3 3) and
B (3 1) are given by
2 3 2 3
10 K 1
A ¼ 4 2 2 0 5 , B ¼ 0 (7.11)
4 5
0 3 3 0
Note that the coefficient K is redefined as ( K). Now calculate the eigenvalues of
matrix A for different values of ‘gain’ K.
The characteristic polynomial is given by
2
3
g λðÞ5λ +6λ +11λ +6+6K (7.12)
For K 5 5, the eigenvalues are: 20.3928 + 2.5980j, 20.3928–2.5980j, 25.2145.
All the eigenvalues have negative real parts, and the system is stable.
For K 5 10, the eigenvalues are: 3.3166j, 2 3.3166j, 2 6.0.
There are two imaginary eigenvalues, and the real eigenvalue is negative. The
system is marginally stable.
For K 5 15, the eigenvalues are: 0.2779 + 3.8166j, 0.2779–3.8166j, 26.5558.
The complex eigenvalues have positive real parts, and the system is unstable.
• If we consider K as a feedback gain, changing the value of K changes the stability
characteristics of the system. This is true even though the system is inherently a
negative feedback system.
• This is true of all negative feedback systems with the overall system order
greater than or equal to three; this is a well-known fact. As the feedback
gain is increased, the system tends towards instability. The phase lag between any
two state variables does not have to be 180° for the system to become unstable.
Consider an open-loop transfer function of order 3 with stable poles and of the form
1
GsðÞ ¼ 2 (7.13)
3
as + bs + cs + d
If the closed-loop system has negative feedback with a simple gain K (as the feed-
back transfer function), the closed-loop transfer function becomes
1
G c sðÞ ¼ 2 (7.14)
3
as + bs + cs + d + K
As the magnitude of gain K increases, the closed loop poles tend to be less negative
and eventually become positive for a limiting value of gain K.