Lecture 3: Stability of dynamical systems.
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Index
1
Stability of dynamical systems
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Stability of dynamical systems
Introduction. Stability of dynamical systems
The first property a control systems must have is stability. In closed loop operation,
it exists the risk that the system become unstable even if in open loop systems were
stable.
The simplest definition is the input-output stability, that is for a bounded
input the system has a bounded output, i.e. the gain is always finite.
In the system of the figure, where G (s) and H(s) are linear or nonlinear,
r(t)
+
e(t)
G(s)
y(t)
-
H(s)
the error e is e = r + H(s)y = r + H(s)G (s)e.
The norm error is given by
kek =
kr k
kr k
≤
k1 − G (s)H(s)k
1 − kG (s)kkH(s)k
If kG (s)kkH(s)k < 1 the system gain is finite and therefore the system is
input-output stable.
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Stability of dynamical systems
Introduction
Small gain theorem Consider the system of previous slide figure. A sufficient
condition for the system be input-output stable is that
kG (s)kkH(s)k < 1.
(1.1)
In the case that the system is linear, it could require a little less demanding
condition, it is sufficient that kG (s)H(s)k < 1.
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Stability of dynamical systems
Classical stability analysis methods
In linear systems, stability is determined by the position of the closed-loop poles.
Consider a system whose transfer function in closed loop is
Qm
(s + zi )
Qi=1
M(s) = K Qn
p
2
2
(s
+
p
)
j
j=1
k=1 (s + 2ζk ωk s + ωk )
The impulse response is
y (t) =
k
X
j=1
Aj e
−pj t
+
p
X
k=1
Bk
1
ωk
e −ζk t sin(ωk t)
For being bounded input-bounded output it is needed both the real poles pj ,
and the real parts of the complex poles ζk are negative.
The sufficient condition for a linear SISO system (Single Input-Single Output)
in continuous time, to be stable, is that all of the poles are located in the left
half-plane (negative real part).
This approach is very difficult to apply when the system is not factorized. In
this case we must apply other methods such as Routh.
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Stability of dynamical systems
Routh’s method
Consider a continuous-time system with characteristic equation
p(s) = an s n + an−1 s n−1 + an−2 s n−2 + . . . + a1 s + a0 = 0
(1.2)
As prerequisites we require that:
1
2
All coefficients have the same sign
No coefficient is zero.
The Routh criterion is based on the construction of a table called Routh table which
is constructed in the following way: The first row is formed by the coefficients
an , an−2 , an−4 . . . and the second row of the remaining an−1 , an−3 , an−5 , . . . . Namely
sn
s
n−1
an
an−1
an−2
an−3
an−4
an−5
...
...
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Stability of dynamical systems
Routh’s method
The remaining rows are constructed as
sn
n−1
s
s n−2
s n−3
...
s1
s0
an
an−1
bn−1
cn−1
...
u1
v1
an−2
an−3
bn−3
cn−3
...
u2
an−4
an−5
bn−5
cn−5
...
...
...
...
...
...
(1.3)
wherein the third row is
an
an−2
an−1 an−3
bn−1 = −
an−1
an
an−4
an−1 an−5
bn−3 = −
an−1
=
1
(an−1 an−2 − an an−3 )
an−1
=
1
(an−1 an−4 − an an−5 )
an−1
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Stability of dynamical systems
Routh’s method
Analogously, the fourth row is
an−1 an−3
bn−1 bn−3
cn−1 = −
bn−1
an−1 an−5
bn−1 bn−5
cn−3 = −
bn−1
and so on down to zero degree.
=
1
(bn−1 an−3 − an−1 bn−3 )
bn−1
=
1
(bn−1 an−5 − an−1 bn−5 )
bn−1
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Stability of dynamical systems
Routh’s method
In developing the table, we can multiply or divide a row by a positive constant
without the result is changed (this is useful to simplify the table).
The Routh criterion states that the number of roots in the right half-plane
coincides with the number of sign changes of the coefficients of the first
column.
That is, the system will be stable if all the signs in the first column are equal.
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Stability of dynamical systems
Special cases
There are a number of remarkable special cases:
There is a zero in the first column.
The value 0 is replaced by a very small positive number ε and the development
of the table continues. To count the number of sign changes in the first
column is necessary to consider that ε goes to zero.
There is a row of zeros. ( Poles symmetrically placed about the origin of the
complex plane).
The coefficients of the row above the row of zeros are considered. With these
coefficients the auxiliary equation is constructed, which will be the grade that
indicates the power of s respectively. The row of zeros is replaced by the
coefficients of the derivative and the method continues.
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Stability of dynamical systems
Example
Normal case
Normal case.
Consider the system with characteristic equation
a0 s 3 + a1 s 2 + a2 s + a3 = 0
with a0 , a1 , a2 y a3 positives.
The corresponding Routh table is
s3
a0
a2
s2
a1
a3
a1 a2 −a0 a3
s1
a1
s0
a3
For all coefficients in the first column be positive, shall be verified that
a1 a2 − a0 a3
>0
a1
Therefore, the system will be stable when a1 a2 > a0 a3 .
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Stability of dynamical systems
Example
Case with a zero in the first column
Case with a zero in the first column.
Consider the system with characteristic equation
p(s) = s 4 + 2s 3 + 4s 2 + 8s + 5
The corresponding Routh’s table is
s4
1
4 5
s3
2
8
s2 0 ≈ ε 5
8ε−10
s1
ε
0
s
5
where the zero value of the first column by the small positive number is
replaced ε.
When ε → 0, sign 8ε−10
= sign −10
= −1. Since there are two sign
ε
ε
changes in the first column, there are two poles in the right half, and therefore
the system is unstable.
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Stability of dynamical systems
Example
Case with a zeros row
Case with a zeros row.
Consider a system with characteristic equation
p(s) = s 4 + 2s 3 + 11s 2 + 18s + 18 = 0.
Solution:
This system has its roots located symmetrically respect to the origin of the
plane s:
±σ ± jω.
Routh’s table is
s4
s3
s2
s1
s0
1
2
2
0→4
18
11
18
18
0→0
18
The auxiliary equation associated with the third row is A(s) = 2s 2 + 18 and its
derivative is dA(s)
= 4s + 0 replaces the old values in third row. Since all the
ds
coefficients of the first column are positive, the system is stable.
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Stability of dynamical systems
Example
Consider the system
r(t) +
-
e(t)
1
s2+2s+5
K
y(t)
1
s+3
Is desired to find the stability of the system as a function of the parameter K .
Solution:
The transfer function in closed loop is
K
Gc (s) =
KG (s)
K (s + 3)
s 2 +2s+5
=
=
K
1
2 + 2s + 5) + K
1 + KG (s)H(s)
(s
+
3)(s
1 + s 2 +2s+5
s+3
Equating the denominator to zero the characteristic equation is obtained,
whose roots are the poles of the closed loop system:
p(s) = (s + 3)(s 2 + 2s + 5) + K = s 3 + 5s 2 + 11s + 15 + K = 0
The precondition requires that all the coefficients are of the same sign:
15 + K > 0 ⇒ K > −15
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Stability of dynamical systems
Example
Routh’s table is
s3
1
11
s2
5
15 + K
40−K
s1
5
s 0 15 + K
For the system to be stable it is necessary that all the coefficients of the first
column be greater than zero:
40 − K
> 0 ⇒ 40 − K > 0 ⇒ K < 40
5
Putting together the two conditions, we find that the system is stable for
−15 < K < 40.
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Stability of dynamical systems
End of Lecture 3
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