Lecture abstract
EE C128 / ME C134 – Feedback Control Systems
Lecture – Chapter 6 – Stability
Alexandre Bayen
Topics covered in this presentation
Department of Electrical Engineering & Computer Science
University of California Berkeley
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Stable, marginally stable, & unstable linear systems
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Relationship between pole locations and stability
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Routh-Hurwitz criterion
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Relationship between stability and eigenvalues
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6 Stability
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6.1 Intro
Chapter outline
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6 Stability
6.1 Introduction
6.2 Routh-Hurwitz criterion
6.3 Routh-Hurwitz criterion: special cases
6.4 Routh-Hurwitz criterion: additional examples
6.5 Stability in state space
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6 Stability
6.1 Introduction
6.2 Routh-Hurwitz criterion
6.3 Routh-Hurwitz criterion: special cases
6.4 Routh-Hurwitz criterion: additional examples
6.5 Stability in state space
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6.1 Intro
Feedback Control Systems
6 Stability
Stability for LTI systems, [1, p. 302]
6.1 Intro
Stability for LTI systems, [1, p. 302]
Total response of a system
c(t) = cforced (t) + cnatural (t)
Stability for LTI systems in terms of pole locations
I Closed-loop TF poles
Stability for LTI systems
I Natural response as t ! 1
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Total response (BIBO)
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Stable: Only in LHP
Unstable: At least 1 in RHP and/or multiplicity greater than 1 on the
imaginary axis
Marginally stable: Only imaginary axis poles of multiplicity 1 and poles
in the LHP
Stable: Every bounded input yields a bounded output
Unstable: Any bounded input yields an unbounded output
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Stable: ! 0
Unstable: Grows without bound
Marginally stable: Neither decays nor grows but remains constant
Marginally stable: Some bounded inputs yield unstable outputs
Stability =) only the forced response remains
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6 Stability
6.2 Routh-Hurwitz criterion
6 Stability
6.2 Routh-Hurwitz criterion
History interlude
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6 Stability
6.1 Introduction
6.2 Routh-Hurwitz criterion
6.3 Routh-Hurwitz criterion: special cases
6.4 Routh-Hurwitz criterion: additional examples
6.5 Stability in state space
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6 Stability
Edward John Routh
I 1831 – 1907
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English mathematician
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1876 – Proposed what became
the Routh-Hurwitz stability
criterion
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6.2 Routh-Hurwitz criterion
6 Stability
History interlude
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6.2 Routh-Hurwitz criterion
Some definitions, [1, p. 305]
Routh-Hurwitz stability criterion
Adolf Hurwitz
I 1859 – 1919
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German mathematician
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1895 – Determined the
Routh-Hurwitz stability
criterion
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Stability information without
the need to solve for the CL
system poles
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How many CL system poles are
in the LHP, RHP, and on the
imaginary axis
2 steps
1. Generate Routh table
2. Interpret the Routh table
Figure: 1905 FIFA World Cup –
Germany vs. England
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6.2 Routh-Hurwitz criterion
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6 Stability
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6.2 Routh-Hurwitz criterion
Generating a basic Routh table, [1, p. 306]
Generating a basic Routh table, [1, p. 306]
Procedure
Procedure
1. Label rows with powers of s
from the highest power of the
denominator of the CLTF down
to s0
2. In the 1st row, horizontally list
every other coefficient starting
with the coefficient of the
highest power of s
4. Remaining row entries are filled
with the negative determinant
of entries in the previous 2
rows divided by entry in the 1st
column directly above the
calculated row. The left-hand
column of the determinant is
always the 1st column of the
previous 2 rows, and the
right-hand column is the
elements of the column above
and to the right.
Figure: Equivalent CL TF
Table: Routh table
3. In the 2nd row, horizontally list
every other coefficient starting
with the coefficient of the next
highest power of s
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Figure: Equivalent CL TF
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Table: Routh table
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6 Stability
6.2 Routh-Hurwitz criterion
6 Stability
6.3 Routh-Hurwitz criterion: special cases
Interpreting a basic Routh table, [1, p. 307]
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The number of roots of the polynomial that are in the RHP is equal
to the number of signs changes in the 1st column of a Routh table
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A system is stable if there are no sign changes in the first column of
the Routh table
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6 Stability
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6 Stability
6.1 Introduction
6.2 Routh-Hurwitz criterion
6.3 Routh-Hurwitz criterion: special cases
6.4 Routh-Hurwitz criterion: additional examples
6.5 Stability in state space
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6.3 Routh-Hurwitz criterion: special cases
6 Stability
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6.3 Routh-Hurwitz criterion: special cases
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Special cases, [1, p. 308]
Zero only in the 1 column, [1, p. 308]
2 procedures
1. Epsilon procedure
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2 special cases
1. Zero only in the 1st column
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If the 1st element of a row is a zero, division by zero would be required
to form the next row
2. Entire row of zeros
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2. Reciprocal roots procedure
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Result of there being a purely even polynomial that is a factor of the
original polynomial
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Root positions to generate
even polynomials
(symmetrical about the
origin)
Even polynomial appears in the
row directly above the row of
zeros
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Figure: Root position to generate even
polynomials: A, B, C, or any
combination
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6.3 Routh-Hurwitz criterion: special cases
Entire row of zeros, [1, p. 313]
1. Symmetrical and real
2. Symmetrical and
imaginary
3. Quadrantal
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6 Stability
Purely even polynomials: Only
have roots that are
symmetrical and real
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A polynomial that has the reciprocal roots of the original polynomial
has its roots distributed the same–RHP, LHP, or imaginary
axis–because taking the reciprocal of the root value does not move it
to another region
The polynomial that has the reciprocal roots of the original may not
have a zero in the 1st column
Replacing s with d1 results in the original polynomial with its
coefficients written in reverse order
6.3 Routh-Hurwitz criterion: special cases
Entire row of zeros, [1, p. 311]
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To avoid this phenomenon, an epsilon, ✏ is assigned to replace zero in
the 1st column
The value ✏ is then allowed to approach zero from either the positive or
the negative side, after which the signs of the entries in the 1st column
can be determined
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Every entry in the table from the even polynomial’s row to the end of
the chart applies only to the even polynomial
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Number of sign changes from the even polynomial to the end of the
table equals the number of RHP roots of the even polynomial
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Even polynomial must have the same number of LHP roots as it does
RHP roots
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Remaining poles must be on the imaginary axis
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The number of sign changes, from the beginning of the table down to
the even polynomial, equals the number of RHP roots
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Remaining roots are LHP roots
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The other polynomial can contain no roots on the imaginary axis
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6 Stability
6.4 Routh-Hurwitz criterion: additional examples
6 Stability
6.4 Routh-Hurwitz criterion: additional examples
Some examples, [1, p. 314]
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Example (Standard)
6 Stability
6.1 Introduction
6.2 Routh-Hurwitz criterion
6.3 Routh-Hurwitz criterion: special cases
6.4 Routh-Hurwitz criterion: additional examples
6.5 Stability in state space
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Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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CLTF
Characteristic equation
Generate Routh table
Interpret Routh table
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6.4 Routh-Hurwitz criterion: additional examples
6 Stability
Some examples, [1, p. 314]
Figure: FB control system
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6.4 Routh-Hurwitz criterion: additional examples
Some examples, [1, p. 314]
Example (Standard)
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Example (Zero in 1st column)
Table: Routh table
Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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CLTF
Characteristic equation
Generate Routh table
Interpret Routh table
T (s) =
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Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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200
s4 + 6s3 + 11s2 + 6s + 200
4
3
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CLTF
Characteristic equation
Generate Routh table
Epsilon method
Interpret Routh table
Figure: FB control system
2
CE(s) = s + 6s + 11s + 6s + 200
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6 Stability
Table: Routh table
Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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+
3s4
6.4 Routh-Hurwitz criterion: additional examples
Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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1
2s5
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Example (Zero in 1st column)
CLTF
Characteristic equation
Generate Routh table
Epsilon method
Interpret Routh table
T (s) =
September 10, 2013
Some examples, [1, p. 314]
Example (Zero in 1st column)
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Feedback Control Systems
6.4 Routh-Hurwitz criterion: additional examples
Some examples, [1, p. 314]
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+
2s3
+
3s2
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+ 2s + 1
CLTF
Characteristic equation
Generate Routh table
Reciprocal root method
Epsilon method
Interpret Routh table
Figure: FB control system
CE(s) = 2s5 + 3s4 + 2s3 + 3s2 + 2s + 1
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6 Stability
6.4 Routh-Hurwitz criterion: additional examples
6 Stability
Some examples, [1, p. 314]
6.4 Routh-Hurwitz criterion: additional examples
Some examples, [1, p. 314]
Example (Zero in 1st column)
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Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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Table: Routh table
Example (Row of zeros)
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CLTF
Characteristic equation
Generate Routh table
Reciprocal root method
Epsilon method
Interpret Routh table
T (s) =
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Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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1
CLTF
Characteristic equation
Generate Routh table
Interpret Routh table
Figure: FB control system
2s5 + 3s4 + 2s3 + 3s2 + 2s + 1
RCCE(s) = s5 + 2s4 + 3s3 + 2s2 + 3s + 2
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6.4 Routh-Hurwitz criterion: additional examples
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6 Stability
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6.5 Stability in state space
Some examples, [1, p. 314]
Example (Row of zeros)
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Problem: Find the number of
poles in the LHP, RHP, and on
the imaginary axis
Solution: On board
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Table: Routh table
1
CLTF
Characteristic equation
Generate Routh table
Interpret Routh table
T (s) =
6 Stability
6.1 Introduction
6.2 Routh-Hurwitz criterion
6.3 Routh-Hurwitz criterion: special cases
6.4 Routh-Hurwitz criterion: additional examples
6.5 Stability in state space
12
s8 + 3s7 + 10s6 + 24s5 + 48s4 + 96s3 + 128s2 + 192s + 128
CE(s) = s8 + 3s7 + 10s6 + 24s5 + 48s4 + 96s3 + 128s2 + 192s + 128
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6.5 Stability in state space
Feedback Control Systems
6 Stability
Some definitions, [1, p. 320]
6.5 Stability in state space
Bibliography
Definition (eigenvalues, eigenvectors, & characteristic equation)
Eigenvalues, , of system matrix, A
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System poles
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Values that permit a nontrivial
solution (other than 0) for
eigenvectors, x, in the equation
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All solutions will be the null
vector except for the
occurrence of zero in the
denominator
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This is the only condition
where elements of x will be
0/0 or indeterminate, it is the
only case where a nonzero
solution is possible
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Solutions of the characteristic
equation det(sI A) = 0, a
polynomial
Ax = x
x = ( I A) 1 0
adj( I A)
=
0
det( I A)
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Norman S. Nise. Control Systems Engineering, 2011.
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