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Stability Concepts
• The concept of stability
• Routh-Hurwitz criterion
• Relative stability
• Stability of state-space systems
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The Concept of Stability
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The Concept of Stability
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Why Stability is Important
Tacoma Narrows Bridge was build
on July 1,1940. The bridge was
found to oscillate whenever the
wind blew. a) as oscillation begins
b) Catastrophic failure
on Moverbmer7,1940.
Why?
Given an example of unstable system in real life.
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S-Plane
and Transient Response
Y ( s ) ∑ Pi ( s )Δ i ( s )
=
,
Δ( s)
R( s)
N
Bk s + Ck
1 M Ai
,
Y (s) = + ∑
+ ∑
s i =1s + σ i k =1 s 2 + 2α k s + (α k2 + ω n2 )
T (s) =
M
y (t ) = 1 + ∑ Ai e
i =1
−σ i t
N
+ ∑ Dk e −σ k t sin(ω k t + θ k ),
k =1
A necessary and sufficient condition for a feedback system
to be stable is that all the poles of the system transfer
function have negative real parts.
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S-Plane
and Transient Response
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Definitions of Stability
• BIBO stability: A system is said to be BIBO
stable if for any bounded input, its output
is also bounded.
• Absolute stability: Stable /Unstable
• Relative stability: Degree of stability (i.e.
how far from instability)
• A stable linear system described by a T.F.
is such that all its poles have negative
real parts
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Feedback and Stability
K=3, stable
K=7, unstable
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Relationship between the coefficients
and roots of the characteristic equation
Consider the simple second-order characteristic equation:
Roots:
For a stable system, the roots of the characteristic equation must
have negative real parts.
b/a>0 and c/a>0
What is the requirements for the coefficients?
Is it possible to find the similar conditions for higher order system?
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Relationship between the coefficients
and roots of the characteristic equation
Consider the nth-order characteristic equation as:
Δ ( s ) = q( s ) = a n s n + a n −1 s n −1 + ... + a1 s + a 0 = 0.
• To ascertain the stability of the system it is necessary to
determine whether any one of the roots of q(s) lies in
the right half of the s-plane.
• Rewrite the above equation in factored form, we have
a n ( s − r1 )( s − r 2 ).....( s − r n ) = 0 ,
where ri is the ith root of the characteristic equation.
q( s) = an s n − an (r1 + r2 + ... + rn )s n−1
+ a n ( r1r2 + r2 r3 + r1r3 + ...) s n − 2
− a n ( r1r2 r3 + r1r2 r4 + ...) s n −3 + ...
+ a n ( −1) n r1r2 r3 ...rn = 0.
a
n −1
= − a
n
( r1 + r 2 + K + ) , a
n − 2
= a
n
( r1 r 2 + . . . . ) , K a
0
= ( − 1 ) n a n r1 . . . . r n
Relationship between the coefficients
and roots of the characteristic equation
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In other words, for an nth order characteristic equation, we obtain
q(s) = an sn − an (sum of all the roots)sn−1
+ an ( sum of the products of the roots taken 2 at a time)s n−2
− an ( sum of the products of the roots taken 3 at a time)s n−3 + ...
+ an (−1)n ( product of all n roots) = 0.
• Note that all the coefficients of the polynomial much have the
same sign if all the roots are in the left-hand of the s-plane.
• It is necessary that all the coefficients for a stable system be
nonzero.
• Those requirements are necessary but not sufficient.
• That is we immediately know the system is unstable if they
are not satisfied; yet if they are satisfied, we must proceed
father to ascertain the stability of the system .
E.g:
q( s ) = ( s + 2)( s 2 − s + 4) = s 3 + s 2 + 2s + 8,
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Relationship between the coefficients
and roots of the characteristic equation
q(s) = (s + 2)(s 2 − s + 4) = (s 3 + s 2 + 2s + 8),
The system is unstable even that the polynomial possesses
all positive coefficients.
• It is desired to obtain a necessary and sufficient criterion for
the stability of linear system with nth order characteristic
equation.
• In the later 1800’s, A. Hurwitz and E.J. Routh published
independently a method of investing the stability of a
linear system.
• The Routh-Hurwitz criterion is a necessary and sufficient
criterion for the stability of linear systems.
• The method is originally developed in terms of determinants
but we shall utilize the more convenient array formulation.
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Routh-Hurwitz Criterion
Edward Routh, 1831 (Quebec)1907 (Cambridge, England)
Adolf Hurwitz, 1859
(Germany)-1919 (Zurich)
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Routh-Hurwitz Criterion
• Consider the polynomial
• Routh-Hurwitz stability criterion is a test to
ascertain without computing the roots, whether
or not all roots of a polynomial have negative
real parts.
• It will be given here without proof
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Routh-Hurwitz Table
Δ(s) = q(s) = an s n + an−1s n−1 + ... + a1s + a0 = 0.
…..
The number of roots of Q(s) with positive real parts is
equal to the number of sign changes in the first column
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Routh-Hurwitz Criterion
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Routh-Hurwitz Criterion
• This criterion requires that there be no changes in sign in
the first column for a stable system. This requirement is
both necessary and sufficient.
• Four distinct cases or configurations of the first column array
much be considered, and each must be treated separately
and requires suitable modifications of the array calculation
procedure:
1. No element in the first column is zero;
2. there is zero in the first column, but some other element of
the row containing the zero in the first column are nonzero;
3. there is a zero in the first column, and other elements of the
row containing the zero are also zero;
4. and as in (3) with repeated roots on the jω-axis.
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Routh Hurwitz Special Cases
• For roots to be in LHP, it is necessary (but sufficient only
in second order case) that all the coefficients be positive
Issues with awkward zero entries:
• Case A: A zero-entry appears in the first column,
but other entries on the row are non-zero
-Solution: Take entry as small value ε >0 and
proceed, taking ε > 0 in subsequent calculations
• Case B: A zero-entry appears in the first column,
and all other entries in that row are also zero
-Solution: Return to the previous row and form the
“Auxiliary Polynomial, Qa(s)”, which will be a divisor
of the original Q(s), divide out Qa(s), and proceed.
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Simple Example
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Polynomials of Degree 2 & 3
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Simple Systems
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Example
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Routh-Hurwitz Criterion
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Tracked Vehicle Turning Control
Want
ess to ramp command to be less than 24%
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Tracked Vehicle Turning Control
Stability:
a
Static error to ramp input:
Ka=42
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A zero-entry appears in the first column, but
other entries on the row are non-zero
-Solution: Take entry as small value ε >0 and proceed,
taking ε >0 in subsequent calculations. After completing
The process let ε approach to zero.
q( s) = s 5 + 2s 4 + 2s 3 + 4s 2 + 11s + 10.
s 5 1 2 11
s 4 2 4 10
s3 ε 6 0
s 2 c1 10 0
s1 d1 0 0
s 0 10 0 0
4ε −12 −12
6c − 10ε
c1 =
=
, d1 = 1
→ 6.
ε
ε
c1
The system is unstable, and two
roots lie in the right half of the
s-plane.
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More example
Consider the characteristic polynomial:
q(s) = s 4 + s 3 + s 2 + s + K ,
To determine the K value that results in the stability.
Routh array:
c1 =
s4
s3
s2
s1
s0
1 1K
1 1 0
ε K0
c1 0 0
K 0 0
ε −K
−K
→
.
ε
ε
System is unstable for all values of gain K.
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A zero-entry appears in the first column, and
all other entries in that row are also zero
-Solution: Return to the previous row and form the
“Auxiliary Polynomial, qa(s)”, which will be a divisor
of the original q(s), divide out qa(s), and proceed.
The auxiliary polynomial is the polynomial immediately
precedes the zero entry in Routh array. The order of the
auxiliary polynomial is always even and indicates the
number of symmetrical roots pair.
Example:
q( s ) = s 3 + 2 s 2 + 4 s + K ,
s3
s2
s1
s0
1
2
8− K
2
K
0 < K < 8.
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K
q(s) / qa (s) ⇒
2s2 + 8 s3 + 2s2 + 4s + 8
+ 4s
s3
qa (s )
2s2
2s2
0
0
When K=8
1/ 2s +1
+8
+8
Marginal stable
qa (s) = 2s 2 + Ks0 = 2s 2 + 8 = 2(s 2 + 4) = 2(s + j 2)(s − j 2).
q( s) = qa ( s)(1 / 2s + 1) = ( s + 2)(s + j 2)(s − j 2)
( when K = 8)
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Repeated roots of the characteristic
equation on the jω-axis
• If the jω-axis roots of the characteristic equation are simple
(first order), the system is neither stable nor unstable; it is
instead called marginal stable, since it has an un-damped
sinusoidal mode.
-If the jω-axis roots are repeated, the system response will be
unstable, with a form t[sin(ωt+Φ)].
-The Routh-Hurwitz criterion will not reveal this form of
instability.
q(s) = (s +1)(s + j)(s − j)(s + j)s − j)= s 5 + s 4 + 2 s 3 + 2 s 2 + s + 1 .
s 5
1 2 1
s 4
1 2 1
s 3 ε ε 0
s 2 1 1
s1
ε 0
s 0
1
s4 + 2s2 +1= (s2 +1)2
⇒ ( s 2 + 1)
Note that the absence of
sign changes, a condition
that falsely indicates that
the system is marginally
stable.
The real response of the system is
increase with time as t[sin(ωt+Φ)].
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Robotic Arm
0<K<25.308
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The Relative Stability of Feedback
Control Systems
• The verification of stability using the Routh-Hurwitz
criterion provides only a partial answer to the question
of stability----whether the system is absolutely stable.
• In practice, it is desired to determine the relative
stability.
- The relative stability of a system can be defined
as the property that is measured by the relative
real part of each root or pair of roots.
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The Relative Stability of Feedback
Control Systems
• Because the relative stability of a system is dictated by
location of the roots of the characteristic equation, we
can extend the Routh-Hurwitz criterion to ascertain relative
stability.
• This can be accomplished by utilizing a change of variable,
which shifts the s-plane vertical axis in order to utilize the
Routh-Hurwitz criterion.
• The correct magnitude of shift the vertical ais must be
obtained on a trial-and-error basis.
• One may determine the real part of the dominant roots
without solving the high order polynomial q(s).
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Example: Axis shift
Consider the simple third -order characteristic equation
First try: let
Second try:
There is no zero in the first column of
Routh array, but with two unstable roots.
We have
The shifting of the s-plane axis to ascertain the relative stability of a system is
a very useful approach, particularly for higher-order system with several pairs
of closed-loop complex conjugate roots.
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Stability of state-space systems
• If the are n states, m inputs, and p outputs, then A is
square (nxn), B is (nxm), C is (pxn) and D is (pxm)
• For a single input, single output system, we have A
square (nxn), B=b and is a (nx1) column vector, C=c is a
(1xn) row vector, and D=d is a (1x1) scalar (often zero)
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Characteristic Equation and Eigenvalues
• Recall that, for a transfer function G(s)=N(s)/D(s),
the roots of the characteristic equation D(s)=0 are
the poles of the system.
• Recall that the denominator of the transfer
function of a state-space representation is det(sI-A)
• The characteristic equation is then det(sI-A)=0
• The roots of this equation are the eigenvalues of
the matrix A
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Eigenvalues
• If the coefficients of A are real, then eigenvalues
are either real or complex conjugate pairs
• The trace of A is the sum of all the eigenvalues
• An eigenvalue of A is also an eigenvalue of AT
• If A is nonsingular with eigenvalues λi , the
−1
eigenvalues of A are 1 / λi
• For the stable system, the real parts of all the
eigenvalues must be negative.
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Example Stabilty region for unstable plant
A jump-jet aircaft has a control system as shown in the Fig.
Goal: Assuming that z>0 and p>0,find a suitable set of K, z,
and p to stabilize the system .
The system is open-loop unstable (without feedback control)
since the characteristic equation of the plant and controller is
Note that since one term within the bracket has a negative
coefficient, the characteristic equation has at least one root
in the right-hand side of the s-plane.
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Example Stabilty region for unstable plant
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Example Stabilty region for unstable plant
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Example Stabilty region for unstable plant
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Example Stabilty region for unstable plant
The characteristic equation of the closed-loop is
The goal is to determine the region of stability for K,p,and z.
The Routh array is
where
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Example Stabilty region for unstable plant
We require that b2 > 0, ( p − 1) > 0, and Kz>0.
b2 =
( p −1)(K − p) − Kz
>0
( p −1)
⇔ ( p −1)(K − p) − Kz > 0
⇔ K [( p − 1) − z ] − p ( p − 1) > 0
p( p −1)
⎧
K
if z > ( p −1)
<
⎪
( p −1) − z
⎪⎪
K[( p −1) − z] − p( p −1) > 0 ⇒ ⇒ ⎨ − p( p −1) > 0 if z = p −1
⎪
p( p −1)
⎪K >
if z < ( p −1)
( p −1) − z
⎪⎩
Consider two cases:
( p −1)(K − p) − Kz > 0
1. z ≥ ( p − 1)
There is no 0 < k < ∞ that leads to stability
p ( p − 1)
, with p > 1 and
( p − 1) − z
z < ( p − 1) will result in stability.
2. z < ( p − 1) : Any K >
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Example Stabilty region for unstable plant
The three-dimensional plot of the stability region for K, p, and
z is shown in the following figure.
z ≥ ( p − 1)
Unstable region.
One acceptable point is z=1, p=10, and K=15
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Disk Drive Read System
Before:
Now:
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Sequential Design Example: Disk
Drive Read System
Chapter 5 results:
Y ( s) =
=
=
5K a
R ( s ),
s( s + 20) + 5K a
5K a
s 2 + 20 s + 5 K a
ω n2
s 2 + 2ζω n s + ω n2
R ( s ),
R ( s ),
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Sequential Design Example: Disk
Drive Read System
?
• The best compromise (Ka=40) still does not meet all the specifications.
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Disk Drive Read System
Without velocity feedback:
The characteristic equation is
or
Condition for stability is
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Disk Drive Read System
With velocity
feedback:
The characteristic equation is
or
Condition for stability is
or
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Disk Drive Read System
Response with K1=0.05 and Ka=100
0.98
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Disk Drive Read System
The system performances is summarized in the following.
The performance specifications are nearly satisfied, and some
iteration of K1 is necessary to obtain the desired 250 ms setting
time.
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Summary
• Routh Hurwitz stability criterion allows
to check for stability without computing
roots of characteristic equation
• Can be used to determine the range of
parameters that guarantees stability