Stability Analysis of Nonlinear Systems Using...
Transcript of Stability Analysis of Nonlinear Systems Using...
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Stability Analysis of
Nonlinear Systems
Using Lyapunov TheoryBy: Nafees Ahmed
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Outline
Motivation
Definitions
Lyapunov Stability Theorems
Analysis of LTI System Stability
Instability Theorem
Examples
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References
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore (NPTEL)
H. K. Khalil: Nonlinear Systems, Prentice Hall, 1996.
H. J. Marquez: Nonlinear Control Systems analysis and
Design, Wiley, 2003.
J-J. E. Slotine and W. Li: Applied Nonlinear Control,
Prentice Hall, 1991.
Control system, principles and design by M. Gopal, Mc
Graw Hill
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Techniques of Nonlinear ControlSystems Analysis and Design
Phase plane analysis: Up to 2nd order or maxi 3rd order system (graphical method)
Differential geometry (Feedback linearization)
Lyapunov theory
Intelligent techniques: Neural networks, Fuzzy logic, Genetic algorithm etc.
Describing functions
Optimization theory (variational optimization, dynamic programming etc.)
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Motivation
Eigenvalue analysis concept does not hold good for nonlinear systems.
Nonlinear systems can have multiple equilibrium points and limit cycles.
Stability behaviour of nonlinear systems need not be always global (unlike linear systems). So we seek stability near the equilibrium point.
Stability of non linear system depends on both initial value and its input (Unlike liner system). Stability of linear system is independent of initial conditions.
Need of a systematic approach that can be exploited for control design as well.
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Idea
Lyapunov’s theory is based on the simple
concept that the energy stored in a stable
system can’t increase with time.
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Definitions
Note:
• Above system is an autonomous (i/p, u=0)
• Here Lyapunov stability is considered only for autonomous system (It
can also extended to non autonomous system)
• We can have multiple equilibrium points
• We are interested in finding the stability at these equilibrium points
• Rn => n dimensions (ie x1,x2 =>n=2 =>two dimensions )
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Definitions
Open Set: Let set A be a subset of R then the set A is open if every point in A
has a neighborhood lying in the set. Or open set means boundary lines are not
included. Mathematically
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Definitions
Open set:
A set 𝐴 ⊂ ℝ𝑛 is called as open, if for each 𝑥 ∈ 𝐴 there exist an 𝜀 > 0 such that
the interval 𝑥 − 𝜀, 𝑥 + 𝜀 is contained in A. Such an interval is often called as
𝜀 -neighborhood of x or simply neighborhood of x.
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Definitions
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1. Starting with a small ball of radius δ(ε) from initial
condition Xo a system will move anywhere around the ball
but will not leave the ball of radius ε
2. Ball δ(ε) is a function of ε.
3. Size of δ(ε) may be larger then ball of radius ε
δ(є)
є
* X0
* Xe
δ(є)
є
δ(є)
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Definitions
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Convergent system: Starting from any initial
condition Xo, system may go anywhere but finally
converges to equilibrium point Xe
* X0
* Xe
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Definitions
Note: System will never leave the ε bound and finally will converge to
equilibrium point Xe.
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Definitions
Conversion : 𝑍 = 𝑋 − 𝑋𝑒 ⇒ 𝑍 = 𝑋 − 𝑋𝑒 ⇒ 𝑍 = 𝑋 𝑋𝑒 = 0 = 𝑓 𝑍 ⇒ 𝑍 = 𝑓(𝑍 + 𝑋𝑒)
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Definitions
A scalar function V : D→R is said to be
Positive definite function: if following condition are
satisfied
(domain D excluding 0)
Positive semi definite function:
Negative define function: (i) condition same, (ii) <
Negative semi define function: (i) condition same, (ii) ≤
Note:
1. Output of function V(x) is a scalar value, hence V(x) is scalar function .
2. Negative define (semi definite) if –V(x) is + definite ( semi definite)
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Note:Condition (i) & (ii) ⇒ V(X) positive definite
Condition (iii) ⇒ 𝑉(𝑋) Negative semi definite
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What about V(X)
There is no general method for selection of V(X).
Some time select V(X) such that its properties are similar to energy i.e.
𝑽 𝑿 =𝟏
𝟐𝑿𝑻𝑿
𝑶𝒓 𝑽 𝑿 = 𝑲𝒊𝒏𝒕𝒆𝒊𝒄 𝑬𝒏𝒆𝒓𝒈𝒚 + 𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝑬𝒏𝒈𝒆𝒓𝒚
𝑶𝒓 𝑽 𝑿 = 𝒙𝟏𝟐 + 𝒙𝟐
𝟐 etc
How to calculate 𝑽(𝑿)
𝑽 𝑿 =𝝏𝑽
𝝏𝒙
𝑻
𝑿 =𝝏𝑽
𝝏𝒙
𝑻
𝒇(𝑿)
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Note:Condition (i) & (ii) ⇒ V(X) positive definite
Condition (iii) ⇒ 𝑉(𝑋) Negative definite
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Radially Unbounded ?
The more and more you go away from the equilibrium point, V(X) will
increase more and more.
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Note: Global⇒ Subset D=R
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NOTE
Here, pendulum with friction should be
asymptotically stable as it comes to an
equilibrium point finally due to friction (⇒ 𝑽(𝑿)should be negative definite not negative semi
definite nsdf)
But we are not able to prove this.
Because
when x2≠ 0, 𝑽(𝑿) will always be –Ve
But when x2= 0 There are multiple equilibrium points
on x1 line.
Negative definite means the movement I go away
from the zero I should get –ve value
x2
x1
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Example: Consider the system described by the equations
𝒙𝟏 = 𝒙𝟐
𝒙𝟐 = −𝒙𝟏 − 𝒙𝟐𝟑
Solution:
Choose 𝑽 𝒙 = 𝒙𝟏𝟐 + 𝒙𝟐
𝟐
Which satisfies following two conditions that is it is positive definite
𝑽 𝟎 = 𝟎 & 𝑽 𝒙 > 𝟎
𝑽(𝒙) = 𝟐𝒙𝟏 𝒙𝟏 + 𝟐𝒙𝟐 𝒙𝟐 = 𝟐𝒙𝟏 𝒙𝟏 + 𝟐𝒙𝟐 −𝒙𝟏 − 𝒙𝟐𝟑 = −𝟐𝒙𝟐
𝟒
𝑽(𝒙) ≤ 𝟎 ⇒ nsdf (similar to pendulum with friction)
So system is stable, we can’t say asymptotically stable
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Analysis of LTI system using Lyapunov
stability
Note: 𝑋 = 𝐴𝑋 ⇒ 𝑋𝑇 = 𝐴𝑋 𝑇 = 𝑋𝑇𝐴𝑇
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Analysis of LTI system using Lyapunovstability…
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Analysis of LTI system using Lyapunovstability….
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Step to solve
Analysis of LTI system using Lyapunov
stability….
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Example: Analysis of LTI system using Lyapunov stability
Determine the stability of the system described by the following equation
𝑥 = 𝐴𝑥 With 𝐴 =−1 −21 −4
Solution:
𝐴𝑇𝑃 + 𝑃𝐴 = −𝑄 = −𝐼
−1 1−2 −4
𝑝11 𝑝12𝑝12 𝑝22
+𝑝11 𝑝12𝑝12 𝑝22
−1 −21 −4
=−1 00 −1
Note here we took p12=p21 because Matrix P will be + real symmetric
matrix
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-2p11+2p12=-1
-2p11-5p12+p22=0
-4p12-8p22=-1
Solving above three equations 𝑃 =𝑝11 𝑝12𝑝12 𝑝22
=
23
60−
7
60
−7
60
11
60
which is seen to be positive definite. Hence this system is asymptotically
stable
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Till now ?
All were Lyapunov Direct
methods
There are some indirect
methods also
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In rough way
In rough way instability theorem state that
if V(X) positive definite
t𝐡𝐞𝐧 𝑽(𝑿) should also be positive definite
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Thanks
?