Stability Analysis of Nonlinear System by
Transcript of Stability Analysis of Nonlinear System by
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Stability Analysis of
Nonlinear System by
Popov’s Stability Criterion
By: Nafees Ahmed
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Introduction
It is the first technique given for nonlinear systems analysis. (1962)
It is in frequency domain technique.
It can be seen in several view points.
We will see it with loop transformation
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Consider the following closed loop
system
Many nonlinear physical systems can be represented as a
feedback connection of a linear dynamical system and a nonlinear element.
G(S)r=0
NLξ σ
σ-ξ
_G(S)
NLξ
σ-ξ
_
σ
T.F. of linear system
G(s)=σ/-ε
NL=> Non linear System
⇒
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Consider following nonlinearity
Let us define the nonlinearity
NL є m[K, ∞) sector
Above nonlinearity can be transformed to new nonlinearity NL1 є m[0, ∞] sector
Slope=K
ε
σ
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Loop transformation
Above sys can be transformed to a new system as shown on next slide keeping same i/p but different o/p for non linear part.
T.F of new linear plat 𝑮𝟏 𝒔 =𝝈
−𝝃𝟏=
𝝈
−𝝃+𝑲𝝈=
𝑮(𝒔)
𝟏+𝑲𝑮(𝒔)
In this way nonlinearity NL is transformed to NL1 i.e. From m[K, ∞) sector to m[0, ∞) sector
Note: here i/p & o/p relationship of G(s) and NL are same as the original one.
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Loop transformation …
G(S)
NLξ
σ
-ξ_
σ
K
ξ1
New Nonlinearity
NL1
Where
ξ1= ξ- σK
Nonlinearity of NL1 є m[0, ∞) sector
K
-ξ1
_
_
New linear Plant
G1(s)
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Loop transformation …
If a transfer function is Passive
=> it is positive real
=> it is asymptotic stable
Interconnection of two passive
systems is also passive system.
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Loop transformation …
Now we can say that if original system is asymptotic
stabile then its transformed sys shown on previous slide
will also be asymptotic stable.
New real plant G1(s) must be positive real i.e 𝑮𝟏 𝒔 =𝑮(𝒔)
𝟏+𝑲𝑮(𝒔)must be a positive real.
New nonlinearity defined by NL1 є m[0, ∞) sector
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Loop transformation…
Same way we can do other loop transformation
Consider following nonlinearity
NL є m[0, K] sector
Above nonlinearity can be transformed to new nonlinearity NL1 є m[0, ∞) sector
Slope=Kε
σ
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Loop transformation…
Above sys can be transformed to a new system as shown on next slide
keeping same o/p but different i/p for non linear part.
T.F of new linear plat 𝑮𝟏 𝒔 =G(s)+1/K
In this way nonlinearity NL is transformed to NL1 i.e. From m[0,K] sector to
m[0, ∞) sector
Note: here i/p & o/p relationship of G(s) and NL are same as the original
one.
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Loop transformation…
G(S)
NL
σ1
-ξ_
ξσ1
1/K
1/K
+
New linear Plant
G1(s)
Where
σ1= σ-(1/K)ξ
New Nonlinearity
NL1
+
Nonlinearity of NL1 є m[0, ∞) sector
σ
σ
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Loop transformation …
Similar to previous
Now we can say that if original system is asymptotic
stabile then its transformed sys shown on previous slide
will also be asymptotic stable.
New real plant G1(s) must be positive real i.e 𝑮𝟏 𝒔 =𝑮 𝒔 + 𝟏/𝑲 must be a positive real.
New nonlinearity defined by NL1 є m[0, ∞) sector
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Loop transformation …
Note: In above two loop transformation one portion is
feed forward and other portion is feed backward with
same sign.
In above loop transformation we introduced a constant
K (=slop of nonlinearity), to change the nonlinearity.
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Loop transformation …
With same nonlinearity NL є m[0, K] sector
The above loop transformation was done
in feedback and feed forward loop, it can
also be done in forward path only as
shown on next slide
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Loop transformation …
G(S)σ1
_-ξ
NL σ1
New linear Plant
G1(s)
+
ξ1/K
σ
+
Where
σ1= Kσ-ξ
New Nonlinearity
NL1
Nonlinearity of NL1 є m[0, ∞) sector
σ
K
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Loop transformation …
Similar to previous
Now we can say that if original system is asymptotic
stabile then its transformed sys shown on previous slide
will also be asymptotic stable.
New real plant G1(s) must be positive real i.e𝑮𝟏 𝒔 = 𝟏 +𝐊𝐆 𝐬 must be a positive real.
New nonlinearity defined by NL1 є m[0, ∞) sector
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Loop transformation …
Now use mix of above two methods
And instead of K we can also use some transfer
function
With same nonlinearity NL є m[0, K] sector
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Loop transformation …
𝑾𝒉𝒆𝒓𝒆 𝝈𝟏 = 𝒂 𝝈 + 𝝈 −𝟏
𝑲𝜻
Nonlinearity of NL1 є m[0, ∞) sector
G(S)σ1
_-ξ
NL σ1
New linear Plant
G1(s)
+
σ
1/(1+as)ξ +
(1+as)
σ
New Nonlinearity
NL1
1/K
1/K
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Loop transformation …
Similar to previous
Now we can say that if original system is asymptotic
stabile then its transformed sys shown on previous slide
will also be asymptotic stable.
New real plant G1(s) must be positive real i.e 𝑮𝟏 𝒔 =𝟏 + 𝐚𝐬 𝐆 𝐬 + 𝟏/𝐊 must be a positive real.
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Interconnection of 2 passive systemes
If G1(s) is passive and NL1 is passive then there interconnection will
also be passive. (Passive=>positive real)
So passivity for new linear system G1(s) 𝑽𝟏 𝑿 ≤ −𝝈𝟏ξ
We have to prove that new nonlinear system NL1 will also be
passive. So that overall interconnection will also be passive.
For this make the differential equation of NL1 assuming NL=f as
shown on next slide
Note: We know Lyapunov function of a passive system will have 𝑽 𝑿 ≤ 𝒖𝒚where u = i/p & y=o/p (passivity in term of Lyapunov function)
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Interconnection of 2 passive
systemes…
Let ξ = 𝒇(𝝈) & f є m[0, K] sector
𝝈𝟏 = 𝒂 𝝈 + 𝝈 −𝟏
𝑲ξ
⇒ 𝝈𝟏 = 𝒂 𝝈 + 𝝈 −𝟏
𝑲𝒇(𝝈)
Now take 𝝈 as state and 𝝈𝟏 as input
So we can write the state equ
𝒂 𝝈 = −𝝈 +𝟏
𝑲𝒇 𝝈 + 𝝈𝟏 state equ
ξ = 𝒇 𝝈 output equ
f σ11/(1+as)ξ +σ
New Nonlinearity
NL1
1/K
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Interconnection of 2 passive
systemes…
Take 𝝈=x and ξ=y & 𝝈𝟏=u sate equ
𝒂 𝒙 = −𝒙 +𝟏
𝑲𝒇 𝒙 + 𝒖 state equ
y= 𝒇 𝒙 output equ
Let us define a Laypunov function for NL1 as
𝑽𝟐 𝒙 = 𝒂 𝟎𝒙𝒇 𝒙 𝒅𝒙 which is always >0 ∀ 𝒙
⇒ 𝑽𝟐 𝒙 = 𝒂𝒇(𝒙) 𝒙
Put the value of 𝒂 𝒙 from above equation
𝑽𝟐 𝒙 = −𝒇 𝒙 𝒙 +𝟏
𝑲𝒇 𝒙 𝟐 + 𝒖𝒇(𝒙)
So 𝑽𝟐 𝒙 ≤ 𝒖𝒇 𝒙 ≤ 𝒖𝒚 ⇒ NL1 is also passive
Slope=Kε=f(x)
σ=x
Note: 𝟎𝒙𝒇 𝒙 𝒅𝒙 is always +ve for above non linearity By: Nafees Ahmed
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Interconnection of 2 passive
systemes…
We know Lyapunov’s function of new linear system (Passive) G1(s)
V1(x)=xTPx with 𝑽𝟏 𝑿 ≤ −𝝈𝟏ξ P is positive definite matrix
And Lyapunov’s function of new nonlinear system (Passive) NL1
𝑽𝟐 𝒙 = 𝒂 𝟎𝒙𝒇 𝒙 𝒅𝒙 𝒘𝒊𝒕𝒉 𝑽𝟐 𝑿 ≤ 𝝈𝟏ξ
So Lyapunov’s function of overall system
V=V1+V2= xTPx+𝒂 𝟎𝒙𝒇 𝒙 𝒅𝒙 with 𝑽 = 𝑽1+ 𝑽2 ≤ −𝝈𝟏ξ+ 𝝈𝟏ξ≤0
Here V>0 and 𝑽 ≤0 and by Lyapunov this is the condition of
Asymptotic stability
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Popov’s criterion
Consider a system shown in following figure
G(S)
NLξ
σ-ξ
_
σ
Nonlinearity of NL1 є m[0, ∞) sector
G(S)σ1
_-ξ
NL σ1
New linear Plant
G1(s)
+
σ
1/(1+as)ξ +
(1+as)
σ
New Nonlinearity
NL1
1/K
1/K
⇒
Nonlinearity of NL1 є m[0, K] sector
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Popov’s criterion …
This system will be asymptotically stable
If 𝑮𝟏 𝒔 = 𝟏 + 𝒂𝒔 𝑮 𝒔 + 𝟏/𝑲 is positive real i.e
Re(G1(s)≥0 ⇒ Re[ 𝟏 + 𝒂𝒔 𝑮 𝒔 + 𝟏/𝑲 ] ≥0
For this G1(s) must be strictly proper [i.e. |deg(num)-deg(den)|≤1], necessary condition
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Popov’s criterion …
How to check condition Re[ 𝟏 + 𝒂𝒔 𝑮 𝒔 + 𝟏/𝑲 ] ≥0
=> Re[ 𝟏 + 𝒋𝒂𝝎 (𝑹𝒆𝑮 𝒋𝝎 + 𝒋𝑰𝐦𝐆 𝐣𝛚 )] + 𝟏/𝑲 ≥0
=> 𝑹𝒆𝑮 𝒋𝝎 − 𝒂𝝎𝑰𝐦𝐆 𝐣𝛚 + 𝟏/𝑲 ≥0
For above draw the following plot called Popov’s plot
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Popov’s criterion …
Re(G(jω))=x
ωIm(G(jω))=y
Popov’s Plot
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Popov’s criterion …
𝑹𝒆𝑮 𝒋𝝎 − 𝒂𝝎𝑰𝒎𝑮 𝒋𝝎 + 𝟏/𝑲 ≥0
⇒x-ay+1/K ≥0
Re(G(jω))=x
ωIm(G(jω))=y
Popov’s Plot
-1/K
Slope=1/a
x-ay+1/K=0
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Popov’s criterion …
Re(G(jω))=x
ωIm(G(jω))=y
Popov’s Plot
-1/K
Slope=1/a
x-ay+1/K=0
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Popov’s criterion …
Hence to check the condition Re[ 𝟏 + 𝒂𝒔 𝑮 𝒔 + 𝟏/𝑲] ≥0 just draw Popov’s plot and draw the line as
shown above, if Popov’s plot is to the right (below) of
line that means above condition is satisfied. So the
overall system will be asymptotically stable.
Note: Difference b/w Popov’s plot and Nyquist plot
Popov’s is Re(G(jω) Vs ωIm(G(jω)
Nyquist plot is Re(G(jω) Vs Im(G(jω)
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Reference
Prof. Harish K Pillai, IIT Bombay, through NPTEL
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Thanks
?By: Nafees Ahmed