lecture 11. stability of numerical integration methods
TRANSCRIPT
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2010-11-19 Slide 1
PRINCIPLES OF CIRCUIT SIMULATIONPRINCIPLES OF CIRCUIT SIMULATION
Lecture 11. Lecture 11. Stability of Numerical IntegrationStability of Numerical Integration
MethodsMethods
Guoyong Shi, [email protected]
School of MicroelectronicsShanghai Jiao Tong University
Spring 2010
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2010-11-19 Lecture 11 slide 2
OutlineOutline• Absolute stability (A.S.)• Convergence problem in transient simulation• Numerical stability of three methods• Region of A.S. for LMS methods
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2010-11-19 Lecture 11 slide 3
Absolute StabilityAbsolute Stability
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2010-11-19 Lecture 11 slide 4
Three Integration FormulasThree Integration Formulas
• •
−−⎛ ⎞− − + =⎜ ⎟⎝ ⎠n n 1n n 1
hy y y y 02
•
−−− − =n 1n n 1y y h y 0
•
−− − =nn n 1y y h y 0
FE
TR
BE
•k m
n ji n i ji j
y h yα β −−= =
+ =∑ ∑0 0
0LMS
All are iteration formulas.
The choice of “h” affects the convergence.
Different methods have different convergence properties.
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2010-11-19 Lecture 11 slide 5
Absolute StabilityAbsolute Stability• “Absolute stability” considers how the choice
of step-size (h) affects the convergence of an integration method.
• Characterized by a convergence region in the complex plane.
• The convergence region is found by a simple test model.
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2010-11-19 Lecture 11 slide 6
-( ) tx t e=
A Simple Test Model A Simple Test Model • Use a scalar model to test how local errors are
accumulated:
(0) 1x =
Test modelTest model
Find the voltage across R:
( ) ( )dx t x tdt
= −The exact solution is:
Initial condition:
RV
inV CR
+ −( )in R Rd V V VC
dt R−
=
(0) 0CV =
(0)R inV V=
R RdV Vdt RC
= −t
RCR inV V e
−=
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2010-11-19 Lecture 11 slide 7
Why Choose a 1D Test Problem?Why Choose a 1D Test Problem?• General nonlinear model (n-dimensional)
dx/dt = F(x) ; x ∈Rn;• Linearization:
dx/dt = Ax, A = ∂F(x0) / ∂x (Jacobian)
• Diagonalization:∃ P, P-1AP = Λ if all λi(A) are distinct;Λ = diag (λ1, λ2, …, λn)dξ/dt = Λ ξ, x = Pξ (state transform)dξi /dt = λi ξi , i = 1, ..., n (scalar models)
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2010-11-19 Lecture 11 slide 8
Test ProblemTest Problem• All n-dimensional non-linear models can be
characterized locally by scalar models:
x x :•
= λ x(0) = 1;
λ∈
x∈
a complex number
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2010-11-19 Lecture 11 slide 9
n 1n n 2x x 2h x•
−−= +n n-2
n 1x - xx
2h
•
− =
x x, x(0) 1•
= − =
Test a numerical methodTest a numerical methodSuppose we use a method called “Explicit Mid-Point (EMP)”for numerical integration;
Use this formula to solve the following test problem:nxnx 2− nx 1−
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2010-11-19 Lecture 11 slide 10
n 1n n 2x x 2h x•
−−= +n n-2n 1
x - xx2h
•
− =
x x, x(0) 1•
= − =
0 0.1 0.2 0.3 9.9 10x 1, x .9, x .82, x .736, ..., x 44.0273186, x 48.6495411= = = = = = −
Diverges ...
Local Error AccumulationLocal Error Accumulation
Choose h = 0.1: xn = xn-2 – 2h xn-1
x1 = x0 + hx’0 (Use Forward Euler for the 1st step)
Exact solution known:
( ) tx t e−=
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2010-11-19 Lecture 11 slide 11
MATLAB SimulationMATLAB Simulation
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
>> h = 0.1;>> t = [0, h]; x = [1, 1-h];>> N = 50; for n = 3:N x(n) = x(n-2) - 2*h*x(n-1); t(n) = t(n-1) + h; end>> plot(t, x, 'b-')>> hold on>> plot(t, exp(-t), 'r:')
Good accuracy at the beginning;
but diverges finally.
What caused the problem?
h = 0.1
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2010-11-19 Lecture 11 slide 12
What if choosing a smaller step ?What if choosing a smaller step ?
0 .01 .1 1 12x 1, x .99, ..., x .3679, ..., x .55, ..., x 12124.17839= = = = =
Still diverges (why?)Choose h = 0.01:
n 1n n 2x x 2h x•
−−= +
01 0 0x x h x (1 h )x•
= + = − (for the 1st step)
Will a smaller “h” make it stable? --- actually not !!
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2010-11-19 Lecture 11 slide 13
0 1 2 3 4 5 6 7 8 9 10-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
MATLAB SimulationMATLAB Simulation>> h = 0.01;>> t = [0, h]; x = [1, 1-h];>> N = 1000; for n = 3:N x(n) = x(n-2) - 2*h*x(n-1); t(n) = t(n-1) + h; end>> clear figure>> f2 = figure;>> plot(t, x, 'b-'); hold on; plot(t, exp(-t), 'r:');
Good accuracy at the beginning;
still diverges eventually.
h = 0.01
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2010-11-19 Lecture 11 slide 14
The Reason ?The Reason ?
Loot at the iteration:
2 12 ,n n nx x h x− −= − ⋅ ( 0)h >
the characteristic equation
2 2 1 0hλ λ+ − =
Find the too roots (characteristic values):
21 1,h hλ = − + + 2
2 1 1h hλ = − − + < −
n n nc c h c2 12λ λ λ− −= − ⋅
Suppose xn = c λn is a solution.
Substitute into the iteration:
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2010-11-19 Lecture 11 slide 15
Check the Characteristic RootsCheck the Characteristic Roots
The general solution is: 1 1 2 2n n
nx c cλ λ= +
where c1 and c2 are constants to be determined by initial conditions.
2 12 ,n n nx x h x− −= − ⋅
21 1,h hλ = − + + 2
2 1 1h hλ = − − + < −
The two characteristic roots determine the convergence of xn !
( 0)h >
(unstable)
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2010-11-19 Lecture 11 slide 16
Plot the rootsPlot the roots
1 1 2 2n n
nx c cλ λ= +
22 1 1h hλ = − − + < −
21 1,h hλ = − + +
1
1−
h
λ
1λ
2λ
unstable
Unless the initial condition makes c2 = 0, the iteration always diverges.
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2010-11-19 Lecture 11 slide 17
(cont(cont’’d)d)
The initial conditions are:
x0 = 1 (given); x1 = (1-h)x0 = 1-h (by F. E.)
But if c2 = 0, we’ll get h = 0 (which is not allowed.)n
nx c1 1λ=
0=h
2 0c =
1 1 2 2n n
nx c cλ λ= +
1 1c =0 1x =
1 1x h= − 1 1 hλ = −
21 1,h hλ = − + +
1 1 hλ = −
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2010-11-19 Lecture 11 slide 18
• Apply Forward Euler with h = 1:
• Apply Forward Euler with h = 3:
x x•
= −
0 1 2 3x 1,x 0,x 0,x 0= = = =
0 1 2 3 4 5x 1,x 2,x 4,x 8,x 16,x 32= = − = = − = = −
(diverges)
h
Numerical BehaviorNumerical Behavior
However, Backward Euler and Trapezoidal Rule would not diverge.
Example:
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2010-11-19 Lecture 11 slide 19
Stability RegionStability Region• Use a simple test model x’ =
λx (λ is complex) to determine a region for the step-size h
• Better if region is larger.
• Stability region can be derived algebraically.
1 2
unstableunstable
q-plane
1xx
1
z 1=x
xx
x x•
= λ
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2010-11-19 Lecture 11 slide 20
Characterization MethodCharacterization Method
1. Choose an integration method with step size “h > 0”.
2. Apply it to the test problem: dx/dt = λ x3. Derive an algebraic characteristic equation.4. Define a quantity: q = λ h (as a complex
number);5. Find a region for q in the C-plane in which
the integration method is stable.• The region is called a “stability region”.
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2010-11-19 Lecture 11 slide 21
Absolute StabilityAbsolute Stability
• An integration method is “absolutely stable”if the stability region contains the point q = 0.
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2010-11-19 Lecture 11 slide 22
Stability of Difference EquationStability of Difference Equation•• Theorem:Theorem: The solutions of the difference equation
are bounded if and only if all roots of the characteristic equation
z1, …, zr (r is the number of distinct roots) are inside or on the complex unit circle { |z| ≤ 1 } and the roots with modulus 1 are of multiplicity 1.
1
z 1=x
xx
k
i k ii
a x0
0−=
=∑
kk i
ii
a z0
0−
=
=∑
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2010-11-19 Lecture 11 slide 23
Forward EulerForward Euler
nn
n n
n n
x h xxh
qxx x
1n 1
1 1
1 1
x
•
−−
− −
− −
= +
= +
= +
λ
z q(1 ) 0− + = z 1 1 q 1≤ ⇔ + ≤
x x•
= λ q h= λ
Char. eqn.Region of Absolute Stability
Region of Absolute StabilityRegion of Absolute Stability
Numerical Stability:
Given λ < 0 (stable model), choose h small enough to have a stable method
Im q
Re q
q-plane1−2−
xx0
q1 1+ ≤
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2010-11-19 Lecture 11 slide 24
Numerical Stability:q = λh lies in the left-half plane for
Re(λ) < 0 (stable model). Hence |q-1| > 1.
Thus, the method is stable for all h > 0 as long as the model is stable.
However, for Re(λ) > 0 (unstable model), the numerical solution may be stable for h large.
nn
n nhx h xx x
n 1
1
x
•
−
−
= +
= + λ z(1 q) 1 0− − = 1z 1 11 q
≤ ⇔ ≤−
Backward EulerBackward Eulerx x•
= λ
1 2
unstableunstable
q-plane
1xx
q h= λ
q 1 1− ≥
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2010-11-19 Lecture 11 slide 25
Trapezoidal RuleTrapezoidal Rule
TR is stable when the model is stable
n nn n
n nn n
hx x x x
x x xh x
11
11
( )2
( )2
• •
−−
−−
+
= + +λ
= + q q1 z 1 02 2
⎛ ⎞ ⎛ ⎞− − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
x x•
= λ
q1 2z 1 1q1 2
+≤ ⇔ ≤
−q h= λ
Im(q)
unstable
Re(q)
stable
xx
2 12
+≤
−
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2010-11-19 Lecture 11 slide 26
Trapezoidal RingingTrapezoidal Ringing
Problem:
If q= iα (pure imaginary), then the root is z = (1+iα)/(1-iα) |z| = 1.
We get “trapezoidal ringing.”
t
Im(q)
unstable
Re(q)
stable x
x
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2010-11-19 Lecture 11 slide 27
(difference equation)
k k
n ii n i ii i
x x0 0
0•
−−= =
α + β =∑ ∑•x = λx
i n i i n ix xh 0− −λα + β =∑
Stability of LMS MethodsStability of LMS Methods
qk
xi i n ii0( )
0α + β =∑ −=
Consider a Linear Multi-Step method
let q = λh
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2010-11-19 Lecture 11 slide 28
Difference EquationDifference Equation
n n n kk kc q z q z q z1
0 0 1 10 ( ) ( ) ... ( )− −⎡ ⎤= α + β + α + β + + α + β⎣ ⎦
k kk kz zq q q1
0 0 1 1( ) ( ) ... ( ) 0−α + β + α + β + + α + β =
nnx c z= ⋅
• Check the stability of this difference equation
0in)xiqβk
0ii(α =−+∑
=
(char. eqn.)
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2010-11-19 Lecture 11 slide 29
Region of Absolute StabilityRegion of Absolute Stability• The region of absolute stabilityregion of absolute stability of an LMS method is the set of q
= λh (complex) such that all solutions of the difference equation
remain bounded as n ∞ .
qk
xi i n ii( ) 0
0α + β =∑ −=
• A method is “absolutely stable” if the stability region contains the point q = 0.
q = λh
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2010-11-19 Lecture 11 slide 30
Region of Absolute StabilityRegion of Absolute Stability
For what values of q do all the k roots of this polynomial lie in the unit disc { |z| ≤ 1 } ?
k kk kz zq q q1
0 1 1(1 ) ( ) ... ( ) 0−+ β + α + β + + α + β =
( )( )
p zqz
= −σ
k kk
k kk
p z z z
z z z
11
10 1
( ) ...
( ) ...
−
−
⎧ = + α + + α⎪⎨σ = β + β + + β⎪⎩
( ) ( )k k k kk kz z z zq1 1
1 0 1... ... 0− −+ α + + α + β + β + + β =
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2010-11-19 Lecture 11 slide 31
Region of Absolute StabilityRegion of Absolute Stability
S q q p z z z{ | ( ) / ( ), 1}= − σ ≤
The “region of absolute stability” is defined by the set
1
z 1=
Mapping btw complex planes
0 qRe
qIm
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2010-11-19 Lecture 11 slide 32
Conformal MappingConformal Mapping
Basic Results from Theory of Complex Variables1. Mapping –p(z) / σ(z) is conformal.2. Region of “left-hand side” (LHS) to Region of LHS.
p(z)(z)
−σ
q-plane
LHSjz e θ=
Z-plane
z 1≤
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2010-11-19 Lecture 11 slide 33
Application to MidApplication to Mid--Point MethodPoint Method•
−−= + n 1n n 2x x 2h x
2z 1 2qz= +
( )j jq z e e jz
1 1 1 sin2 2
θ θ θ−⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
( )( )
p zqz
= −σ
1
z-plane
The stability region is just the interval [-j +j] on the jω axis.
Hence, the mid-point method is inherently unstable !
jz e ϑ=
1-1
1
-1
q-plane
unstable
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2010-11-19 Lecture 11 slide 34
εε AnalysisAnalysis
1(1
1 121 1 (1 )2 11 1 cos ()
11 ) sin
2 1
j j
q zz
e e
j
θ θεε
θ ε θεε ε
−
⎡ ⎤+
⎛ ⎞= −⎜ ⎟⎝ ⎠⎧ ⎫= + −⎨ ⎬+⎩ ⎭⎧ ⎫⎡ ⎤= + + +−⎢ ⎥+⎣⎨ ⎬⎢ ⎥+⎦ ⎣ ⎦⎩ ⎭
1
z-planeje ϑ
(1 ) je ϑ+ ε0ε > (1 ) je ϑ+ ε
0ε <
0; if 00; if 0
εε
> >< <
[ ... ] always > 1
1-1
1
-1
q-plane
The vertical line [-1, 1] is at the LHS of the blue ellipse and at the RHS of the redellipse.
0ε → ±
z to q
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2010-11-19 Lecture 11 slide 35
Interpretation Interpretation -- 11• For any point outside
of the interval jsinθ in the q-plane, there exist two curves passing that point, one is mapped from a circle |z| > 1, the other from a circuit |z| < 1.
1-1
1
-1
q-plane
2z 1 2qz= +
Poles: 1 2 1ρ ρ⋅ = −Both inside & outsize of |z| = 1 mapped to the region outside of the interval line.
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2010-11-19 Lecture 11 slide 36
Interpretation Interpretation -- 22
1
z-planeje ϑ
(1 ) je ϑ+ ε0ε > (1 ) je ϑ+ ε
0ε <
1-1
1
-1
q-plane
Both inside and outside of the unit circle are mapped to the region outside of the interval [ jsinθ ].
outside
inside