1 ee 616 computer aided analysis of electronic networks lecture 3 instructor: dr. j. a. starzyk,...
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EE 616 Computer Aided Analysis of Electronic Networks
Lecture 3
Instructor: Dr. J. A. Starzyk, ProfessorSchool of EECSOhio UniversityAthens, OH, 45701
09/12/2007
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Review and Outline
Review of the previous lecture* Network scaling
* Thevenin/Norton Analysis
* KCL, KVL, branch equations
* Sparse Tableau Analysis (STA)
* Nodal analysis
* Modified nodal analysis
Outline of this lecture* Network Equations and Their Solution -- Gaussian elimination -- LU decomposition(Doolittle and Crout algorithm) -- Pivoting
-- Detecting ILL Conditioning
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A is n x n real non-singular X is nx1; B is nx1;
bxM
SeY
Problems:
Direct methods: find the exact solution in a finite number of steps
-- Gaussian elimination, LU decomposition, Crout, Doolittle)
Iterative methods: produce a sequence of approximate solutions hopefully converging to the exact solution
-- Gauss-Jacobi, Gauss-Seidel, Successive Over Relaxation (SOR)
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Gaussian Elimination Basics
Reminder by 3x3 example
11 12 13 1 1
21 22 23 2 2
3 331 32 33
M M M x b
M M M x b
x bM M M
11 1 12 2 13 3 1M x M x M x b
21 1 22 2 23 3 2M x M x M x b
31 1 32 2 33 3 3M x M x M x b
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Gaussian Elimination Basics – Key idea
Use Eqn 1 to Eliminate x1 from Eqn 2 and 3
11
13
11
132
11
121 M
bx
M
Mx
M
Mx
11 1 12 2 13 3 1M x M x M x b
21 21 2122 12 2 23 13 3 2 1
11 11 11
M M MM M x M M x b b
M M M
31 31 3132 12 2 33 13 3 3 1
11 11 11
M M MM M x M M x b b
M M M
Multiply equation (*) by –M21 and add to eq (2)
Eq.1 divided by M11
(*)
Multiply equation (*) by –M31 and add to eq (3)
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GE Basics – Key idea in the matrix
11
11 12 13
21 21 2122 12 23 12 2 2 1
11 11 11
31 3132 12 33 12
311 113 3
0
0
xb
M M M
M M MM M M M x b b
M M M
M MM M M M
MM M x b
1
111
bM
Pivot
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GE Basics – Key idea in the matrix
11 12 13
21 2122 12 23 12
11 11
3132 12
1131 2133 12 23 12
11 112122 12
11
0
0 0
M M M
M MM M M M
M M
MM M
MM MM M M M
M MMM M
M
11
212 1
11
2
3132 12
1131 213 1 2 1
11 112122 12
113
x b
Mb b
Mx
MM M
MM Mb b b b
M MMM M
Mx
Continue this step to remove x2 from eqn 3
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11
11 12 13
21 21 2122 12 23 12 2 2 1
11 11 11
31 3132 12 33 12
311 113 3
0
0
xb
M M M
M M MM M M M x b b
M M M
M MM M M M
MM M x b
1
111
bM
22M 23M
32M33M
3b2b
GE Basics – Simplify the notation
Remove x1 from eqn 2 and eqn 3
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22 23 2
32 3233 23 3 2
111 12 13
2 2
1
2
3
2 2
0
0 0
x
M M b
M MM M b b
M M
bM
x
M
x
M
GE Basics – Simplify the notation
Remove x2 from eqn 3
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GE Basics – GE yields triangular system
AlteredDuring
GE
11 12 13 1 1
22 23 2 2
33 3 3
0
0 0
U U U x y
U U x y
U x y
11
2
3
22
11 12 1
2
3
2
3 3
3
30
0 0
x
M M b
b
M
M M M
x
x b
~ ~
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11
33
33
yx
U
2 23 32
22
y U xx
U
1 12 2 13 31
11
y U x U xx
U
11 12 13 1 1
22 23 2 2
33 3 3
0
0 0
U U U x y
U U x y
U x y
GE Basics – Backward substitution
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1
1
212 12
11
3 322
2
13 3 1
11 2
by
Mb by
M
M Mby b
M Mb
32
1 1
212 2
113 3
31
11 22
1 0 0
1 0
1
y bM
y bM
y bMM
M M
GE Basics – RHS updates
3
2
1
3
2
1
3231
21
1
01
001
b
b
b
y
y
y
LL
L
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GE basics: summary
(1) M x = b
U x = y Equivalent systemU: upper triangle
(2) Noticed that:Ly = b L: unit lower triangle
(3) U x = yLU x = b M x = b
GE
Efficient way of implementing GE: LU factorization
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Solve M x = bStep 1
Step 2 Forward Elimination
Solve L y = bStep 3 Backward Substitution Solve U x = y
=M = L U
Gaussian Elimination Basics
Note: Changing RHS does not imply to recompute LU factorization
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LU decomposition
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LU decomposition
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LU decomposition – Doolittle example
5052
1183
1241
6442
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LU factorization (Crout algorithm)
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LU factorization (Crout algorithm)
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Properties of LU factorization
Now, let’s see an example:
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LU decomposition - example
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Relation between STA and NA
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Pivoting for Accuracy:
Example 1: After two steps of G.E. MNA matrix becomes:
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Pivoting for Accuracy:
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Pivoting for Accuracy:
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Pivoting for Accuracy:
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Pivoting Strategies
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Error Mechanism
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Detecting ILL Conditioning
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Detecting ILL Conditioning