Post on 17-Jan-2016
Circuits TheoryCircuits TheoryExamplesExamples
Newton-Raphson Method
Formula for one-dimensional case:
Series of successive solutions:
If the iteration process is converged , the limit is the solution of the equationf(x)=0.
)k()k()k()k( xfx'fxx11
,...x,x,x )()()( 210
0)x(f
0xf )(Multidimensional case:
)k()k()k()k( xfxJxx11
where:
n
nnn
n
n
x)(f
x)(f
x)(f
x)(f
x)(f
x)(f
x)(f
x)(f
x)(f
xxx
xxx
xxx
xJ
21
2
2
2
1
2
1
2
1
1
1
)()k(
kxxxJxJ
JACOBIANMATRIX
ALGORALGORITHMITHM
STEP 0 )o(0k x STARTING POINT
STEP 1 )k()k( , xJxfCalculate
STEP 2 Solve the equation:
)()()()1()1()()(
)()1()(
,, kk xfbxxyxJA
byA
kkkkk
kkk
STEP 3 find )()1()1( kkk xyx
check STOP conditions
If the current solution is not acceptable:1kk
GO TO 1
EXAMPLE of STOP PROCEDUREEXAMPLE of STOP PROCEDURE
1
1 )( ky
2
1 )( kxf
NNoo
NoNo
k=k+1GOTO 1
YesYes
YesYes
*)1k( xx STOP
• Stop condition parameter 1
1
212
1
1
1
1 )k(
n
k
n
)k(kk xx...xxy
• Stop condition parameter
2
2121
11 k
nn
k xf...xf1kxf
2
Numerical EXAMPLESNumerical EXAMPLES
Example 1
Solve the following set of nonlinearequation using the Newton’s Method:
02
0143
01023
2
3
3
21
3
2
2
3
1
32
2
1
xxx
xxx
xxx
T321 xxx,)( x0xf
2
143
1023
2
3
3
213
3
2
2
3
12
32
2
11
xxx)(f
xxx)(f
xxx)(f
x
x
x
Starting point (first approximation):
T)0( 111x
T)0( 194)(f xCalculate:
)()(' 0xx
0 xJxf
1x1x1x
322
221
1
3
2
1x2x31
3x2x3
12x6
231
323
126
)0()1()( xfyxJ 0 where: )0()1()1( xxy
1
9
4
231
323
126
)1(3
)1(2
)1(1
y
y
y
123
9323
426
)1(3
)1(2
)1(1
)1(3
)1(2
)1(1
)1(3
)1(2
)1(1
yyy
yyy
yyy (1a)
(1b)
(1c)
3)0(
333)0(
232)0(
131
2)0(
323)0(
222)0(
121
1)0(
313)0(
212)0(
111
byayaya
byayaya
byayaya
(1a)
(1b)
(1c)
Let us assume )0(yy
3333232131
2323222121
1313212111
byayaya
byayaya
byayaya
(1a)
(1b)
(1c)
Gauss elimination computer scheme
STEPSTEP 1 ELIMI 1 ELIMINATE NATE y y11 fromfrom b i c b i c:
1y2y3y
9y3y2y3
4yy2y6
321
321
321
Multiply by
and add to 1b63
aa
11
21
7y25
y
9y3y2y3
2y21
yy3
32
321
321
1y2y3y
9y3y2y3
4yy2y6
321
321
321
Multiply by
and add to 1c61
61
a
a
11
31
31
y6
11y
310
1y2y3y32
y61
y31
y
32
321
321
New set : )()( 22 byA
)2(33
)2(332
)2(32
)2(23
)2(232
)2(22
)2(13
)2(132
)2(121
)2(11
byaya
byaya
byayaya
(2a)
(2b)
(2c)
31
y6
11y
310
7y25
y1
4yy2y6
32
32
321
(2a)
(2b)
(2c)
31
y6
11y
310
7y25
y1
4yy2y6
32
32
321
(2a)
(2b)
(2c)
Elimination scheme repeat for equations 2b i 2c:
Multiply by
add o 2c1
3/10a
a)2(
22
)2(32
370
y325
y310
32
371
y661
3
)3()3( byA
)3(33
)3(33
)3(23
)3(232
)3(22
)3(13
)3(132
)3(121
)2(11
bya
byaya
byayaya
(3a)
(3b)
(3c)
371
y661
7y25
y1
4yy2y6
3
32
321
(3a)
(3b)
(3c)
Back substitution part:
328.261
142ab
y )3(33
)3(3
3
Setting y3 to 3b:
61142
y
7y25
y1
4yy2y6
3
32
321
Multiply by
add to 3b
25
a )3(23
61142
y
6172
y
4yy2y6
3
2
321
1a )3(13
2a )3(12
328.2
180.1
115.0
y
y
y
100
010
001
3
2
1
3283
1802
8150
0
3
1
3
0
2
1
2
0
1
1
1
.
.
.
xy
xy
xy
)()(
)()(
)()(
)( 1x
Because )()()( 011 xxyy
It is the first calculated approximation of the solution.Next iterations form a converged series:
006.3
010.2
002.1)2(x
3
2
1)3(x *)4(
3
2
1
xx
ExampleExample 2 2
Nonlinear circuit having two variables (node voltages)
R 3
VS 3
R2
j1 j
4
5i
6iv
5v
6
1
2
i 3
i2
e1
e2
Data:
1)(
)(
6666
25
35555
kvedvgi
cbvavvgi
VkAd
AcV
Ab
V
Aa
VvAjAj
RR
S
11,1
,1,1,1
,3,4,1
,3,2
23
341
32
R 3vS 3
R2
j1 j
4
5i
6iv
5v
6
1
2
Nodal equations:
013
312215
2
1
jR
veeeeg
R
e S1
2 04
3
31226215
j
R
veeegeeg S
013
312215
2
1
jR
veeeeg
R
e S
043
31226215
j
R
veeegeeg S
Jacobian matrix:
2
)(2)(31
)(2
)(31
)(2)(31
)(2)(3
11
)(
212
213
21
221
3
212
213
212
21
32
kedke
eebeeaR
eeb
eeaR
eebeeaR
eebeea
RR
eJ
We choose starting vector:
0
0)0(e
4
1)( )0(ef
Calculate:
333.1333.0
333.0833.0)( )0(eJ
Applying N-R scheme:
)0()1()( efyeJ 0 where: )0()1()1( eey
4
1
333.1333.0
333.0833.0)1(
2
)1(1
y
y
hence:
6673
66721
2
1
1
.
.
y
y)(
)(
667.3
667.2)0(
2)1(
2
)0(1
)1(1
)1(2
)1(1
ey
ey
e
e
STOP CRITERIA not satisfied:
1211
1 100010 .,.,y )(
455.34
0)( )1(ef
k=k+1:
455.40333.1
333.1833.1)( )1(eJ
Second NR iteration
)1()2()1( efyeJ
where:)1()2()2( eey
46311
10
455403331
333183312
2
2
1
.
.
y
y
..
..)(
)(
hence:
8730
63502
2
2
1
.
.
y
y)(
)(
794.2
032.2)1(
2)2(
2
)1(1
)2(1
)2(2
)2(1
ey
ey
e
e
for k=7: )6()7()6( efyeJ where: )6()7()7( eey
002.0
003.0
105.6885.0
718.0225.1)7(
2
)7(1
y
y
hence:
47
2
7
1
101811
0010
.
.
y
y)(
)(
629.1
807.1)6(
2)6(
2
)6(1
)6(1
)7(2
)7(1
ey
ey
e
e
Because:
6
6)(
10653.2
10689.2)( 7ef
629.1
807.1*2
*1
)7(2
)7(1
e
e
e
e
2
6
272
712
272
711
7
10777.3
,,
eefeefef
Briefly about:Briefly about:
Iterative models of nonlinear elements
Iterative NR model of nonlinear resistor (voltage Iterative NR model of nonlinear resistor (voltage controled)controled)
vfi i i
v v
circuit
11 '' kkkkkk vvfvvfii
ki~ kG kkkk vvfii '~ kk vfG '
From NR method:From NR method:
Model iterowany opornika (Model iterowany opornika (66))
11 ~ kkkk vGii
ki~
kG 1ki 1kv
ExampleExample 3 3
Newton-RaphsonNewton-Raphson
Iterative model method
R 3
VS 3
R2
j1 j
4
5i
6iv
5v
6
1
2
i 3
i2
e1
e2
Data:
1)(
)(
6666
25
35555
kvedvgi
cbvavvgi
VkAd
AcV
Ab
V
Aa
VvAjAj
RR
S
11,1
,1,1,1
,3,4,1
,3,2
23
341
32
Scheme for (k+1) iterationScheme for (k+1) iteration
v S3
R 2
R3
j1 j4
v 5(k+1)
i5(k+1)
G5
(k)
(k)i5~ (k)
i6~ G 6
(k)
v 6(k+1)
i 6(k+1)
1
2
11
ke 12
ke
0~51
3
31
11
2
5
12
11
2
11
k
Skk
k
kkk
ij
R
vee
R
ee
R
e1
v S3
R 2
R3
j1 j4
v 5(k+1)
i5(k+1)
G5
(k)
(k)i5~ (k)
i6~ G 6
(k)
v 6(k+1)
i 6(k+1)
1
2
11
ke 12
ke
2
0~~654
3
31
11
2
6
12
5
12
11
kk
Skk
k
k
k
kk
iij
R
vee
R
e
R
ee
v S3
R 2
R3
j1 j4
v 5(k+1)
i5(k+1)
G5
(k)
(k)i5~ (k)
i6~ G 6
(k)
v 6(k+1)
i 6(k+1)
1
2
11
ke 12
ke
0~51
3
31
11
2
5
12
11
2
11
k
Skk
k
kkk
ij
R
vee
R
ee
R
e1
2
0~~654
3
31
11
2
6
12
5
12
11
kk
Skk
k
k
k
kk
iij
R
vee
R
e
R
ee
3
351
53
12
352
11
~
11111
R
vij
RRe
RRRe
Sk
kk
kk
1
2
3
3
654
356
12
53
11
11
11111
R
v
RRj
RRRe
RRe
Skk
kkk
kk
• For starting vector:
0
00
2
0
10
v
v)(v
0' 02
015
05 eegG
02
01
05 eev
1'~ 05
05
05
05 vvfii
055
05 vgi
• We calculate parameters of the models:
0' 026
06 egG
02
06 ev
0'~ 06
066
06
06 vvgii
066
06 vgi
• For nonlinear element g6:
Linear equations for the first approximationLinear equations for the first approximation::
4
1
333.1333.0
333.0833.01
2
11
e
e
667.3
667.2)1(
2
)1(1
e
e
Solution for k=1=i5
x1y11
Second stepSecond step
371.110
0
468.40333.1
333.1833.12
2
21
e
e
794.2
032.2)2(
2
)2(1
e
e
Solution for k=2
Briefly about:Briefly about:
Forward Euler Method (Explicit)
Backward Euler Method (Implicit)
Forward Euler Method (Explicit)
),( 111 kkkk txfhxx
Backward Euler Method (Explicit)
),(1 kkkk txfhxx
Backward Euler Method (Explicit) is based on the following Taylor series expansion
2
1
hdt
dxhtx
htxx
ktk
kk
),(1 kkkk txfhxx
EX A M PLE .
E
L
CR
i (t)L
t=0
Cu (t)v (t)
Cvs
HL
FC
R
VvVv CS
25.0
1
200
200)0(,100
eudt
diL C
L (1)
R
u
dt
duCi CC
L (2)
State vector:
2
1
L
C
x
x
i
ux (3)
f r o m ( 1 ) a n d ( 2 ) :
SCL
LCc
vL
vLdt
di
iC
vRCdt
du
11
11
( 4 )
o r :
SvLx
x
L
CRC
1
0
01
11
2
1x ( 5 )
,.
L
CRC
04
101050
01
1164
A
( 6 )
VvvL
S 100,4
010
B ( 7 )
I n i t i a l c o n d i t i o n s :
5.0
2001
200
0
0
0
0
2
1
Rvi
u
x
x
sL
Cox . ( 8 )
F E M :
11
11 )(
kk
kkk
h
fh
xx
xxx
Dla n=1, korzystając z wzoru (9) i uwzględniając, że V200u0u )0(CC i A5.0i)0(i )0(LL otrzymamy:
eL
1u
L
1hii
iC
1u
RC
1huu
)0(C)0(L)1(L
)0(L)0(C)0(C)1(C
(11)
F o r t h e s t e p h = 1 0 - 4 :
46.0400800105.0
15010
5.0
10200
20010200
4)1(
664
)1(
L
C
i
v
N e x t s t e p k = 2 :
44.04006001046.0
12110
46.0
10200
15010150
4)2(
664
)2(
L
C
i
v
vC(tk)
iL(tk)
Table for 10 iterations n 0 1 2 3 4 5 6 7 8 9 vC
200 150 121 104.5 95.41 90.685 88.506 87.789 87.891 88.429
iL 0.5 0.46 0.44 0.4316 0.4298 0.4316 0.4354 0.4399 0.4448 0.4497
B E M :
n1n
n1nn
h
)(fh
xx
xxx
SkCkLkL
kLkCkCkC
vL
vL
hii
iC
vRC
huv
11
11
)()1()(
)()()1()(
O r i n m a t r i x f o r m :
)e(h
)(h
n1n
nn1nn
BAxx
BuAxxx
f o r k = 1 , s e t t i n g : Vvv CC 2000 )0(
Aii LL 5.0)0( )0(
4004105.0
101020010200
)1(4
)1(
6)1(
6)1(4
)1(
CL
LCC
vi
ivv
solving in terms of )1(Cv and )1(Li :
935.164)1( Cv 474.0i )1(L
h=0.0001.
vc(tk)
iL(tk)
Example with nonlinear capacitorExample with nonlinear capacitor
• FEM
211
11
RRv
R
v
dt
dqc
S
0)0(,2,1,10 221 qqvRRVv cS
2410 qq
)410( 211
kqhqqkk
FEM stepsFEM steps
110*1.0)410( 2001
qhqq
6.1)410(1.01)410( 2112
qhqq
5811.1)410( 2889
qhqq
BEM step 1BEM step 1
)410( 2101
qhqq
0
)410(
11
11011
2
qfequationnonlinearofsolution
qqhqqf
0
)410(
1
01
2
xfequationnonlinearofsolution
xxhqxf
7655644.01
q
00 x
Using N-R method with starting point Using N-R method with starting point
)1(1)1()1()(11
' kkkk xfxfxx
1101.010 1)1( x
12)1(
011)1()( 4101.018.0 kkkkk xxqxxx
7777.0)161.0(18.01 1)2( x
7656.0)4( x
00 x
BEM step 2BEM step 2
)410( 2212
qhqq
0
)410(
22
22122
2
qfequationnonlinearofsolution
qqhqqf
0
)410(
2
12
2
xfequationnonlinearofsolution
xxhqxf
1947.12
q
after N-R procedure with new starting point
Using N-R method with starting point Using N-R method with starting point
2403.1)1( x
1947.1)4( x
7656.00 x
02)0(1
10)0()1( 410(1.018.0 xxqxxx
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6