Large Change Sensitivities
Sensitivities based on the first derivatives are good
tool to study network response when changes in its
parameters are small. When some elements change
significantly, this approximated result may be
unacceptable. A specialized approach called large
change sensitivity method is used. It is useful for
fault diagnosis, symbolic analysis, even for analysis
with decomposition.
Consider a nominal network described by equation
T0 X 0  W
when T0 is NxN matrix, and the output function is
F0 =
dT X0
Denote the change in a nominal value by h i  h i 0   i .
The equations for the modified system are
T
(i) TX  T0  P  Q  X  W
F=
dT X0
where P and Q are parts of the incidence matrices
corresponding to the changed elements, and
  diag ( i mm ) (usually m << n).
From (i) we can obtain the solution for X = X0 + X
1
(ii)
X  T0 1 W  T0 P Q T X


X0

z

let us introduce a new vector z equal to z = QTX,
then
QTX = -1z and substitute (ii) for X
QTT0-1W - QTT0-1Pz = -1z
So the problem of finding an updated solution of the
modified system have been reduced to the solution of
mxm system for z
(iii) 
1
 Q T T0 P  z  Q T T0 W
1
1
after which the original variables X can be obtained
from (ii).
Example
G6=1
1
G1=1
2
3
+
G4=1
G6=1
G5=1
V2
+
g3v2
g3=1
G2=1
Vout
J=1
-
-
Consider the network shown and assume that we
want to investigate the effect of change in g3 from 1
to 2 (i.e. 3 = 1). The nominal admittance matrix and
its inverse are
 2 1 0 
T0   1 3  1 ,
 0 0 2 
6 2 1 
1
1
T0  2 4 2 ,
10
0 0 5
det T0  10
we have
  1 
QT  0 1 0
So from (iii)
PT  0 0 1
W T  1 0 0

1


 2 
1
1
 Q T To P z   1     z  Q T T0 W
10  

12
2
1
z
 z
10
10
6
substitute it for z in (ii) to obtain
6 / 10 1
 35 
 7 / 12 
1 1
1
1
1
X  T0 W  T0 P z  2 / 10  2     10    2 / 12 

   10 6   60 

 0  5
 5
 1 / 12
Nominal response
Change in the response due to  3
Analysis of Decomposed Networks Based on
Large Change Sensitivities:
Consider a network decomposed into s subnetworks
by means of ideal switches.
f1
f8
f2
f3
f3
f5
f7
f6
The entire network is described by its system
equations TX = W
where
T1


T


 F1
1 
 2 
 

s 
F  I
T2

Ts
Fs
F2
where Fi contain switch variables (0 or 1) and i are
parts of the incidence matrix which correspond to
switches in different subnetworks. When all
switches are open Fi = 0, and the coefficient matrix
becomes block triangular
T1


T0  




T2

Ts
1 
 2 
 

s 
 I
In order to use the results of the large change
sensitivity analysis we will call T0 a "nominal"
matrix. It corresponds to the "nominal" system
equations in which subnetworks are disconnected:
T0X0 = W
Where
T0
1
1
T11
T1 1

1
1
T
T
2
2 2




1
1
Ts
Ts s


I




  X 0  T0 1W




So, each block can be solved (factorized) separately
and X0 calculated. To calculate X we use large scale
sensitivity approach (ii)
X = X0 - T0-1Pz
where z is obtained from
1
1
 1  QT T0 P z  QT T0 W
The only changes we consider to represent
connection of blocks in a single network are the
changes in switch from 0 to 1, therefore  = I and the
size m of the auxiliary system is equal to the number
of switches.


Example
Use the large-scale sensitivity based decomposition
approach to solve the following network.
Assume that s = 1
1
C1=1F
4
G1=1S
2
F1
6
G10=1S
8
+
G9=1S
G4=1S
G8=1S
1*Vx
Vx
-
E=1V
C2=1F
G2=1S
3
G3=1S
F2
5
G6=1S
7
9
G5=1S
C3=2F
We have two subsystems separated by two switches
 G4
 sC 1
0  v2   EG 1 
G1  G4  sC 1

 v   EG 

G
G

G

G

sC
0

sC
4
2
3
4
2
2
 3    2 
T1 X 10  
  sC 1
0
sC 1
0  v 4   0 

  

v
0

sC
0
sC
0


2
2  5 

We know v2 = v4 and v3 = v5 (try to see why) so it is
enough to invert 2 x 2 matrix but we will work on
matrix 4 x 4 anyway.
So the nominal solution of the first subsystem is
3
 1
T1 X 10  
 1

0


1 
T1 



0.6
0.2
0.6
0.2
1
4
0
1
1
0
1
0
0.2
0.4
0.2
0.4
0
 1
0

1
0.6
0.2
1.6
0.2
v 2   1 
v   1 
 3  
v 4   0 
   
 v5   0 
0.2 
 1  0.8
   
0.4 
1  1  0.6
X 10  T1

 0  0.8
0.2 

   
1.4 
 0  0.6
and the nominal solution of the second subsystem is
0
G 8
0   v6   0 
G 8  G 9
 0
v   0 
G

G
0

G
5
6
6
  7   
T2 X 20  
 1 G8
0
G 8  G 10
 1   v8   0 

   
0

G
0
G

sC
6
6
3   v9   0 

2
0
T2  
0

0
0
2
0
1
1
0
2
0
0
1
1
3

0.5

0

1
, T2  

0



0
0.05 0.25
0.6 0
0.1 0.5
0.2 0
0.1 
0 
0 
0.2 
, X 20   
0 
0.2 


0.4 
0 
The nominal solution vector is as follows
X 0  [ X10T , X 20T , I F1 , I F 2 ]T  0.8 0.6 0.8 0.6 0 0 0 0 0 0
T
The coefficient matrix of the complete system is


T1




0 0
T 
0 0
0 0

0 0
0 0

0 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
F1
0
 F1
0
0
0
0
F2
0
 F2
0
0
T2



1
0 

0
1 
1
0 

0
1 
0
0 

0
0 
F1  1
0 

0
F2  1
0
0
0
0
So we can identify the following incidence matrices
0
0
1  
1

0
0
 1 0 
 0  1
0

2  


0
0 0



1
0 0
and topological matrices describing switch location
0
0

0

0
0
P
0

0
0

1

0
0
0
0

0
0
0

0
0
0

1
0 0
0 0


1 0


 0 1   
 1 0   1 
Q
   2 
0

1

  
0 0 I 


0
0


1 0


 0 1 
1 0
 

0 1 
To get the corrector vector z we use

1
 Q T T0 P  z  Q T T0 W

1
1
X0





 

 0  
 1 0
1
T
 Q T0  ~  


1 0 
 0 1 
0 1 

 

last 2 columns of T0 1 

 z1 
T
z   Q X 0
 2
and solve the reduced size equation

 1 0
T



 1
 0 1



T2
 T111  
 1    z1 
1
1
I T2 2      1T T1 1  T2 T2 2
  I    z2 




 zz   00..68
1
 2 
where


0 0 1 0  
1
T
 1 T1  1  

0 0 0 1  


0.6 0.2 0.6 0.2 
0.2 0.4 0.2 0.4 
0.6 0.2 1.6 0.2 

0.2 0.4 0.2 1.4 
0
0

1

0
0
0.6
0 0 0 1 0 0.2

0 0 0 0 1 1.6


1
0.2
and
0.5

  1 0 0 0  0
T 1
 2 T2  2  

 0  1 0 0  0

0
0.05 0.25 0.1   1
0.6 0 0.2   0
0.1 0.5 0.2   0

0.2 0 0.4   0
0
 1 0.5 0.05

,
0   0 0.6 

0
0.2
0.4 1.6 0.2

0.2 0.2 1.4 

1.4 

Then we can compute z using
 2.1 0.25
0.2 2 


 z1  0.8
 z   0.6
 2  
and the corrector vector z is equal to:
 z1   0.3494
 z    0.2651

 2 
The final solution vector is obtained by combining
the nominal solution vector X0
X 0  [ X10T , X 20T , I F1 , I F 2 ]T  0.8 0.6 0.8 0.6 0 0 0 0 0 0
T
with the correction term as follows
 0.6 0.2 

 0.2 0.4 




 1.6 0.2 




 0.2 1.4 



- 0.5 - 0.05  0.3621 
T01 Pz  


0
0.6
0.2638


 



0 - 0.1



0 - 0.2 


 -1 0 






0 - 1 

0.2627 

0.1759 



0.6120


0.4410 


- 0.1880
1
 X  X 0  T0 Pz  
- 0.1590


- 0.0265


- 0.0530



- 0.3494



- 0.2651
,
0.5373
0.4241
0.1880

0.1590
0.1880

0.1590
0.0265

0.0530
0.3494

0.2651
Symbolic analysis
The main advantage of symbolic analysis is the
insight it provides and accuracy of the solution. For
any fixed frequency we may consider a transfer
function of parameter deviations i.
N 1 ,  2 , ... ,  m 
F 1 ,  2 , ... ,  m  

D1 ,  2 , ... ,  m 

m
m
m
m
m
m
i
m
i i j
m m
i
m
i  j j k
m m
i
i
i j
i
i  j j k
a   ai i   aij i j   aijk i j k  ...
b   bi1   bij i j   bijk i j k  ...
where
a  N  k 0
ai 
b  D  k 0
N
 i
bi 
 k 0
2 N
aij 
 i  j
k  1,... , m
D
 i
2D
bij 
 1 1
 k 0
 k 0
 k o



First, we can assume that all nominal parameter
values are equal to 1, so
hi 0  1
 in actual circuit  i  hi  1 where hi is
the original network parameter.
Formulas for derivatives of D and N w.r.t. variable
parameters are derived in pp 312-316, and are as
follows:
LD
 det T0 det F̂  j1 , j2 , ..., j L 
 j1 j 2 ...  jL
and
LN
 det T0 det Fˆ  j1 , j 2 , ..., j L , m  1
 j1 j 2 ...  jL
T
T
Q T  1
Q
T
P
Q
T
W



0
0
Fˆ   T
1
   d T   T0 P,
T
d T P d T0 W   
1
where
1
1
0
X

Example
Consider a network described by the nodal equations
(was analyzed before)
1
G1=1
2
3
+
1
G1=1
2
+
3
+
G4=1
G5=1
V2
+
g3v2
g3=1
G2=1
Vout
J=1
-
-
Using nodal analysis we have
G1  G4
T    G1
 0
 G1

 G6 
G2  G6 
0
G1  G5  G6
g 3  G6
From direct symbolic analysis we have:
V3 
G1 (1  g3 )
*J
1  2G1  2G2  g3  3G1G2  G1g3
where we can observe dependence on three symbolic
values.
We will develop the equivalent result by using large
change sensitivity approach with parameters
G1 , G2 and g3 deviating from their nominal values.
We have the parameter location indicated by
 1 0 0
1  1 0
P   1 0 0, QT  0 0 1, and
 0 1 1
0 1 0
0
d  0
1
where d is a selector of the output V3 . As before we
have
 2 1 0 
T0   1 3  1
 0 0 2 
T0
1
6 2 1 
1
 2 4 2 det T0  10
10
0 0 5
 4 1 1
1
T01P   2 2 2
10
 0 5 5
and
6 
1
1
X  T0 W  2
10
0
so the matrix needed to compute symbolic
coefficients is as follows:
QT  1
F   T  T0 P
d 
^

 6 1 1

5 5
1 0
X 
10  2 2 2

5 5
0
4
0
2

0

we have
a  N  det T0 det F ( 4)  0
a1 
n
10 6 4
 det T0 det F (1,4 ) 
0
G1
100 0 0
a2 
N
10 5 0
 det T0 det F ( 2,4) 
0
G 2
100 5 0
a3 
N
10 2 2
 det T0 det F (3,4) 
 1
g 3
100 5 0
6 1 4
a12 
N
10
 det T0 det F (1,2,4) 
0
G1G2
1000
0
6
a13 
N
10
 det T0 det F (1,3,4) 
2
G1g3
1000
0
5
0 0
5
0
1 4
2
2
5
0
1
( 40  60)  1
100
a 23
5
N
10

 det T0 det Fˆ ( 2,3,4) 
2
G 2 g 3
1000
5
5
2
5
0
2 0
0
a123
6
0
N
10

 det T0 det Fˆ (1,2,3,4) 
G1 G 2 g 3
10000  2
0
1 1
5
5
2
2
5
5
4
0
0
2
0
b  D  det T0 det Fˆ (0)  det T0  10
D
10
 det T0 det Fˆ (1) 
6 6
G
10
D
10
b2 
 det T0 det Fˆ ( 2) 
5 5
G 2
10
b1 
b3 
D
10
 det T0 det Fˆ (3) 
2 2
g 3
10
b12 
D
10 6
 det T0 det Fˆ (1,2) 
G1 G 2
100 0
b13 
D
10 6
 det T0 det Fˆ (1,3) 
G1 g 3
100  2
b23 
D
10 5
 det T0 det Fˆ (2,3) 
G 2 g 3
100 2
b123
1
3
5
1
1
2
5
0
2
6
D
10

 det T0 det Fˆ (1,2,3) 
0
G1 G 2 g 3
1000
2
1 1
5
5 0
2
2
as a result we obtain a transfer function in terms of
changes in G1, G2 and g3.
V3 
  3  1 3
J
10  61  5 2  2 3  31 2  1 3
we can obtain the original function if we substitute
1  G1  1,  2  G 2  1,  3  g 3  1.
Fault Diagnosis
The equations for a faulty system are


TX  T0  PQ T X  W
(i)
So we can evaluate the deviations from the nominal
circuit response using
1
X  T0 1W  T0 PQ T X



(ii)
z
or simply
1
X  T0 PQT X
(iii)
The problem in fault diagnosis is to determine
changes  from measurements in selected test points.
X is unknown solution vector in the faulty circuit.
Since the fault location is unknown so matrices P and
Q are also not known.
Consider first the single fault location. If the faulty
element f is located on intersection of rows i and j
and columns k and l then using the unit vector
notation P = ei - ej and QT = (ek - el)T and we can
1
T
modify (iii) using 0 P  si  s j , where s i  s j is a
difference of two columns of the inverse of the
1
coefficient matrix S  T0 . We also have QT X = xk
– xl so the equation (iii) can be simplified to
X  ( si  s j ) f ( xk  xl )  ( si  s j ) f
This result can be easily extended to multiple faults
in which case we can show that
 f 1 
 
 f 2

X  ( si1  s j1 ) ( si 2  s j 2 ) ... ( sif  s jf )
  ...   S 
f
f
 
  ff 
(iv)
In equation (iv), coefficients  fv represent f faulty
parameters and are related to the deviations of
parameter values from their nominal through scalar
equations
 fv ( xkv  xlv )   fv where v  1,..., f
(v)
So, in order to solve fault diagnosis problem, we
must first construct a (nxf) matrix Sf, which is
obtained from (nxn) matrix S by subtracting columns
of S which correspond to row numbers indicating
location of individual circuit parameters. Therefore,
(iv) represents (hopefully unique) expansion of
X into a linear combination of the column vectors
from the matrix Sf. In order for the expansion to be
unique, we must have the number of faults f less than
the number of measurements m. If m<n (most often)
then we only consider the corresponding rows of X
and matrix Sf. We will get
X m  S mf  f
A simple way of finding which columns of matrix Sm
should be selected, can be obtained by eliminating
X m from matrix Sm using Gauss elimination step
and then finding ambiguity groups of the resulting
matrix. A QR factorization approach can be used to
accomplish the task of finding ambiguity groups.
After finding the correct set of faulty parameters
which satisfy (iv), we can obtain all parameters  fv .
Using (iv) with full matrix Sf we can determine the
full vector X and subsequently determine the actual
solution vector X. Finally the deviations of
parameters can be obtained from:
  f1 


 f 1   ( x k 1  x l1 ) 
     f 2 
 f2  

 ...   ( x k 2  x l 2 ) 
...
  


 ff     ff 
 (x  x ) 
lf 
 kf
Example
The nominal circuit shown has all conductances
expressed in Siemens.
G6=3
G4=1
1
+
V2
2
G7=2
3
4
-
+
G2=2
J=1
G1=1
G3=2
g3v2
g3=4
G5=2
G8=1
Vout
The inverse of the nominal coefficient matrix in the
circuit shown is as follows:
S  T01
 0.18
 0.23

  0.11

 0.15
0.23 0.11 0.15
0.34 0.16 0.22
0.16 0.30 0.18

0.22 0.18 0.34
The response of this circuit was measured at nodes 1
and 4 and the change of the response vector was
found to be proportional to difference of column 2
and 4 of the nominal coefficient matrix inverse
 0.008 
X m  
 0.1 ( s2 m  s4 m )

 0.012
where s2 and s4 are columns of S and m are selected
measurement nodes (1 and 4). Assume that there is
only one fault in the network.
From equation for X m we can determine that the
fault is located between nodes 2 and 4 and that the
fault diagnosis variable λ =0.1.
Since
 0.18 
 0.23
1

X 0  T0 W  
  0.11


 0.15
and
 0.008 
 0.012 

X  0.1( s2  s4 )  
 0.002


 0.012
Then the actual solution vector X is
 0.18   0.008   0.188 
 0.23  0.012   0.218



X  X 0  X  
  0.11  0.002  0.112

 
 

 0.15  0.012  0.162
and the element deviation δ is


 0.1

 1.786
x2  x4  0.056
So, in this case the element G6 is faulty and its actual
value is 4.786 [S], however uniqueness of this
solution depends whether there is a single element
who effects currents between nodes 2 and 4.