A.51 becomes very cumbersome if A is of an order larger than 3 by 3.2 Today the digital computer eliminates the drudgery of having to find the inverse of a matrix in numerical applicatio
Trang 1If we let C = adj A • A and use the technique illustrated in Section A.7,
we find the elements of C to be
Therefore
C =
=
cu
C\2
^13
^21
c 2 2
c 2 3 C31 C32
^33
~ - 8
0
0 Jet A
= 9
-= 18
= 27
21 + 4 = - 8,
- 14 - 4 = 0,
- 7 - 20 = 0,
= - 1 6 + 24 - 8 = 0,
= - 3 2 + 16 + 8 = - 8 ,
= - 4 8 + 8 + 40 = 0,
= 5
-= 10
= 15
0
- 8
0
•u
9 + 4 = 0,
- 6 - 4 = 0,
- 3 - 20 = - 8
0"
0
- 8 _
= - 8
" l 0
0 1 _0 0
0
0
1
A square matrix A has an inverse, denoted as A ', if
A- 1 A = AA_1 = U (A.50) Equation A.50 tells us that a matrix either premultiplied or postmultiplied
by its inverse generates the identity matrix U For the inverse matrix to exist, it is necessary that the determinant of A not equal zero Only square matrices have inverses, and the inverse is also square
A formula for finding the inverse of a matrix is
A-! =
The formula in Eq A.51 becomes very cumbersome if A is of an order larger than 3 by 3.2 Today the digital computer eliminates the drudgery
of having to find the inverse of a matrix in numerical applications of matrix algebra
It follows from Eq A.51 that the inverse of the matrix A in the previ-ous example is
-1 _
A- 1 = -1/8
9
16
5
- 7
8
- 3
- 4
8
~4_
-1.125 0.875
2 - 1 -0.625 0.375
- 1 A _ 1 i
You should verify that A l A = AA = U
0.5
- 1 0.5
2 You can learn alternative methods for finding the inverse in any introductory text on
matrix theory See, for example, Franz E Hohn, Elementary Matrix Algebra (New York:
Macmillan, 1973)
Trang 2A.9 Partitioned Matrices
It is often convenient in matrix manipulations to partition a given matrix into submatrices The original algebraic operations are then carried out in terms of the submatrices In partitioning a matrix, the placement of the partitions is completely arbitrary, with the one restriction that a partition must dissect the entire matrix In selecting the partitions, it is also neces-sary to make sure the submatrices are conformable to the mathematical operations in which they are involved
For example, consider using submatrices to find the product
C = AB, where
A =
1
5
1
0
0
2
4
0
1
2
3
3
2
- 1
1
4
2
- 3
0
- 2
5
1
1
1
0
and
B
2
0
- 1
3
0 Assume that we decide to partition B into two submatrices, Bn and
B 2 j ; t h u s
B B21
Now since B has been partitioned into a two-row column matrix, A must be partitioned into at least a two-column matrix; otherwise the multiplication cannot be performed The location of the vertical partitions of the A matrix will depend on the definitions of Bn and B2i For example, if
then An must contain three columns, and A12 must contain two columns Thus the partitioning shown in Eq A.52 would be acceptable for execut-ing the product AB:
C =
1 2
5 4 -1 0
0 1
0 2
3
3
2
- 1
1
4 5~
2 1
3 1
0 1
2 0_
2~
0
- 1
3 0_
(A.52)
Trang 3If, on the other hand, we partition the B matrix so that
Bn - 2 .0 and B?i =
- 1
3
0 then A n must contain two columns, and A12 must contain three columns
In this case the partitioning shown in Eq A.53 would be acceptable in exe-cuting the product C = AB:
C =
1
5
1
0
0
2
4
0
1
2
3
3
2
- 1
1
4
2
- 3
0
- 2
5~
1
1
1 0_
2
0
- 1
3
0
(A.53)
For purposes of discussion, we will focus on the partitioning given in
Eq A.52 and leave you to verify that the partitioning in Eq A.53 leads to the same result
From Eq A.52 we can write
C = [ An A12]
B„ = AnBn + A12B 21- (A 54)
It follows from Eqs A.52 and A.54 that
A,,B l l * » 1 l
1 2
5 4 -1 0
0 1
0 2
3~
3
2
1
1
2~
0 _ - ! _
=
~ - l "
7
- 4
1 _ - ! _
A12B21
-4 5~
2 1
- 3 1
0 1 _ ~ 2 0_
"3"
.0 =
~ 12"
6
- 9
0 _ - 6 _
and
11
13
- 1 3
1
- 7
The A matrix could also be partitioned horizontally once the vertical partitioning is made consistent with the multiplication operation In this simple problem, the horizontal partitions can be made at the discretion of
Trang 4the analyst Therefore C could also be evaluated using the partitioning shown in Eq.A.55:
1
5
1
0
0
?
4
0
1
2
3
3
2
- 1
1
5
1
1
1 0_
2~
0
- 1
3 0_
(A.55)
From Eq A.55 it follows that
A n A21
A12
A,?
B,i
B„ LC 2 1
(A.56)
where
You should verify that
Cj! - AuBn + AI 2B2|,
C2i = A21B11 + A22B21
C„ = 1 2 3
5 4 3 + 4 5 2 1
- 1
7 +
12
6
= 11
13
c„ =
-1 0
0 1
0 2
2~
1 1_
2~
0 _ - ! _ +
" - 3
0 _ - 2
r
1 0_
"3*
.0
- 4
1
- 1 +
- 9
0
- 6
=
- 1 3
1
- 7
and
C =
11
13
- 1 3
1
- 7
We note in passing that the partitioning in Eqs A.52 and A.55 is conformable with respect to addition
Trang 5720 The Solution of Linear Simultaneous Equations
A.10 Applications
The following examples demonstrate some applications of matrix algebra
in circuit analysis
Example A.l
Use the matrix method to solve for the node
volt-ages V\ and v 2 in Eqs 4.5 and 4.6
Solution
The first step is to rewrite Eqs 4.5 and 4.6 in matrix
notation Collecting the coefficients of «, and v 2
and at the same time shifting the constant terms to
the right-hand side of the equations gives us
1.7«! - 0.5¾ = 10,
(A.57) -0.5«] + 0.6«2 = 2
It follows that in matrix notation, Eq A.57 becomes
1.7
- 0 5
•0.5 0.6
10
2
or
where
AV = I,
(A.58)
(A.59)
A
A —
V =
I =
1.7 .-().5
V
'10"
2
•
•
- 0 5 "
0.6
To find the elements of the V matrix, we
pre-multiply both sides of Eq A.59 by the inverse of
A; thus
or
A_ 1AV = A_ 1I
Equation A.60 reduces to
UV = A "T,
V = A " I
(A.60)
(A.61)
(A.62)
It follows from Eq A.62 that the solutions for
V\ and v 2 are obtained by solving for the matrix product A- 11
To find the inverse of A, we first find the cofactors of A Thus
AH = (-1)2(0.6) = 0.6, A12 = ( - l )3( - 0 5 ) = 0.5,
A2i = ( " l )3( - 0 5 ) = 0.5, A22 = (-l)4(l-7) = 1.7
The matrix of cofactors is
(A.63)
B 0.6
0.5
and the adjoint of A is
adj A = B7 = The determinant of A is
0.5 1.7
0.6
L0.5
0.5 1.7
(A.64)
(A.65)
d e t A 1.7
- 0 5
-0.5 0.6 (1.7)(0.6) - (0.25) = 0.77
(A.66)
From Eqs A.65 and A.66, we can write the inverse
of the coefficient matrix, that is,
(A.67) A"1 1
0.77
0.6 .0.5
0.5 1.7 Now the product A- 11 is found:
A - I = ^
77
0.6 0.5"
.0.5 1.7
loor 7"
77 U.4 _ "
"10 2
" 9.09"
.10.91
It follows directly that
V
v 2
-' 9.09-' .1 3.91 '
(A.68)
(A.69)
or vj = 9.09 V and v 2 = 10.91 V
Trang 6A.10 Applications 7 2 1
Example A.2
Use the matrix method to find the three mesh
cur-rents in the circuit in Fig 4.24
Solution
The mesh-current equations that describe the
cir-cuit in Fig 4.24 are given in Eq 4.34 The constraint
equation imposed by the current-controlled voltage
source is given in Eq 4.35 When Eq 4.35 is
substi-tuted into Eq 4.34, the following set of
equations evolves:
25/,- - 5/2 - 20/3 = 50,
- 5 / , - 4z2 + 9/3 = 0
In matrix notation, Eqs A.70 reduce to
AI = V, where
A =
(A.70)
(A.71)
and
25 - 5
- 5 10
- 5 - 4
'*]
h h_
1
V =
r 5 0 "
0
0
- 2 0
- 4
9
It follows from Eq A.71 that the solution for I is
We find the inverse of A by using the relationship
A-' = _ adjA
d e t A (A.73)
To find the adjoint of A, we first calculate the
cofac-tors of A Thus
An = (-1)2(90 - 16) = 74,
A12 = ( - l )3( - 4 5 - 20) = 65,
A13 = (-1)^(20 + 50) = 70,
A21 = ( - l )3( - 4 5 - 80) = 125,
A22 =
A 23 =
A 3 1 = I
A 3 2 = <
A33 = (
;-l)4(225 - 100) = 125,
; - l )5( - 1 0 0 - 25) = 125,
; - l )4( 2 0 + 200) = 220,
- 1 )5( - 1 0 0 - 100) = 200, -1)6(25() - 25) = 225
The cofactor matrix is
B
74 65 70
125 125 125
220 200 225
(A.74)
from which we can write the adjoint of A:
adj A = B r
74
65 _70 The determinant of A is
d e t A =
25 - 5 - 2 0
- 5 10 - 4
- 5 - 4 9
125
125
125
220
200
225
(A.75)
= 25(90 - 1 6 ) + 5 ( - 4 5 - 80) - 5(20 + 200) = 125
It follows from Eq A.73 that
-1 _ 1
125
74 125 220
65 125 200
70 125 225
(A.76)
The solution for I is
1
125
74 125
65 125
70 125
220
200
225
50
0
0
=
29.60 26.00 28.00 (A.77)
The mesh currents follow directly from Eq A.77 Thus
(A.78)
or /j = 29.6 A, i 2 = 26 A, and /3 = 28 A Example A.3 illustrates the application of the matrix method when the elements of the matrix are complex numbers
h
h
_*3_
=
"29.6"
26.0 _28.0_
Trang 7722 The Solution of Linear Simultaneous Equations
Example A.3
Use the matrix method to find the phasor mesh
cur-rents I, and I2 in the circuit in Fig 9.37
Solution
Summing the voltages around mesh 1 generates
the equation
(1 + /2)1, + (12 - /16)(1, - I2) = 150/0" (A.79)
Summing the voltages around mesh 2 produces
the equation
(12 - /16)(I2 - Ii) + ( 1 + /3)I2 + 39IV = 0.(A.80)
The current controlling the dependent voltage
source is
I, = (Ii - I2)- (A.81) After substituting Eq A.81 into Eq A.80, the
equations are put into a matrix format by first
collect-ing, in each equation, the coefficients of I, and I2: thus
(13 - /14)1, - (12 - /16)I2 = 150/0°,
(27 + /16)1, - (26 + /13)¾ = 0
Now, using matrix notation, Eq A.82 is written
(A.82)
where
A =
I
AI = V,
13 - /14 - ( 1 2 - /16) [27 + /16 - ( 2 6 + /13)
(A.83)
and V 150/0
0
It follows from Eq A.83 that
The inverse of the coefficient matrix A is found
using Eq A.73 In this case, the cofactors of A are
^ n = ( - l )2( - 2 6 - / 1 3 ) = - 2 6 - / 1 3 ,
A12 = ( - l )3( 2 7 + /16) = - 2 7 - / 1 6 ,
A2i = ( - 1 )3( - 1 2 + /16) = 12 - /16,
A22 = (-1)4(13 - /14) = 13 - /14
The cofactor matrix B is
B = The adjoint of A is adj A = B7 = The determinant of A is
d e t A =
( - 2 6 - /13) ( - 2 7 - /16) ( 1 2 - / 1 6 ) ( 1 3 - / 1 4 )
( - 2 6 - /13) (12 - /16)
L ( - 2 7 - / 1 6 ) ( 1 3 - / 1 4 ) J
(A.85)
(A.86)
(13 - /14) (27 + /16)
(12 - /16) (26 + /13)
= - ( 1 3 - /14)(26 + /13) + (12 - /16)(27 + /16)
= 60 - /45 (A.87) The inverse of the coefficient matrix is
A~! =
( - 2 6 - / 1 3 ) ( 1 2 - / 1 6 ) L(-27 - / 1 6 ) (13 - / 1 4 ) J
(60 - /45) Equation A.88 can be simplified to
(A.88)
60 + /45
5625
( - 2 6 - /13) (12 - /16) ( - 2 7 - /16) (13 - /14)
1
375
- 6 5 - /130 96 - /28
- 6 0 - /145 94 - /17 (A.89) Substituting Eq A.89 into A.84 gives us
1
375
( - 6 5 - /130) (96 - /28) ( - 6 0 - /145) (94 - /17)
150/0C
0 ( - 2 6 - /52)
( - 2 4 - /58)
It follows from Eq A.90 that
I, = ( - 2 6 - /52) = 58.14/-116.57° A,
h = ( - 2 4 - /58) = 62.77/-122.48° A
(A.90)
(A.91)
In the first three examples, the matrix elements have been numbers—real numbers in Examples A.l and A.2, and complex numbers in Example A.3 It
is also possible for the elements to be functions Example A.4 illustrates the use of matrix algebra in a circuit problem where the elements in the coeffi-cient matrix are functions
Trang 8A.10 Applications 723
Example A.4
Use the matrix method to derive expressions for
the node voltages V x and V 2 in the circuit in Fig A 1
Solution
Summing the currents away from nodes 1 and 2
generates the following set of equations:
R + VxsC + (K, - V2)sC = 0,
f + (½ ~ VfisC + (V 2 - V g )sC = 0
(A.92)
Letting G = \/R and collecting the coefficients of
V] and V2 gives us
(G + 2sC)Vi - sCV 2 = GVp
-sCVi + (G + 2sC)V 2 = sCV R
Writing Eq A.93 in matrix notation yields
AV = I, where
(A.93)
(A.94)
G + 2sC -sC
v2 , and
—5 C
G + 2sC_
I
-_sCV
i
V =
It follows from Eq A.94 that
V - A ' l (A.95)
As before, we find the inverse of the coefficient
matrix by first finding the adjoint of A and the
determinant of A The cofactors of A are
An = (-1) 2[G + 2sC] = G + 2sC,
A12 = (-l)\sC) = sC,
A2i = {-l)\sC) - sC,
A22 = (~1)4[G + 2sC] = G + 2sC
The cofactor matrix is
G + 2sC
B
-sC
sC
G + 2sC (A.96)
and therefore the adjoint of the coefficient matrix is
I G + 2sC sC
sC G + 2sC
Figure A.l • The circuit for Example A.4
The determinant of A is
G + 2sC sC
d e t A = „ _ „ _
sC G + 2sC = G2 + 4sCG + 3.v2C2
(A.98) The inverse of the coefficient matrix is
A~l =
G + 2sC sC
sC G + 2sC
(G 2 + AsCG + 3.v2 C 2 ) (A.99)
It follows from Eq A.95 that
G + 2sC sC
sC G + 2sC sCV a
{G z + AsCG + 3s l C l )
(A.100)
Carrying out the matrix multiplication called for in Eq.A.100 gives
V 2 J (G 2 + 4.vCG + 3.v 2 C 2 )
(G2 + 2sCG + s 2 C 2 )V g (2sCG + 2s*C 2 )V n
(A.101)
Now the expressions for V\ and V 2 can be written directly from Eq A 101; thus
(G2 + 2sCG + s2C2)VK V\ = ,w> ^ „ ^ » ((G A-102)
2 + 4sCG + 3s2 C 2 ) '
and
2 ^
l C z )V<,
(G2 + 4sCG + 3s2C2) (A.103)
Trang 9In our final example, we illustrate how matrix algebra can be used to analyze the cascade connection of two two-port circuits
Example A.5
Show by means of matrix algebra how the input
variables V x and Iy can be described as functions of
the output variables V 2 and I 2 in the cascade
con-nection shown in Fig 18.10
Solution
We begin by expressing, in matrix notation, the
relationship between the input and output variables
of each two-port circuit Thus
(A.104) and
(A.105)
Vi
/J
v\
/',
«il
.«21
r«u
k i
-«12
~«22 « 1 2 ]
~«22 J
v2
L/2
\v2
lh
Now the cascade connection imposes the constraints
V' 2 = V\ and 1' 2 = -J\ (A.106)
These constraint relationships are substituted into
Eq A.104 Thus
«h
«21
«ii
«21
- « 1 2
- « 2 2
«12
«22 I'u
L/'I
(A.107)
The relationship between the input variables (V h /j)
and the output variables (V 2 , J 2 ) is obtained by
substituting Eq A.105 into Eq A.107 The result is
«11
«23
«12
«22
«11 L« 2 'i
- « 1 2
"«22J
v2
L/2 (A.108) After multiplying the coefficient matrices, we have
V 2
Vi
LA
(«il«ll + «12«2l) («21«11 + «22«2l)
- ( « 1 1 « ' ] 2 + «12«22) -(«21«12 + «22«22)J L/2
(A.109) Note that Eq.A.109 corresponds to writing Eqs 18.72 and 18.73 in matrix form
Trang 10Appendix
Q Complex Numbers
Complex numbers were invented to permit the extraction of the square roots
of negative numbers Complex numbers simplify the solution of problems
that would otherwise be very difficult The equation x 2 + 8x + 41 = 0,
for example, has no solution in a number system that excludes complex
numbers These numbers, and the ability to manipulate them algebraically,
are extremely useful in circuit analysis
B.l Notation
There are two ways to designate a complex number: with the cartesian, or
rectangular, form or with the polar, or trigonometric, form In the
rectangular form, a complex number is written in terms of its real and
imaginary components; hence
n = a + jb, (B.l)
where a is the real component, b is the imaginary component, and ; is by
definition V - l 1
In the polar form, a complex number is written in terms of its
magni-tude (or modulus) and angle (or argument); hence
where c is the magnitude, 6 is the angle, e is the base of the natural
loga-rithm, and, as before, j = V - T In the literature, the symbol /6° is
fre-quently used in place of e jB\ that is, the polar form is written
n = c/6° (B.3)
Although Eq B.3 is more convenient in printing text material, Eq B.2 is of
primary importance in mathematical operations because the rules for
manipulating an exponential quantity are well known For example, because
(77> = y*»,then(e^)" = e jn6\ because v"v = l / y \ then e'10 = l / e ^ ; a n d
so forth
Because there are two ways of expressing the same complex number,
we need to relate one form to the other The transition from the polar to
the rectangular form makes use of Euler's identity:
1 You may be more familiar with the notation i = y/^ In electrical engineering, / is used
as the svmbol for current, and hence in electrical engineering literature,/ is used to denote