Download ECE 3144 Lecture 4

ECE 3144 Lecture 9
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Reminder from Lecture 8
• Exam 1 on Feb 6, Wednesday
• Chapter 1: Basic concepts
– Charge (q), current (i), and their relationship
– Voltage (v), work/energy (w), power (p), and their
relationships.
– Passive sign convention
• The positive reference of voltage v(t) is at the same
terminal the the current variable i(t) is entering.
• How to determine absorbing or supplying energy?
– Active elements and passive elements
– Active elements
• Independent sources
• Dependent sources.
2
Reminder from Lecture 8
• Chapter 2: Resistive Circuits
–
–
–
–
–
–
–
–
–
–
Resistor, Resistance, Resistivity
Ohms’law V = I*R: only linear resistors satisfy Ohm’s law.
Power P = V*I= V2/R = I2R
Kirchoff’s current law (KCL): The algebraic sum of the currents
leaving (entering) a node is zero.
Kirchoff’s voltage law: The algebraic sum of the voltages around
any loop path is zero.
Single loop circuit: resistors in series and voltage divider
Single node circuit: resistors in parallel and current divider
Circuits containing a single source and a series-parallel
interconnection of resistors
Wye-delta or Delta-to-Wye transformations
Circuits containing dependent sources
3
Chapter 3: Nodal and Loop Analysis
Techniques
• We are going to learn a systematic manner in this chapter so that we
can calculate all current and voltages in circuits with multiple nodes
and loops.
• Matrix Construction
• Nodal analysis
• Loop analysis
• Operational amplifier
• Circuits with operational amplifier
4
Matrix Construction
Given a linear system of n linearly independent equations with n unknowns:
a11 x1  a12 x 2  ...  a1n x n  b1
a 21 x1  a 22 x 2  ...  a 2 n x n  b2
.
.
.
.
.
.
.
.
.
.
.
.
a n1 x1  a n 2 x 2  ...  a nn x n  bn
Its equivalent matrix form is:
AX = B
With
 a11
a
21

A .

 .
a n1
a12
a 22
.
.
an2
... a1n 
... a 2 n 
... . 
... . 
... a nn 
 b1 
b 
2

B .
 
.
bn 
 x1 
x 
2

X  . 
 
 . 
 x n 
5
Matrix construction
If the n equations are linearly independent, then the linear system given has
the unique solution. In another word, if A  0 , then the system has one
and only one solution as follows:
X  A 1 B
Where A-1 is the inverse matrix of A.
A
A*

A
1
|A| is the determinant of A.
A* is the adjoint matrix of A. A* is defined as follows: Let A=[aij] be an n by n
matrix. Let Mij be the (n-1) by (n-1) submatrix of A obtained by deleting the ith row
and jth column of A. Then the cofactor Aij of element aij is defined as
Aij  (1)(i  j ) M ij
A 
n
a
j 1
ij
Aij
6
Matrix operation
Then the adjoint matrix A* is the matrix whose (i,j)th element is the
cofactor Aij of aij.
 A11
A
12
*

A  .

 .
 A1n
A21
A22
.
.
A2 n
...
...
...
...
...
An1 
An 2 
. 
. 
Ann 
7
Example 1: matrix operation
2x+z
x+y
3x+2y+z
=
=
=
2
3
1
2
A 
1

3
0
1
2
1
0

1

,
2
B  3
1 
2 0 1
A  1 1 0  2 *1 *1  0 * 0 * 3  1 * 2 *1  1 *1 * 3  3 * 0 * 0  1 *1 * 2  2  0  4  3  2  1
3 2 1
2
1
A*   1  1
 1  4
 1
1 
2 
 1 2  1
A
1
A 
  1  1 1 
| A|
 1  4 2 
*
,
2  1 2  2  6  1   7 
 x
1
 y   A 1 B   1  1 1  * 3    2  3  1     4 
 

   
 

 z 
 1  4 2  1  2  12  2  12
8
Cramer’s Rule
Given a linear system of n linearly independent equations with n unknowns:
a11 x1  a12 x 2  ...  a1n x n  b1
a 21 x1  a 22 x 2  ...  a 2 n x n  b2
.
.
.
.
.
.
.
.
.
.
.
.
a n1 x1  a n 2 x 2  ...  a nn x n  bn
If |A| 0
The unique solutions is
X  A 1 B
Cramer’s rule:
x1  
A1
A
, x2 
A2
A
,...., x n 
An
A
Where Ai is the matrix obtained from A replacing the ith column of A by B.
9
Example 2
2x+z
x+y
3x+2y+z
=
=
=
2
A 
1

3
2
3
1
0
1
2
1
0

1

,
2
B  3
1 
2 0 1
A1  3 1 0  2 *1 *1  0 * 0 *1  2 * 3 *1  1 *1 *1  1 * 3 * 0  2 * 0 * 2  7
1 2 1
2 2 1
A2  1 3 0  2 * 3 *1  2 * 0 * 3  1 *1 *1  1 * 3 * 3  1 *1 * 2  1 * 0 * 2  4
3 1 1
2 0 2
A3  1 1 3  2 *1 *1  2 * 2 *1  3 * 3 * 0  2 *1 * 3  1 *1 * 0  2 * 3 * 2  12
3 2 1
x1 
A1
A
 7, x2 
A2
A
 4, x3 
A3
A
 12
10
Homework for Lecture 9: Due Feb 4
• Problem 1: solve x1, x2, x3
2 x1  x 2  3 x3  1
4 x1  2 x 2  5 x3  4
2 x1  2 x3  6
Lecture 8 notes,
• Problem 2: solve x1, x2, x3, x4
x1  2 x2  3 x3  2 x4
2 x1  x2  2 x3  3 x4
3 x1  2 x2  x3  2 x4
2 x1  3 x2  2 x3  x4
6
8
4
 8
11
12