1. ECE 3144 Lecture 10
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University

2. Reminder from Lecture 9 , Matrix construction AX = B
If A-1 exists, X=A-1B
Method 1:
Cramer’s rule: ,
Where Ai is the matrix obtained from A replacing the ith column of A by B.

3. Example: Cramer’s Rule
2x+z = 2
x+y 3
3x+2y+z 1 ,

4. Example: cont’d

5. Nodal Analysis
Simple circuits in Chapter 2 =>can be reduced to one containing only two nodes =>can be solved by a single equation for the one unknown quantity, typically the voltage between the two nodes.
In nodal analysis
A circuit contains n (n > 2) nodes
Circuit variables: node voltage
Any one node from these n nodes can be selected as the reference node. Usually the ground node or the node to which the largest number of branches are connected is selected. The voltage for the reference node is zero.
The node voltages are defined with respect to the reference node. Usually select the node voltages as positive with respect to the reference node. If some of the node voltages are actually negative, the analysis will indicate it.
With the selection of the reference node, the node voltages for other nodes, say a, b, c, etc. can be represented as va, vb, vc, etc. So voltage at node a is va, which is really the voltage at node a with respect to the ground node.

6. How to relate the node voltages with voltage across resistors and current through resistors?b is + - vc + - + - R2 R3 A
Notice that the voltage across R2 is equal to the node voltage at node a and the voltage across R3 is equal to the node voltage at node b.
Apply KVL to loop A: va+vc+vb = 0 => vc = va – vb
The equation expresses the voltage across R1 as a function of node voltages, va and vb. Thus the label for the voltage across R1 is va – vb instead of vc. We already know that voltage across R2 is node voltage va, voltage across R3 is node voltage vb.
Applying Ohm’s law to each resistor expresses the current in each resistor as a function of node voltages and resistances.
Applying KCL to get the node equations at node a and node b. ,

7. Given:
va = 12 V: the voltage at node a is 12V with respect to the reference node e
vb = 3 V: the voltage at node b is 3V with respect to the reference node e
vc = 3/2 V: the voltage at node c is 3/2V with respect to the reference node e
vd = 3/8 V: the voltage at node c is 3/8V with respect to the reference node e
ve = 0 V a v1 b v3 c v5 d I1 I3 I5
9k
3k
9k +
6k
4k
3k - I2 I4 e
v1 = va – vb = 12-3 = 9 V
v3 = vb – vb = 3-3/2 = 3/2 V
v5 = vc – vd = 3/2-3/8 = 9/8 V
I1 = v1/9k = 9V/9K = 1 mA
I3 = v3/3k = 1.5V/3K = 0.5 mA
I5 = v5/9k = 9/8V/9K = mA
I2 = (vb-ve)/6k = 3V/6K = 0.5 mA
I4 = (vc-ve)/4k = 3/2/4K = mA

8. General Rule
In general, if the node voltages are known, using Ohm’s law can calculate the
current through the resistors
Node m i R
Node n + - + - vm vn

9. Remember: It is very important to specify the reference node
Remember: It is very important to specify the reference node. Without the reference node, there is no meaning for the voltage at a node.
va = 4 V
vb = -2 V R2 a b R1 R3 c
There are three 3 nodes in this network, a, b, c. Node c is selected as the reference node.
Node a has a voltage 4 V with respect to reference node c.
Node b has a voltage –2 V with respect to reference node c.
However the voltage at node a is 6 V with respect to node b and the voltage at node b is –6 V with respect to node a.
If node b is selected as the reference node, what are the voltages at node a, b,c?

10. Nodal analysis
For a N-node network, since one node is selected as the ground node (v= 0), the number of linearly independent KCL equations is (N-1)
This set of (N-1) linearly independent equations, when solved, will yield to (N-1) unknown node voltages.
Based on the node voltages, the current through each resistor can be calculated by using the Ohm’s law.

11. Nodal analysis with independent current sourcesva vb a b
Given iA, iB, and resistors R1, R2, R3, find the node voltages va, vb, vc i2 R2 i3 iA iB R3 R1 i1 c
There are 3 nodes, thus only 2 independent equations are needed.
Node c is selected as the reference node => vc = 0.
At node a applying KCL: -iA + i1 + i2 = 0 =>
-iA + G1(va – vc) + G2(va-vb) = (1)
At node b applying KCL: -i2 + iB + i3 = 0 =>
-G2(va-vb) + iB + G3(vb – vc) = (2)
(1) + (2) => -iA + G\(va-vc) + iB + G3(vb – vc) = (3)
Equation (3) is the KCL equation at reference node c.
Any two of the three equations can be used to derive the third one. Thus there are only 2 linearly independent equations for this 3-node circuit.
In general, for a circuit with N nodes, only (N-1) linearly independent equations are needed to solve the (N-1) unknowns va and vb.

12. Nodal analysis with independent current sources
In the matrix format: GV = I
If G-1 exists, V = G-1I. The matrix operations we have learned from Lecture 9 can be used to solve the problem.
It is not difficult to find that the formula shows here is the Ohm’s law in the matrix format. Later on, we will explain a systematic way on how to get the Ohm’s law in matrix format.
One more example will be shown in order to get the general procedures for determining the formats for matrices G, V and I.

13. Example: nodal analysis with only independent current sourcesiA iB R3 R2 R1 R4 R5 i3 i5 i4 i2 i1 v1 v2 v3 1 2 3
In node 1 applying KCL:
i1-iA+i2-i3 = 0 or
G1v1 – iA + G2(v1-v2) –G3(v3-v1) = 0 or
(G1+G2+G3)v1 –G2v2-G3v3 = (1)
In node 2 applying KCL:
-i2 + i4 –i5 = or
-G2(v1-v2) + G4v2-G5(v3-v2) = 0 or
-G2v1 + (G2+G4+G5)v2 – G5v3 = (2)
In node 3 applying KCL:
i3 + i5 +iB = or
G3(v3-v1) + G5(v3-v2) + iB= or
-G3v1 – G5v2 + (G3+G5)v3 = (3)

14. Example: cont’d
Three equations for three unknowns. The equations can be transformed into the matrix format:
The G matrix is in symmetrical format.
Actually for all the circuits containing only resistors and independent current sources, the G matrix is in this symmetrical form.
In next lecture, we are going to provide more interpretations of the coefficients in the matrices and also explanations on why the G matrix is symmetric.
GV = I

15. Homework for lecture 10 Due Feb 4 Problem 3.1, 3.2, 3.3, 3.4, 3.7
You do not need to provide the matlab result for Problem 3.7.