294
Practical MATLAB
® Applications for Engineers
ANALYTICAL Solution
The two node equations are
for node V 1 : (2 − j)V 1 + jV 2 = 5 ∠0°
for node V 2 : jV 1 + (4 − j4)V 2 = −6 ∠−90°
The matrix node equation is given by
2
4
4
5 0
6
90
1
2
Ϫ
Ϫ
ϭ
Њ
Ϫ Ϫ Њ
j
j
j
j
V
V
∠
∠
*
The aforementioned matrix equation can be expressed as [Y] * [V] = [I], where
Y
j
j
j
j
ϭ
Ϫ
Ϫ
2
4
4
and
I ϭ
Њ
Ϫ Ϫ Њ
5 0
6
90
∠
∠
where the phasor current sources are converted to complex numbers as
5 ∠0° = 5 and −6 ∠−90° = −(−j6) = j6
MATLAB Solution
% Script file: nodes
disp(‘*****************************************************’)
disp(‘
System Matrices
’)
disp(‘*****************************************************’)
disp(‘The admittance matrix Y is given by:’)
Y = [2-j j;j 4-4*j]
disp(‘The current matrix I is given by:’)
I = [5;j*6]
disp(‘******************************************************’)
disp(‘
Node voltages
’)
disp(‘*****************************************************’)
2
4
V 1
V 2
5
6
j 3
−j4
−j3
j
−j
0°A
−90°A
FIGURE 3.73
Network of Example 3.15.
CRC_47760_CH003.indd 294
CRC_47760_CH003.indd 294
7/23/2008 1:27:50 PM
7/23/2008 1:27:50 PM
Practical MATLAB
® Applications for Engineers
ANALYTICAL Solution
The two node equations are
for node V 1 : (2 − j)V 1 + jV 2 = 5 ∠0°
for node V 2 : jV 1 + (4 − j4)V 2 = −6 ∠−90°
The matrix node equation is given by
2
4
4
5 0
6
90
1
2
Ϫ
Ϫ
ϭ
Њ
Ϫ Ϫ Њ
j
j
j
j
V
V
∠
∠
*
The aforementioned matrix equation can be expressed as [Y] * [V] = [I], where
Y
j
j
j
j
ϭ
Ϫ
Ϫ
2
4
4
and
I ϭ
Њ
Ϫ Ϫ Њ
5 0
6
90
∠
∠
where the phasor current sources are converted to complex numbers as
5 ∠0° = 5 and −6 ∠−90° = −(−j6) = j6
MATLAB Solution
% Script file: nodes
disp(‘*****************************************************’)
disp(‘
System Matrices
’)
disp(‘*****************************************************’)
disp(‘The admittance matrix Y is given by:’)
Y = [2-j j;j 4-4*j]
disp(‘The current matrix I is given by:’)
I = [5;j*6]
disp(‘******************************************************’)
disp(‘
Node voltages
’)
disp(‘*****************************************************’)
2
4
V 1
V 2
5
6
j 3
−j4
−j3
j
−j
0°A
−90°A
FIGURE 3.73
Network of Example 3.15.
CRC_47760_CH003.indd 294
CRC_47760_CH003.indd 294
7/23/2008 1:27:50 PM
7/23/2008 1:27:50 PM