Alternating Current Analysis
289
ANALYTICAL Solution (series case)
|Z()| = |R + j(X L − X C )|
|Z()| = √
_______________
R
2
+ (X L − X C )
2
The magnitude equation is given by
Z
R
L
C
( )
ϭ
Ϫ
Ϫ
2
2
1

 

 








The phase equation is given by
∠

 

 
Z
X
X
R
L
C
( ) arc tan
ϭ
Ϫ
ANALYTICAL Solution (parallel case)
Y
R j L
j C
( )
ϭ ϩ
ϩ ր
1
1
1
1
The magnitude equation is given by
Y
R
C
L
( )
ϭ
ϩ
Ϫ
1
1
2

 

 

 

 
2
The phase equation is given by
∠

 

 












Y
C
L
R
( ) arc tan
ϭ
Ϫ
1
1
MATLAB Solution
% Script file: impedance _ plots
clf
R = 10;
L = 5e-3;
C = 12.5e-6;
w = [100:100:10000];
Z = R+ j*(w*L-1./(C*w));
% series case
subplot(2,2,1);
plot(w,abs(Z));
title(‘Mag. [Z(w)] vs w (series case)’);
ylabel(‘Mag[Z(w)] in Ohms’);
grid on;
subplot(2,2,2);
plot(w,angle(Z)*180/pi); grid on;
title(‘Phase[Z(w)] vs w (series case)’);
ylabel(‘Phase angle in degrees’);
CRC_47760_CH003.indd 289
CRC_47760_CH003.indd 289
7/23/2008 1:27:49 PM
7/23/2008 1:27:49 PM