Page 27 - Circuit Analysis II with MATLAB Applications

The Parallel GLC Circuit

i t + i t + i t = i t
C
L
G
S
or
1
dv
Gv + --- ³ t vt + I + C------ = i t ! 0
d
L 0 0 dt S
By differentiation,
dv 2 dv v di
S
C -------- + G------ + --- = ------- t ! 0 (1.40)
dt 2 dt L dt
To find the forced response, we must first specify the nature of the excitation , that is DC or AC.
i
S
If is DC ( =constant), the right side of (1.40) will be zero and thus the forced response compo-
i
v
S
S
nent v = 0 . If is AC (i = Icos Zt + T , the right side of (1.40) will be another sinusoid and
i
f
S
S
therefore v = Vcos Zt + M . Since in this section we are concerned with DC excitations, the right
f
side will be zero and thus the total response will be just the natural response.
The natural response is found from the homogeneous equation of (1.40), that is,
dv 2 dv v
C -------- + G------ + --- = 0 (1.41)
dt 2 dt L
whose characteristic equation is
2
--- =
Cs + Gs + 1 0
L
or
i
2
----s +
s + G ------- = 0
C LC
from which
G G 2 1
s s = – ------- r --------- – ------- (1.42)
1
2
2C 4C 2 LC
and with the following notations,

1
G
D = ------- Z = ----------- E = D P 2 – Z 2 0 Z nP = Z – D P 2
2
0
P
2C
P
0
LC
­ ° ® ° ¯ ­ ° ° ® ° ° ¯ ­ ° ° ® ° ° ¯ ­ ° ° ° ® ° ° ° ¯ (1.43)
D or Damping Resonant Beta Damped Natural
Coefficient Frequency Coefficient Frequency
where the subscript stands for parallel circuit, we can express (1.42) as
p

1-15 Circuit Analysis II with MATLAB Applications
Orchard Publications