ABCD Parameters of Two Parallel Transmission Line or Networks
Let us consider
that the two-transmission line or network is in parallel. In this theory, we
find out equivalent A, B, C and D parameters of parallel networks or
transmission line
Let
VS =
Sending end voltage to neutral
VR = Receiving end voltage to neutral
IS = Sending end current
IR =
Receiving end current
IS1 =
Sending end current of 1st transmission network
IS2 =
Sending end current of 2nd transmission network
IR1 =
Receiving end current of 1st transmission network
IR2 =
Receiving end current of 2nd transmission network
A1, B1, C1 and D1
– Transmission line parameters of 1st transmission network
A2, B2, C2 and D2
– Transmission line parameters of 2nd transmission network
The sending end
current
IS = IS1
+ IS2 …… ( 1 )
The receiving end currents
IR = IR1 + IR2 …… ( 2 )
Generalized
Constant of the transmission line is given by sending end voltage and sending
end current equation
VS = AVR
+ BIR……. ( 3 )
IS = CVR
+ DIR……....( 4 )
First network
VS = A1VR + B1IR1…..….
( 5 )
IS1 = C1VR + D1IR1……....(
6 )
Second network
VS = A2VR + B2IR2…..….
( 7 )
IS2 = C2VR + D2IR2……....(
8 )
Compare equation (
5 ) and ( 7 )
A1VR +
B1IR1 = A2VR + B2IR2
A1VR – A2VR
= B2IR2 – B1IR1
( A1 – A2 ) VR = B2IR2
– B1IR1….. ( 9 )
As IR =
IR1 + IR2
So IR2 =
IR – IR1
substitute value
of IR2 into equation ( 9)
( A1 – A2 ) VR
= B2 ( IR – IR1 ) –
B1IR1
( A1 – A2 ) VR
= B2 IR – B2 IR1 –
B1IR1
( A1 – A2 ) VR
= B2 IR – IR1 ( B1 + B2 )
IR1 (
B1 + B2 ) = B2 IR – ( A1 – A2 ) VR
Therefore
IR1 = B2 IR – ( A1 – A2
) VR / ( B1 + B2 ) …. ( 10 )
Substitute value
of IR1 into equation ( 5 )
VS = A1VR
+ B1IR1
= A1VR + B1{ B2 IR –
( A1 – A2 ) VR / ( B1 + B2 ) }
= A1VR + B1 B2 IR –
B1( A1 – A2 ) VR / ( B1 + B2 ) }
= A1VR
+ ( B1 / B1 + B2 ){ B2 IR – ( A1 – A2 )
VR }
= ( A1 – { B1 ( A1
– A2 ) / B1 + B2 }VR +
( B1B2 / B1 + B2 ) IR
VS
= {A1B2 + A2B1 / B1 + B2 }VR + ( B1B2 / B1 + B2 ) IR …( 11
)
Compare equation (
11 ) with equation ( 3 )
VS = AVR
+ BIR
A = {A1B2 + A2B1 / B1 + B2 }
B = ( B1B2 / B1 + B2 )
Combine equation (
6 ) and ( 8 )
IS = IS1
+ IS2
IS = C1VR
+ D1IR1 + C2VR + D2IR2
IS = (
C1 + C2 ) VR + D1IR1 + D2IR2
Substitute IR2
= IR – IR1
IS = ( C1
+ C2 ) VR + D1IR1 + D2 ( IR – IR1 )
IS = ( C1
+ C2 ) VR + D1IR1 + D2 IR – D2 IR1
IS = ( C1
+ C2 ) VR + ( D1 – D2 )IR1 + D2 IR
Substitute value
of IR1 from equation ( 10 )
IS = (
C1 + C2 ) VR +
( D1 – D2 )( B2 IR – ( A1 – A2
) VR / ( B1 + B2 ) +D2 IR
Combine VR
and IR
IS ={
( C1 + C2 ) – ( D1 – D2 )( A1 – A2 ) / ( B1 + B2 ) } VR +
B2( D1 – D2 ) / ( B1 + B2 ) IR +
D2 IR
Simplify
IS ={
( C1 + C2 ) – ( D1 – D2 )( A1 – A2 ) / ( B1 + B2 ) } VR +
{
B2D1 – B2D2 + B1D2 + B2D2 ) / ( B1 + B2 )} IR
IS ={
( C1 + C2 ) – ( D1 – D2 )( A1 – A2 ) / ( B1 + B2 ) } VR +
{ B1D2 + B2D1 / ( B1 + B2 )} IR
Compare this
equation with equation ( 4 )
IS =
CVR + DIR
C = { ( C1 + C2 ) – ( D1 – D2 )( A1 – A2 )
/ ( B1 + B2 ) }
D = { B1D2 + B2D1 / ( B1 + B2 )}
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