Open circuit and Short circuit impedance of a two-port network in terms of ABCD parameters
From the generalized ABCD parameter equation
V1=AV2 – BI2
I1=CV2 – DI2
Applying Open circuit at port-2; I2=0 we get
V1= AV2; I1= CV2
Z1O = V1/I1= A/C ——————————(i)
Applying the short circuit at port-2; V2=0, we get
V1= – BI2
I1=-DI2
Or
Z1S = V1/I1 = B/D ————————–(ii)
Now looking from port and applying:
Open circuit port-1, I1=0
0 = CV2 – DI2
CV2= DI2
Z2O = V2/I2 = D/C ————————–(iii)
Now short circuiting the port-1, V1=0, we get
0 = AV2 – BI2
AV2 = BI2
Z2s = V2/I2 = B/A —————————(iv)
Ratio of open circuit to short circuit ratio
Z1O/ Z1S = (B/D)(C/A) ————————–(v)
To find ABCD parameters
Z2O – Z2S = D/C – B/A = (AD –BC)/CA
But for the two network to be reciprocal, the condition is AD-BC =1
Therefore:
Z2O – Z2S = 1/CA ———-(vi)
CA = 1/ (Z2O – Z2S )
From eqn(i) Z1O = A/C
Multiplying eqn(i) and eqn(v)
CA(A/C) = 1/ (Z2O – Z2S )* Z1O
A2 = Z1O / (Z2O – Z2S )
Or
A= √( Z1O / (Z2O – Z2S ) ——————–(vii)
From eqn(i) i.e. Z1O = A/C C = A/ Z1O
Therefore from eqn (vii):
C = √( Z1O / (Z2O – Z2S ) /Z1O
C = √( 1/ Z1O (Z2O – Z2S ) ———–(viii)
From short circuit equations (iv) Z2s = B/A
B = AZ2s
= √( Z1O / (Z2O – Z2S )* Z2s
From short circuit equations (ii) Z1s = B/D
D = B/ Z1s
= √( Z1O / (Z2O – Z2S )* Z2s/ Z1s
= A* Z2s/ Z1s ——————————–(ix)
= Z20*√( 1/ Z1O (Z2O – Z2S )——————(x)
Summery, the ABCD parameters in terms of open and short circuit impedance:
A = A= √( Z1O / (Z2O – Z2S ) B = AZ2s = √( Z1O / (Z2O – Z2S )* Z2s
C = √( 1/ Z1O (Z2O – Z2S ) D = A* Z2s/ Z1s = Z20*√( 1/ Z1O (Z2O – Z2S )
Image Impedance in terms of ABCD parameters
In a two port network, if the impedance at input port with ouptput port impedance as Zi2 is Zi1 and simultaneously, the output impedance with input impedance being Zi1 is Zi2, then the impedances Zi1 and Zi2 are called as the image impedances.
Let us again write the general ABCD parameter equations
V1= AV2 – BI2
I1= CV2 – DI2
V1/I1 = (AV2-BI2) / (CV2 – DI2)
But V2/(-I2) = Zi2 or V2 = -Zi2I2 —–(i)
Therefore:
Zi1 = V1/I1 = (-AZi2 – BI2) / (-Zi2C – DI2)
Or
Zi1 = (-AZi2 – B) / (-Zi2C – D) —————– (ii)
Looking inwards from port-1 Zi1 = V1/I1 Zi2 = V2/(-I2)
Looking inwrds from port-2 Zi1 = V1/(-I1) Zi2 = V2/I2
Now the general equations:
V1 = AV2 – BI2 and I1= CV2 – DI2
AV2 = V1 + BI2 and CV2 = I1 + DI2
Rewriting by balancing, and adding
AV2 = -Zi1I1 + BI2
Zi1CV2 = Zi1I1 + Zi1DI2
———————————————–
V2(A+Zi1C) = (B + Zi1D)I2
Or
Zi2 = V2/I2 = (B + Zi1D) / (A+Zi1C) ————-(iii)
Substituting the value of Zi2 from eqn(iii) in eqn(ii) and simplifying we get
Zi1 = (-A((B + Zi1D) / (A+Zi1C)) – B) / (-((B + Zi1D) / (A+Zi1C))C – D) —————– (ii)
or CDZi12 = AB
Zi1 = √((AB) / (CD) ————————————(iv)
Similarily we can substitute Zi1 from eqn(ii) in eqn (iii) for Zi2 and get
Zi2 = √(BD)/ (AC) ————————————(v)
Image Transfer Parameters
Again looking at the general equation of ABCD parameter
I2 = – V2/Zi2
V1= AV2 – BI2
I1= CV2 – DI2
V1 = AV2 + BV2/Zi2
= (A + B/Zi2) V2
V1/V2 = ( A + B√(AC/BD) because Zi2 = √(BD)/ (AC)
= ( A + √(ACBBD/BDD)
= ( A + √(ABCD)/D ) ———————————(vi)
Also Converting I1 equation only in terms of I2
I1 = CV2 – DI2
= -Zi2CI2 – DI2
= -(-(√(BD)/ (AC) )C + D)I2
= – (D + √(ABCCD)/ (AAC)I2
= – (D + √(ABCD)/ A)I2
-I1/I2 = (D + √(ABCD)/ A) ——————————–(vii)
Multiplying eqn vi and eqn vii, to get
-V1/V2*I1/I2 = (AD + √ABCD)2 / AD2
= (√(AD) + √(BC))2
√((-V1/V2)* (I1/I2) = √(AD) + √(BC) from reciprocity condition AD – BC = 1 or BC = AD-1
√((-V1/V2)* (I1/I2) = √(AD) + √(AD – 1)
Now
√(AD) = coshθ θ = cosh-1√(AD)
√(BC) = √(AD – 1)=sinh θ
cosh θ + sinhθ = eθ = √((-V1/V2)* (I1/I2)
taking antilog
θ = log √((-V1/V2)* (I1/I2)
= log(√((-Z0I1/Z0I2)* (I1/I2)
= log (I1/I2)
