## 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 output 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

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)  + √(BC)                        from reciprocity condition AD – BC  =  1  or BC = AD-1

Now

√(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)

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