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

Z_1O   =   V₁/I₁= A/C   ——————————(i)

Applying the short circuit at port-2; V₂=0, we get

V₁= – BI₂

I₁=-DI₂

Or

Z_1S = V₁/I₁ = B/D                ————————–(ii)

Now looking from port and applying:

Open circuit port-1, I₁=0

0 = CV₂ – DI₂

CV₂= DI₂

Z_2O = V₂/I₂ = D/C                               ————————–(iii)

Now short circuiting the port-1, V₁=0, we get

0 = AV₂ – BI₂

AV₂  =  BI₂

Z_2s   =  V₂/I₂ = B/A             —————————(iv)

Ratio of open circuit to short circuit ratio

Z1O/ Z1S = (B/D)(C/A)     ————————–(v)

To find ABCD parameters

Z_2O   –   Z_2S = D/C – B/A     = (AD –BC)/CA

But for the two network to be reciprocal, the condition is AD-BC =1

Therefore:

Z_2O   –   Z_2S    =  1/CA                           ———-(vi)

CA   =   1/ (Z_2O   –   Z_2S  )

From eqn(i)        Z_1O   = A/C

Multiplying eqn(i) and eqn(v)

CA(A/C)   =   1/ (Z_2O   –   Z_2S  )* Z_1O

A² = Z_1O / (Z_2O   –   Z_2S  )

Or

A= √( Z_1O / (Z_2O   –   Z_2S  )  ——————–(vii)

From eqn(i) i.e. Z_1O   = A/C           C = A/ Z_1O

Therefore from eqn (vii):

C    =  √( Z_1O / (Z_2O   –   Z_2S  ) /Z_1O

C      =     √( 1/ Z_1O (Z_2O   –   Z_2S  )      ———–(viii)

From short circuit equations (iv)                Z_2s   = B/A

B   =  AZ_2s

= √( Z_1O / (Z_2O   –   Z_2S  )* Z_2s

From short circuit equations (ii)                 Z_1s   = B/D

D          =     B/ Z_1s

                                =   √( Z_1O / (Z_2O   –   Z_2S  )* Z_2s/ Z_1s

= A* Z_2s/ Z_1s            ——————————–(ix)

= Z₂₀*√( 1/ Z_1O (Z_2O   –   Z_2S  )——————(x)

Summery,  the ABCD parameters in terms of open and short circuit impedance:

A     =  A= √( Z_1O / (Z_2O   –   Z_2S  )                     B   =        AZ_2s               =    √( Z_1O / (Z_2O   –   Z_2S  )* Z_2s

C      =     √( 1/ Z_1O (Z_2O   –   Z_2S  )                      D   =     A* Z_2s/ Z_1s        =    Z₂₀*√( 1/ Z_1O (Z_2O   –   Z_2S  )

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) = Z_i2                                or            V₂ = -Z_i2I₂        —–(i)

Therefore:

Z_i1  =  V₁/I₁           = (-AZ_i2   –   BI₂) / (-Z_i2C   – DI₂)

Or

Z_i1    = (-AZ_i2   –   B) / (-Z_i2C   – D)   —————–    (ii)

Looking inwards from port-1       Z_i1 = V₁/I₁                             Z_i2 = V₂/(-I₂)

Looking inwrds from port-2         Z_i1 = V₁/(-I₁)                        Z_i2 = 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   =    -Z_i1I₁   + BI₂

Z_i1CV₂ =     Z_i1I₁  + Z_i1DI₂

———————————————–

V₂(A+Z_i1C)  =  (B + Z_i1D)I₂

Or

Z_i2    =    V₂/I₂ =   (B + Z_i1D) / (A+Z_i1C)   ————-(iii)

Substituting the value of Zi2 from eqn(iii) in eqn(ii) and simplifying we get

Z_i1    = (-A((B + Z_i1D) / (A+Z_i1C))   –   B) / (-((B + Z_i1D) / (A+Z_i1C))C   – D)        —————–    (ii)

or            CDZ_i1²   = AB

Z_i1           =             √((AB) / (CD)      ————————————(iv)

Similarily we can substitute Z_i1 from eqn(ii) in eqn (iii) for Z_i2 and get

Z_i2 = √(BD)/ (AC)                               ————————————(v)

Image Transfer Parameters

Again looking at the general equation of ABCD parameter

I₂ = – V₂/Z_i2

V₁= AV₂ – BI₂

I₁=  CV₂  – DI₂

V₁  =  AV₂    +  BV₂/Z_i2

       =  (A     +  B/Z_i2) 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

                                =  -Z_i2CI₂   –  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 =  e^θ =   √((-V₁/V₂)* (I₁/I₂)

taking antilog

θ             =  log  √((-V₁/V₂)* (I₁/I₂)

                =  log(√((-Z₀I₁/Z₀I₂)* (I₁/I₂)

= log (I₁/I₂)