General Terminologies of a TwoPort networks:
Network: A network is an interconnection of passive elements and dependent sources. There are no active or independent sources inside the box. The network is treated to be part of the square box shown in figure.
Terminal: A terminal is the end point of the conductor which is connected to the node of the network which is brought out of the designated square box representing the network
Port: port is a pair of terminal which is used to connect to the sources of energy i.e. independent sources and or the excitation.
A network may be oneport in which there is only one driving force or multiport network to which multiple sources of energy are connected. We consider a 2port network for our study.
In a multiport network of specifically in our study a 2port network, one port may be connected to driving force and other port may be used for measurement, for connecting to load or for interconnecting to the input of some other networks.
Driving Point: The ports which are used for connecting sources or load and for interconnecting networks is called driving points of a network.
Network functions:
Transform impedance: It is written as Z(s) = V(s) /I(s) is defined as the ratio of the voltage transform and the current transform at the same port. They are also called as driving point impedance.
Transform admittance: It is written as Y(s) and equals the inverse of transform impedance i.e. Y(s) = 1/Z(s) is defined as the ration of current transform to the voltage transform at the same port.
Transform /Driving Point  Port1 voltage and current  Port2 voltage and current 
Impedance  Z_{11} = V_{1}(s) / I_{1}(s)  Z_{22} = V_{2}(s) / I_{2}(s) 
Admittance  Y_{11} = I_{1}/V_{1}  Y_{22} = I_{2 }/V_{2} 
Transfer Function: Transfer function is defined as the ratio of the voltage or current transform at the output to the transform of voltage or current at input port:
Transfer function ( Z, Y, G, α)


Transfer impedance (Z_{21})  Ratio of Voltage transform at output port (V_{2}(s) ) to the transform of current (I_{1}(s))at the input port  Z_{21} = V_{2}(s) / I_{1}(s) 
Transfer admittance (Y_{21 })  Ratio of current transform (I_{2}(s)) at output port to the transform of Voltage (V_{1}(s) ) at the input port  Y_{21} = I_{2}(s) / V_{1}(s) 
Voltage Transfer Ratio (G_{21})  Ratio of Voltage transform at output port (V_{2}(s) ) to the voltage transform (V_{1}(s)) at the input port  G_{21} = V_{2}(s) / V_{1}(s) 
Current Transfer Ratio (α_{21})  Ratio of Current transform at output port (I_{2}(s) ) to the current transform (I_{1}(s)) at the input port  α_{21} = I_{2}(s) / I_{1}(s) 
When solving for the transfer function, the degree of the polynomial may be found to be greater than the degree of the denominator polynomial, in which case we can reduce by simple division to make the degree of the numerator less than or equal to degree of denominator polynomial.
For determining the above quantities, we should be well conversant with the series and parallel combination of the components. Using the transform of the R, L and C components it is convenient and easy to determine as the transform allow us find these quantities as if we are calculating the values for series and parallel combination of resistance values. It should be clear from the following examples.
Example1: For the following oneport network find the transform impedance and admittance
Solution:
In figure(a) applying transform to the components R_{1} –> R_{1} L_{1} –> sL_{1} The network is a series connection of R1 and L1. The driving point impedance of the network is: Z(s) = R_{1} + sL_{1} The driving point admittance is: Y(s) = 1/Z(s) = 1/( R_{1} + sL_{1}) 
Solution:
From figure(b), applying the transform to the network components. R2 –> R2 L2 –> sL2 Impedance of parallel network = (Z_{A}Z_{B})/(Z_{A} + Z_{B}); therefore we get the driving point impedance of the parallel network : = (R_{2}.sL_{2}) / (R_{2} + sL_{2}) The driving point admittance = 1 / Z = (R_{2} + sL_{2}) / (R_{2}.sL_{2}) 
Practice Problem:
Examples of determining the driving point impedance, driving point admittance and the transfer function of a 2port network.
Ex2 Determine the transfer impedance of the network
Solution: The transfer impedance of the network is V_{2}(s) / V_{1}(s)
Writing the network components in the transform form R_{1} –> R_{1}; R_{2} –> R_{2}; and L –> sL; V_{1} –> V_{1}(s) and V_{2} –> V_{2}(s) and I_{1} –> I_{1}(s)
We can write down the KVL equation with the port2 as open circuited. With port2 open the current I2=0. Therefore the two loop equations are:
V_{1}(s) = I_{1}(s)(R_{1} + R_{2})
V_{2}(s) = I_{1}(s)R_{2}
The transfer impedance
G_{21} = V_{2}(s) / V_{1}(s)
= I_{1}(s)R_{2 }/ I_{1}(s)(R_{1} + R_{2})
= R_{2 }/ (R_{1} + R_{2})
The transform / driving point impedance
Z_{11} = V_{1}(s) / I_{1}(s)
V_{1}(s) = I_{1}(s)(R_{1} + R_{2})
Z_{11 } = V_{1}(s) / I_{1}(s) = (R_{1} + R_{2})
Ex3: determine the transfer functions and the driving point impedance for the following network
Solution:
Writing the network components in transform form
L_{1 }–> sL_{1}; L_{2 }–> sL_{2}; C_{1} — > 1/sC_{1}; V_{1 }–> V_{1}(s);
V_{2} –> V_{2}(s); I_{1 } –> I_{1}(s)
With output port open (I_{2}=0) the loop equations gives:
V_{1}(s) = ( sL_{1} + R_{1} +1/sC_{1})I_{1}(s)
= {(s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)/sC_{1}} I_{1}(s)
V2(s) = (R_{1}+1/sC_{1})I_{1}
_{ }={(1+sR_{1}C_{1})/sC_{1}}I_{1}
The transfer function G_{21}
= V2(s) /V1(s)
=(1+sR_{1}C_{1})/sC_{1}} / {(s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)/sC_{1}}
=(1+sR_{1}C_{1}) / (s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)
Z_{11} = V_{1}(s) /I_{1}(s)
= (s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)/sC_{1}
Example4: Determine the driving point impedance Z_{11} and the transfer ratio G_{21 for the following network.}