Question 17.1: Find the impedance parameters of the resistive circuit in Fi......

We start with an open circuit at port 2 (I_{2} = 0). The resistance looking in at port 1 is

z_{11} = 50  ||(125 + 75) = 40  Ω

To find the forward transfer impedance, we use current division to express the current through the 75-Ω resistor in terms of I_{1}.

I_{75} = \frac{50}{50  +  125  +  75}I_{1} = 0.2I_{1}

By Ohm’s law the open-circuit voltage at port 2 is V_{2} = I_{75} × 75. Therefore, the forward transfer impedance is

z_{21} =\frac{ V_{2}}{I_{1}}|_{I_{2}=0}=\frac{ (0.2I_{1} )  ×  75}{I_{1}}= 15  Ω

Next we assume that port 1 is open (I_{1} = 0). The resistance looking in at port 2 is

z_{22} = 75||(125 + 50) = 52.5  Ω

To find the reverse transfer impedance, we first express the current through the 50-Ω resistor in terms of I_{2}. Using current division again

I_{50} =\frac{ 75}{50  +  125  +  75}I_{2} = 0.3I_{2}

By Ohm’s law the open-circuit voltage at port 1 is V_{1} = I_{50} × 50. Therefore, the reverse transfer impedance is

z_{12} = \frac{V_{1}}{I_{2}}|_{I_{1}=0}= \frac{(0.3I_{2})  ×  50}{I_{2}}= 15  Ω

Note that since z_{12} = z_{21} = 15  Ω, the two-port network is reciprocal.