The input and output resistances $R_{in}$ and $R_{out}$, as well as the voltage gain $A_{oc}$ of a two-port network can be obtained experimentally. First, connect an ideal voltage source $v_s$ (a new battary with very low internal resistance) in series with a resistor $R_s$, and then connect load $R_L$ of two different resistances to the output port. Now the three parameters can be derived from the known values of $v_s$, $R_s$ and the two measurements of the load voltage $v_{out}$, corresponding to the two resistance values used.

Assume $v_s=1.5V$, $R_s=5 k\Omega$, and the input voltage is measured to be $v_{in}=1.25 V$; also, assume the two different load resistors used are $R_1=150 \Omega$ and $R_2=200 \Omega$ respectively, with the two corresponding output voltage $v_1=18.75V$ and $v_2=20$. Find $R_{in}$, $R_{out}$ and $A_{oc}$.

Solution:

First consider the voltage $v_{in}$ of the input port:

$$v_{in}=v_s \frac{R_{in}}{R_s+R_{in}} =1.5 \frac{R_{in}}{5 k\Omega +R_{in}}=1.25 V$$

Solving this equation for $R_{in}$, we get $R_{in}=25 k\Omega$

Next consider the voltage of the output port:

$$v_{out}=A v_{in} \frac{R_L}{R_L+R_{out}}$$

i.e.,

$$R_L A v_{in}-v_{out} R_{out} = v_{out} R_L$$

Using the values of $R_L$ and $V_L$ of the two experiments, we get

$$\left\{ \begin{array}{l} 150 Av_{in}-18.75 R_{out} = 2812.5 \\ 200 Av_{in}-20.00 R_{out} = 4000 \end{array} \right.$$

Solving these two equations we get

$$\left\{ \begin{array}{l} R_{out}=50 \Omega Av_{in}=25 V \end{array} \right.$$

But we know $v_{in}=1.25 V$, we get $A=20$.