Two Port Networks Questions and Answers

Answer: C

For a two-port network described by impedance (z) parameters, the voltage equations are \(V_1 = Z_{11}I_1 + Z_{12}I_2\) and \(V_2 = Z_{21}I_1 + Z_{22}I_2\). A network is reciprocal when the transfer impedances are equal, \(Z_{12} = Z_{21}\); for a passive network of resistors, capacitors and inductors this is always true.

The condition \(Z_{11} = Z_{22}\) is the condition for a symmetrical network, not for reciprocity. The conditions \(Z_{12} = Z_{22}\) and \(Z_{11} = Z_{21}\) are not general conditions on any network. The answer is option C.

Answer: B

The y-parameter \(y_{22}\) is the output short circuit admittance, found by shorting the input terminals and applying a source at the output.

\[ y_{22} = \frac{I_2}{V_2}\bigg|_{V_1 = 0} \]

With the input shorted, the 4 ohm resistor is shorted out and does not carry current. The 2 ohm series resistor goes from the output terminal to the shorted input, so it is in parallel with the 8 ohm shunt resistor.

\[ \begin{aligned} y_{22} = \frac{1}{2} + \frac{1}{8} = 0.5 + 0.125 \\ &= 0.625\ \text{S} \end{aligned} \]

The value 0.5 S leaves out the 8 ohm branch and 0.75 S wrongly includes the 4 ohm resistor. The answer is option B.

Answer: B

The z-parameter \(Z_{11} = V_1 / I_1\) with the output port open circuited (\(I_2 = 0\)).

With port 2 open, no current flows in the 3 kilohm resistor, so the input sees the 5 kilohm resistor in series with the 12 kilohm shunt resistor.

\[ Z_{11} = 5 + 12 = 17\ \text{k}\Omega \]

The 3 kilohm resistor does not enter because it carries no current. The value 15 kilohm adds the 12 and 3 and leaves out the 5, 16 kilohm and 12 kilohm do not match any combination of the series arms with the shunt. The answer is 17 kilohm, option B.