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.

Answer: A

For a passive network, \(h_{12} = -h_{21}\).

A passive, linear, bilateral network is reciprocal. In terms of the other parameter sets reciprocity means \(z_{12} = z_{21}\), \(y_{12} = y_{21}\) and \(AD - BC = 1\). Converting the z-condition into hybrid parameters gives \(h_{12} = -h_{21}\), with a minus sign because \(h_{12}\) is a voltage ratio and \(h_{21}\) a current ratio defined in opposite senses.

\(h_{12} = h_{21}\) misses the sign. \(h_{11}h_{22} - h_{12}h_{21} = 1\) is the condition for a symmetrical network, not a reciprocal one, and \(h_{11} = h_{22}\) equates an impedance with an admittance.

Answer: B

The impedance is \(Z(s) = \dfrac{2(s+3)}{s^{2}+2s+2}\).

The figure shows poles (crosses) at \(s = -1 \pm j1\), and the mark at \(-3\) on the real axis is the zero. So

\[ \begin{aligned} Z(s) &= \frac{K(s + 3)}{(s + 1 - j)(s + 1 + j)} \\ &= \frac{K(s + 3)}{s^{2} + 2s + 2} \end{aligned} \]

Setting \(s = 0\): \(Z(0) = 3K/2 = 3\), so \(K = 2\).

The denominator \(s^{2} + 2s + 3\) would put the poles at \(-1 \pm j\sqrt{2}\), not \(\pm j1\). The options with \(s - 3\) and \(s^{2} - 2s - \ldots\) place the zero and poles in the right half-plane, which is not what the plot shows.

Answer: B

A two port network consists of two pairs of terminals.

A port is a pair of terminals at which current enters at one terminal and leaves at the other. A two port network, drawn as a box, has an input port and an output port, that is, two pairs of terminals, and it is described by the relations between the voltages and currents at them through z, y, h or ABCD parameters.

'Two pairs of ports' would be four ports, an n-pairs description is a general n-port network, and two terminals make only a single port (a one-port network).

Answer: C

For a symmetrical two-port network the image impedances are equal.

The image impedances Z_I1 and Z_I2 are the pair of impedances such that terminating each port in its own image impedance makes the impedance seen at the other port equal to its image impedance. They are given by \(Z_{I1} = \sqrt{Z_{oc1}Z_{sc1}}\) and \(Z_{I2} = \sqrt{Z_{oc2}Z_{sc2}}\).

In a symmetrical network the two ports look the same from either side, so the open- and short-circuit impedances at port 1 equal those at port 2, and the two image impedances are equal; the common value is the characteristic impedance. They are not zero or infinite in general, and 'not equal' describes an asymmetrical network.

Answer: B

For a reciprocal network, \(Z_{12} = Z_{21}\).

A network built only of linear passive bilateral elements, such as this T of resistors, obeys the reciprocity theorem: the voltage at port 2 per unit current at port 1 equals the voltage at port 1 per unit current at port 2. For the T network the shared shunt arm gives both transfer impedances directly:

\[ Z_{12} = Z_{21} = 3\ \Omega \]

The other conditions do not hold here: \(Z_{11} = 1 + 3 = 4\ \Omega\) and \(Z_{22} = 2 + 3 = 5\ \Omega\), so the network is not symmetrical, and \(Z_{11} = Z_{12}\) or \(Z_{12} = Z_{22}\) are not conditions of reciprocity.

Answer: B

The input impedance with the output open (\(I_2 = 0\)) is \(Z_{11} = E_1/I_1\), which is \(\frac{6}{11}\) ohm (option B). With the output open, the loop holds the 2 ohm and 4 ohm resistors and the dependent source, whose plus terminal is at the bottom.

\[ E_1 - 2I_1 - 4I_1 + 10E_1 = 0 \]

\[ 11E_1 = 6I_1 \]

\[ Z_{11} = \frac{6}{11}\ \Omega \]

The value \(-\frac{6}{11}\) (A) comes from taking the source with the opposite polarity, which would give \(1 - 10 = -9\) in the denominator, not 11. The other two values do not follow from the loop.

Answer: D

The determinant of the admittance matrix is 1/5, because the admittance matrix is the inverse of the impedance matrix and the determinant of an inverse is the reciprocal of the determinant. Here the impedance matrix is

The value 5 is the determinant of the impedance matrix and not of the admittance matrix, which is the likely slip. The values 0 and 1 do not arise from these figures, since 0 would mean the matrix cannot be inverted. The answer is 1/5.