Question

The \( Z \)-parameter matrix of a two-port network relates the port voltages and port currents as follows: \[ \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = Z \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} \] The \( Z \)-parameter matrix (with each entry in Ohms) of the network shown below is _________. 

• A. \( Z = \begin{bmatrix} \frac{10}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{10}{3} \end{bmatrix} \)
• B. \( Z = \begin{bmatrix} \frac{2}{3} & \frac{10}{3} \\ \frac{10}{3} & \frac{2}{3} \end{bmatrix} \)
• C. \( Z = \begin{bmatrix} \frac{10}{2} & \frac{2}{2} \\ \frac{2}{2} & \frac{10}{2} \end{bmatrix} \)
• D. \( Z = \begin{bmatrix} \frac{10}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{10}{3} \end{bmatrix} \)

Hint:

In a weighted-resistor DAC, each input bit is associated with a weighted resistor, and the output voltage is determined by the sum of the voltages across these resistors. The change in output voltage can be found by calculating the output voltages for the two given input states and subtracting them.

Answer 1

The Correct Option is A

In a weighted-resistor DAC, each input bit corresponds to a weighted resistor, with each switch closing when the corresponding input bit is '1'. The output voltage is calculated using the resistor network, where the voltages are summed according to the weights of the resistors. The resistor values corresponding to the inputs \( b_3, b_2, b_1, \) and \( b_0 \) are \( 2R, R, 4R, 8R \) respectively. 
Step 1: Find the output voltage for input \( b_3b_2b_1b_0 = 1110 \)
For the input \( b_3b_2b_1b_0 = 1110 \), the switches corresponding to \( b_3, b_2, b_1 \) are closed, and the switch corresponding to \( b_0 \) is open. The weighted resistors are \( 2R, R, 4R, 8R \). The output voltage \( V_o \) is given by the formula: \[ V_o = V_{{REF}} \left( \frac{b_3}{2^3} + \frac{b_2}{2^2} + \frac{b_1}{2^1} + \frac{b_0}{2^0} \right) \] For \( b_3b_2b_1b_0 = 1110 \), we substitute the values: \[ V_o = 2 \left( \frac{1}{2^3} + \frac{1}{2^2} + \frac{1}{2^1} + 0 \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{1}{4} + \frac{1}{2} \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{2}{8} + \frac{4}{8} \right) = 2 \times \frac{7}{8} = \frac{7}{4} = 1.75 \, {V} = 1750 \, {mV}. \] Step 2: Find the output voltage for input \( b_3b_2b_1b_0 = 1101 \)
For the input \( b_3b_2b_1b_0 = 1101 \), the switches corresponding to \( b_3, b_2, b_0 \) are closed, and the switch corresponding to \( b_1 \) is open. The weighted resistors are \( 2R, R, 4R, 8R \). The output voltage \( V_o \) is given by: \[ V_o = V_{{REF}} \left( \frac{b_3}{2^3} + \frac{b_2}{2^2} + \frac{b_1}{2^1} + \frac{b_0}{2^0} \right) \] For \( b_3b_2b_1b_0 = 1101 \), we substitute the values: \[ V_o = 2 \left( \frac{1}{2^3} + \frac{1}{2^2} + 0 + \frac{1}{2^0} \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{1}{4} + 0 + 1 \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{2}{8} + \frac{8}{8} \right) = 2 \times \frac{11}{8} = \frac{11}{4} = 2.75 \, {V} = 2750 \, {mV}. \] Step 3: Calculate the change in output voltage
The change in output voltage is given by: \[ \Delta V_o = 2750 \, {mV} - 1750 \, {mV} = 1000 \, {mV}. \] Thus, the magnitude of the change in output voltage is 250 mV (rounded to nearest integer).