Question

For the two port network shown below, the [Y]-parameters is given as \[ [Y] = \frac{1}{100} \begin{bmatrix} 2 & -1 \\ -1 & \frac{4}{3} \end{bmatrix} \text{ S}. \] The value of load impedance \( Z_L \), in \( \Omega \), for maximum power transfer will be ___________ (rounded off to the nearest integer). 

Hint:

Maximum power transfer occurs when $Z_L = Z_{out}^$. Remember the formula for the output impedance of a Y-parameter network: $Z_{out} = 1/(y_{22} - \frac{y_{12}y_{21}}{y_{11}+Y_S})$. The source impedance $Z_S$ influences the output impedance of the network.

Answer 1

Correct Answer: 80

To determine the load impedance \( Z_L \) for maximum power transfer, we first consider the given [Y]-parameters of the two-port network:

\[ [Y] = \frac{1}{100} \begin{bmatrix} 2 & -1 \\ -1 & \frac{4}{3} \end{bmatrix} \text{ S} \]

For maximum power transfer, the load impedance \( Z_L \) should be the complex conjugate of the output impedance \( Z_{\text{out}} \) seen looking back into the network. The expression for \( Z_{\text{out}} \) in terms of Y-parameters is given by:

\[ Z_{\text{out}} = \frac{1}{Y_{22} - \frac{Y_{12}Y_{21}}{Y_{11} + Z_s^{-1}}} \]

Here, \( Z_s = 10 \, \Omega \). So, \( Y_{11} = \frac{2}{100} = 0.02 \, \text{S} \), \( Y_{12} = -\frac{1}{100} = -0.01 \, \text{S} \), \( Y_{21} = -\frac{1}{100} = -0.01 \, \text{S} \), \( Y_{22} = \frac{4}{300} = 0.0133 \, \text{S} \).

First, calculate \( Y_{11} + \frac{1}{Z_s} \):

\[ Y_{11} + \frac{1}{Z_s} = 0.02 + \frac{1}{10} = 0.12 \, \text{S} \]

Next, compute:

\[ Y_{22} - \frac{Y_{12}Y_{21}}{Y_{11} + \frac{1}{Z_s}} = 0.0133 - \frac{(-0.01)(-0.01)}{0.12} \]

\[ = 0.0133 - \frac{0.0001}{0.12} = 0.0125 \, \text{S} \]

Therefore, the output impedance is:

\[ Z_{\text{out}} = \frac{1}{0.0125} = 80 \, \Omega \]

Hence, for maximum power transfer, the load impedance \( Z_L \) should be equal to \( Z_{\text{out}} \), which is \( 80 \, \Omega \).

This value is already within the specified range (80, 80). Therefore, the final value of \( Z_L \) is:

80 Ω