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

For maximum power transfer to the load, the load impedance $Z_L$ must be the complex conjugate of the output impedance of the network, $Z_{out}$.
$Z_L = Z_{out}^$.
The output impedance of a two-port network described by Y-parameters, when driven by a source with admittance $Y_S = 1/Z_S$, is given by:
$Z_{out} = \frac{1}{Y_{out}} = \frac{1}{y_{22} - \frac{y_{12}y_{21}}{y_{11}+Y_S}}$.
From the problem statement, we have:
Source voltage $V_S = 120$ V and source resistance $Z_S = 10 \Omega$.
Source admittance $Y_S = \frac{1}{Z_S} = \frac{1}{10} = 0.1$ S.
The Y-parameters are:
$y_{11} = \frac{2}{100} = 0.02$ S
$y_{12} = \frac{-1}{100} = -0.01$ S
$y_{21} = \frac{-1}{100} = -0.01$ S
$y_{22} = \frac{4/3}{100} = \frac{4}{300}$ S
Now, we calculate the output admittance $Y_{out}$:
$Y_{out} = y_{22} - \frac{y_{12}y_{21}}{y_{11}+Y_S} = \frac{4}{300} - \frac{(-0.01)(-0.01)}{0.02 + 0.1}$
$Y_{out} = \frac{4}{300} - \frac{0.0001}{0.12} = \frac{4}{300} - \frac{1}{1200}$
Using a common denominator of 1200:
$Y_{out} = \frac{16}{1200} - \frac{1}{1200} = \frac{15}{1200} = \frac{1}{80}$ S.
The output impedance is $Z_{out} = \frac{1}{Y_{out}} = 80 \Omega$.
Since $Z_{out}$ is purely real, for maximum power transfer, $Z_L = Z_{out}^ = Z_{out} = 80 \Omega$.