Question

4. Find the S-parameters of a 25-12 series resistor connected to a shunt 75-Q resistor. Assume Z0-50 Ω. Verify using NI Microwave Office.

Answer 1

The S-parameter of a system having the ABCD-parameters is given by

The S-parameter of a system having the ABCD-parameters is given by S_11 = A + B/Z_0 - C Z_0 - D/A + B/Z_0 + C Z_0 + D S_12 = 2 (A D - B C)/A + B/Z_0 + C Z_0 + D S_21 = 2/A + B/Z_0 + C Z_0 + D S_22 = -A + B/Z_0 - C Z_0 + D/A + B/Z_0 + C Z_0 + D The ABCD parameter of a series element is [A C, B D] = [1 0, Z 1] = [1 0, R 1] The ABCD parameter of a shunt element is [A C, B D] = [1 Y, 0 1] = [1 1/R, 0 1] So, the ABCD parameters of the series resistor of 25 Ohm is [A_1 C_1, B_1 D_1] = [1 0, 25 1] The ABCD parameters of the shunt resistor 75 Ohm is [A_2 C_2, B_2 D_2] = [1 0.0133, 0 1]

$S_{12}=\frac{2(AD-BC)}{A+\frac{B}{Z_o}+CZ_o+D}$

$S_{21}=\frac{2}{A+\frac{B}{Z_o}+CZ_o+D}$

$S_{22}=\frac{-A+\frac{B}{Z_o}-CZ_o+D}{A+\frac{B}{Z_o}+CZ_o+D}$

The ABCD parameter of a series element is

$\begin{bmatrix} A &B \\ C&D \end{bmatrix}=\begin{bmatrix} 1 &Z \\ 0& 1 \end{bmatrix}=\begin{bmatrix} 1 &R \\ 0& 1 \end{bmatrix}$

The ABCD parameter of a shunt element is

$\begin{bmatrix} A &B \\ C&D \end{bmatrix}=\begin{bmatrix} 1 &0 \\ Y& 1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 1/R& 1 \end{bmatrix}$

So, the ABCD parameters of the series resistor of $25\Omega$ is

$\begin{bmatrix} A_1 &B_1 \\ C_1&D_1 \end{bmatrix}=\begin{bmatrix} 1 &25 \\ 0& 1 \end{bmatrix}$

The ABCD parameters of the shunt resistor $75\Omega$ is

$\begin{bmatrix} A_2 &B_2 \\ C_2&D_2 \end{bmatrix}=\begin{bmatrix} 1 &0 \\ 0.0133& 1 \end{bmatrix}$

The ABCD parameters of the combined series resistor of $25\Omega$ and shunt resistor of $75\Omega$ is

$\begin{bmatrix} A_{eq} &B_{eq} \\ C_{eq}&D_{eq} \end{bmatrix}=\begin{bmatrix} A_1 &B_1 \\ C_1&D_1 \end{bmatrix}\begin{bmatrix} A_2 &B_2 \\ C_2&D_2 \end{bmatrix}=\begin{bmatrix} 1 &25 \\ 0& 1 \end{bmatrix}\begin{bmatrix} 1 &0 \\ 0.0133& 1 \end{bmatrix}=\begin{bmatrix} 1.3325 &25 \\ 0.0133& 1 \end{bmatrix}$

The S-parameters of the system with a load impedance of $Z_o=50\Omega$ is

$S_{11}=\frac{1.3325+\frac{25}{50}-(0.0133)50-1}{1.3325+\frac{25}{50}+(0.0133)50+1}=\frac{0.1675}{3.4975}=0.0479$

$S_{12}=\frac{2((1.3325)(1)-(25)(0.0133))}{1.3325+\frac{25}{50}+(0.0133)50+1}=\frac{4.665}{3.4975}=1.3338$

$S_{21}=\frac{2}{1.3325+\frac{25}{50}+(0.0133)50+1}=\frac{2}{3.4975}=0.5718$

$S_{22}=\frac{-1.3325+\frac{25}{50}-(0.0133)50+1}{1.3325+\frac{25}{50}+(0.0133)50+1}=\frac{-0.4975}{3.4975}=-0.1422$

So, the S-parameter matrix is

$S=\begin{bmatrix} 0.0479 &1.3338 \\ 0.5718 &-0.1422 \end{bmatrix}$