Solved Numerical 2

Given S-Parameter Matrix

$ [S] = \begin{bmatrix} 0.2\angle0^\circ & 0.6\angle90^\circ \\ 0.6\angle90^\circ & 0.1\angle0^\circ \end{bmatrix} $

The matrix contains four scattering parameters:

• S11 = 0.2∠0°
• S12 = 0.6∠90°
• S21 = 0.6∠90°
• S22 = 0.1∠0°

Part (a): Verify Reciprocity and Losslessness

Step 1: Check Reciprocity Condition

A two-port microwave network is reciprocal if the scattering matrix is symmetric.

$ S_{12}=S_{21} $

Substituting the given values:

$ 0.6\angle90^\circ = 0.6\angle90^\circ $

Since both parameters are identical, the scattering matrix is symmetric.

$ [S]=[S]^T $

Result: The network is reciprocal.

Step 2: Check Lossless Condition

A lossless microwave network must satisfy the unitary condition:

$ [S]^\dagger[S]=[I] $

For a reciprocal two-port network, one of the required conditions is:

$ |S_{11}|^2+|S_{21}|^2=1 $

Substituting the given values:

$ |S_{11}|^2+|S_{21}|^2 = (0.2)^2+(0.6)^2 $ $ = 0.04+0.36 $ $ = 0.40 $

For a lossless network, this sum must equal 1.

$ 0.40 \neq 1 $

The power conservation condition is violated.

Result: The network is not lossless.

Conclusion for Part (a)

• The network is reciprocal because S12 = S21.
• The network is not lossless because the unitary condition is not satisfied.

Part (b): Return Loss at Port 1 for a Short-Circuited Port 2

Step 1: Reflection Coefficient of the Load

For a short-circuit termination:

$ \Gamma_2=-1 $

Step 2: Input Reflection Coefficient Formula

The input reflection coefficient looking into Port 1 is:

$ \Gamma_1 = S_{11} + \frac{S_{12}S_{21}\Gamma_2} {1-S_{22}\Gamma_2} $

Since the network is reciprocal:

$ S_{12}=S_{21}=0.6e^{j\pi/2} $

Substituting all values:

$ \Gamma_1 = 0.2 + \frac{(0.6e^{j\pi/2})^2(-1)} {1-0.1(-1)} $

Step 3: Simplify the Numerator

$ 0.6e^{j\pi/2}=j0.6 $

Squaring the complex quantity:

$ (j0.6)^2 = j^2(0.6)^2 $ $ = (-1)(0.36) $ $ = -0.36 $