Solved numerical 3

Given Data

$ S_{11}=0.1\angle0^\circ $ $ S_{12}=0.8\angle-45^\circ $ $ S_{21}=0.8\angle45^\circ $ $ S_{22}=0.2\angle0^\circ $

Part (a): Determine Whether the Network is Reciprocal and Lossless

Step 1: Check Reciprocity Condition

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

$ S_{12}=S_{21} $

Substituting the given values:

$ S_{12}=0.8\angle-45^\circ $ $ S_{21}=0.8\angle45^\circ $

Since the phase angles are different:

$ 0.8\angle-45^\circ \neq 0.8\angle45^\circ $

The matrix is not symmetric.

$ [S]\neq[S]^T $

Result: The network is not reciprocal.

Step 2: Check Lossless Condition

For a lossless network, the scattering matrix must satisfy the unitary condition:

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

One of the required conditions is:

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

Substituting the given values:

$ |S_{11}|^2+|S_{21}|^2 = (0.1)^2+(0.8)^2 $ $ = 0.01+0.64 $ $ = 0.65 $

Alternatively:

$ \sum_{n=1}^{2}|S_{ni}|^2 = 0.65 \neq 1 $

Since the power conservation condition is not satisfied, the network cannot be lossless.

Result: The network is not lossless.

Conclusion for Part (a)

• The network is not reciprocal.
• The network is not lossless.

Part (b): Return Loss at Port 1 for a Matched Load at Port 2

When Port 2 is terminated with a matched load:

$ \Gamma_2=0 $

Under matched conditions, the input reflection coefficient becomes:

$ \Gamma_1=S_{11} $

Therefore:

$ |\Gamma_1|=|S_{11}|=0.1 $

Return loss is given by:

$ RL_1 = -20\log_{10}|\Gamma_1| $ $ = -20\log_{10}(0.1) $ $ = 20\ \text{dB} $

Answer: The return loss at Port 1 with a matched load at Port 2 is 20 dB.

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

When Port 2 is short-circuited, the load reflection coefficient becomes:

$ \Gamma_2=-1 $

This means any wave reaching Port 2 is reflected back with a phase reversal of 180°.

Step 1: Start with the Two-Port S-Parameter Equations

$ [b]=[S][a] $

Expanding the matrix equation:

$ b_1=S_{11}a_1+S_{12}a_2 $ $ b_2=S_{21}a_1+S_{22}a_2 $

Since Port 2 is short-circuited:

$ b_2=-a_2 $

Substituting into the first equation:

$ b_1=S_{11}a_1-S_{12}b_2 $

Similarly, substituting into the second equation:

$ b_2=S_{21}a_1-S_{22}b_2 $

Step 2: Solve for b₂

Rearranging:

$ b_2+S_{22}b_2=S_{21}a_1 $ $ b_2(1+S_{22})=S_{21}a_1 $ $ b_2=\frac{S_{21}a_1}{1+S_{22}} $

Dividing both sides by a₁:

$ \frac{b_2}{a_1} = \frac{S_{21}}{1+S_{22}} $

Step 3: Calculate the Input Reflection Coefficient

Divide the first network equation by a₁:

$ \frac{b_1}{a_1} = S_{11} - S_{12}\frac{b_2}{a_1} $

Since the input reflection coefficient is defined as:

$ \Gamma_1 = \frac{b_1}{a_1} $

Substituting the result obtained in Step 2:

$ \Gamma_1 = S_{11} - S_{12} \left( \frac{S_{21}} {1+S_{22}} \right) $

Step 4: Substitute the Given S-Parameter Values

$ \Gamma_1 = 0.1 - \frac{ (0.8\angle -45^\circ) (0.8\angle 45^\circ) } {1+0.2} $

First multiply the transmission coefficients:

$ (0.8\angle -45^\circ) (0.8\angle 45^\circ) = 0.64\angle 0^\circ $

Substituting:

$ \Gamma_1 = 0.1 - \frac{0.64}{1.2} $ $ = 0.1 - 0.5333 $ $ = -0.4333 $

Therefore, the magnitude of the input reflection coefficient is:

$ |\Gamma_1| = 0.4333 $

Using the rounded value from the textbook solution:

$ |\Gamma_1| \approx 0.43 $

Step 5: Calculate Return Loss

Return loss is given by:

$ RL_1 = -20\log_{10}|\Gamma_1| $

Substituting the rounded value:

$ RL_1 = -20\log_{10}(0.43) $ $ = 7.33\ \text{dB} $

Final Answer

The return loss at Port 1 when Port 2 is short-circuited is:

$ RL_1 = 7.33\ \text{dB} $