Solved Numerical 5

Step 1: Verify Reciprocity

A microwave network is reciprocal if its scattering matrix is symmetric:

$ S_{ij}=S_{ji} $

Checking the off-diagonal elements:

$ S_{12}=S_{21}=0.6\angle45^\circ $

$ S_{13}=S_{31}=0.4\angle45^\circ $

$ S_{23}=S_{32}=0.3\angle(-45^\circ) $

Since all corresponding elements are equal:

$ [S]=[S]^T $

Therefore, the network satisfies the reciprocity condition.

Result:

$ \boxed{\text{The network is reciprocal}} $

Step 2: Test Whether the Network is Lossless

A lossless microwave network must satisfy the unitary condition:

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

One quick test is to verify that the sum of the squared magnitudes of every row equals unity.

Check Row 1

$ |S_{11}|^2+|S_{12}|^2+|S_{13}|^2 $

$ =(0.178)^2+(0.6)^2+(0.4)^2 $

$ =0.0317+0.36+0.16 $

$ =0.5517 $

Since:

$ 0.5517 \neq 1 $

the first row alone violates the lossless condition.

Check Row 2

$ |S_{21}|^2+|S_{22}|^2+|S_{23}|^2 $

$ =(0.6)^2+0+(0.3)^2 $

$ =0.36+0.09 $

$ =0.45 $

Again:

$ 0.45 \neq 1 $

The network clearly does not satisfy power conservation.

Result:

$ \boxed{\text{The network is not lossless}} $

The network is therefore a lossy 3-port network.

Step 3: Calculate Return Loss at Port 1

Return loss is determined directly from the reflection coefficient at Port 1:

$ RL=-20\log_{10}|S_{11}| $

Substituting:

$ RL=-20\log_{10}(0.178) $

$ RL=14.99\text{ dB} $

Therefore:

$ \boxed{RL \approx 15\text{ dB}} $

Step 4: Calculate Insertion Loss Between Ports 2 and 3

Insertion loss is obtained from the transmission coefficient between the two ports.

Given:

$ |S_{23}|=0.3 $

The insertion loss is:

$ IL=-20\log_{10}|S_{23}| $

Substituting:

$ IL=-20\log_{10}(0.3) $

$ IL=10.46\text{ dB} $

Therefore:

$ \boxed{IL \approx 10.46\text{ dB}} $

Step 5: Determine Transmission Phase and Phase Delay

The transmission coefficient between Ports 2 and 3 is:

$ S_{23}=0.3\angle(-45^\circ) $

Transmission Phase

The transmission phase is simply the phase angle of the S-parameter:

$ \angle S_{23}=-45^\circ $

Therefore:

$ \boxed{\text{Transmission Phase}=-45^\circ} $

Phase Delay

Phase delay is defined as the negative of the transmission phase:

$ \theta_d=-\angle S_{23} $

$ \theta_d=-(-45^\circ) $

$ \theta_d=45^\circ $

Therefore:

$ \boxed{\text{Phase Delay}=45^\circ} $

Final Answers

• Reciprocal: Yes
• Lossless: No (Lossy Network)
• Return Loss at Port 1: 15 dB
• Insertion Loss (Port 2 to Port 3): 10.46 dB
• Transmission Phase: −45°
• Phase Delay: 45°

Exam Tip

Many students incorrectly test losslessness using:

$ |S_{11}|^2+|S_{22}|^2+|S_{33}|^2 $

This is not a valid lossless condition.

For any lossless 3-port network:

$ \sum_{j=1}^{3}|S_{ij}|^2=1 $

for every row (or every column), along with the orthogonality conditions:

$ \sum_{k=1}^{3}S_{ki}S_{kj}^{*}=0 \qquad (i\neq j) $

Always start with these conditions when solving  problems involving 3-port S-parameter networks.