Solved Numerical 10
Consider the following two-port scattering matrix:
$ [S] = \begin{bmatrix} 0.15\angle0^\circ & 0.85\angle-45^\circ\\ 0.85\angle45^\circ & 0.2\angle0^\circ \end{bmatrix} $
(a) Reciprocal Network Check
A two-port network is reciprocal if its scattering parameters satisfy the symmetry condition:
S12 = S21
Given the scattering parameters from your data:
- S12 = 0.85 ∠ -45°
- S21 = 0.85 ∠ 45°
Although the magnitudes are equal, the phase angles are different (-45° ≠ 45°). Therefore, the scattering matrix is not symmetric:
[S] ≠ [S]T
Consequently:
Conclusion: The network is not reciprocal.
(b) Lossless Network Check
For a lossless network, the sum of squared magnitudes in every row or column must equal unity.
Checking the first column:
$|S_{11}|^2+|S_{21}|^2$
$=(0.15)^2+(0.85)^2$
$=0.0225+0.7225$
$=0.745$
Since:
$0.745\neq1$
the network does not satisfy the lossless condition.
Therefore:
$\boxed{\text{The network is lossy}}$
(c) Return Loss at Port 1 When Port 2 is Matched
When Port 2 is terminated with a matched load:
$\Gamma_L=0$
The input reflection coefficient becomes:
$\Gamma_{in}=S_{11}$
$\Gamma_{in}=0.15\angle0^\circ$
Return loss is:
$RL=-20\log_{10}|\Gamma_{in}|$
$RL=-20\log_{10}(0.15)$
$RL=16.48\text{ dB}$
Therefore:
$\boxed{RL\approx16.5\text{ dB}}$
(d) Return Loss at Port 1 When Port 2 is Short-Circuited
For a short-circuit termination:
$\Gamma_L=-1$
The general input reflection coefficient formula is:
$ \Gamma_{in} = S_{11} + \frac{S_{12}S_{21}\Gamma_L} {1-S_{22}\Gamma_L} $
Substituting \(\Gamma_L=-1\):
$ \Gamma_{in} = S_{11} - \frac{S_{12}S_{21}} {1+S_{22}} $
Step 1: Calculate S₁₂S₂₁
$ S_{12}S_{21} = (0.85\angle-45^\circ) (0.85\angle45^\circ) = 0.7225 $
Step 2: Calculate 1 + S₂₂
$ 1+S_{22} = 1+0.2 = 1.2 $
Step 3: Calculate the Fraction
$ \frac{S_{12}S_{21}} {1+S_{22}} = \frac{0.7225}{1.2} = 0.602 $
Step 4: Calculate the Input Reflection Coefficient
$ \Gamma_{in} = 0.15-0.602 = -0.452 $
Step 5: Calculate Return Loss
$ RL = -20\log_{10}(0.452) = 6.90\text{ dB} $
Therefore:
$\boxed{RL\approx6.9\text{ dB}}$
Detailed Derivation Using Wave Equations
The scattering equations are:
$V_1^- = S_{11}V_1^+ + S_{12}V_2^+$
$V_2^- = S_{21}V_1^+ + S_{22}V_2^+$
For a short circuit:
$V_2^+ = -V_2^-$
Substituting into the second equation:
$V_2^- = S_{21}V_1^+ - S_{22}V_2^-$
$V_2^-(1+S_{22}) = S_{21}V_1^+$
$V_2^- = \frac{S_{21}}{1+S_{22}}V_1^+$
Substituting into the first equation:
$ V_1^- = S_{11}V_1^+ - S_{12} \left( \frac{S_{21}}{1+S_{22}} V_1^+ \right) $
Dividing by \(V_1^+\):
$ \Gamma_{in} = \frac{V_1^-}{V_1^+} = S_{11} - \frac{S_{12}S_{21}} {1+S_{22}} $
Exam Takeaway
- Reciprocal networks satisfy: $S_{12}=S_{21}$
- Lossless networks satisfy: $\sum |S_{ij}|^2=1$
- For a matched load: $\Gamma_{in}=S_{11}$
- For a short-circuit load: $\Gamma_L=-1$
- Return loss is calculated using: $RL=-20\log_{10}|\Gamma|$