Solved Numeriacl 9 T-Attenuator Network

Method 1: Using ABCD Parameter Conversion

The T-attenuator consists of two series resistors of 8.56Ω and one shunt resistor of 141.8Ω.

Series Element ABCD Matrix

$ \begin{bmatrix} A & B\\ C & D \end{bmatrix}_{series} = \begin{bmatrix} 1 & 8.56\\ 0 & 1 \end{bmatrix} $

Shunt Element ABCD Matrix

$ \begin{bmatrix} A & B\\ C & D \end{bmatrix}_{shunt} = \begin{bmatrix} 1 & 0\\ \frac{1}{141.8} & 1 \end{bmatrix} $

Overall ABCD Matrix

Multiplying the three matrices gives the complete network representation:

$ \begin{bmatrix} A & B\\ C & D \end{bmatrix} = \begin{bmatrix} 1.060 & 17.63\\ 0.0071 & 1.060 \end{bmatrix} $

Converting ABCD Parameters to S-Parameters

Applying the standard ABCD-to-S conversion formulas for a 50Ω system:

$S_{11}=0$

$S_{21}=0.707$

$S_{12}=0.707$

$S_{22}=0$

The attenuator is perfectly matched at both ports and exhibits equal forward and reverse transmission.

Method 2: Direct Impedance Analysis

A second method verifies the result using input impedance calculations.

With Port 2 terminated in a matched 50Ω load:

$ Z_{in}' = 8.56+ \left[ 141.8 \parallel (8.56+50) \right] $

Substituting values:

$Z_{in}'=50\Omega$

Since the input impedance equals the characteristic impedance:

$S_{11}=0$

Because the network is symmetric:

$S_{22}=0$

Calculating Transmission Coefficient

The output voltage is:

$ V_2 = V_1 \left( \frac{41.44}{41.44+8.56} \right) \left( \frac{50}{50+8.56} \right) $

$V_2=0.707V_1$

Therefore:

$ S_{21} = \left. \frac{V_2^-}{V_1^+} \right|_{V_2^+=0} = 0.707 $

Since the network contains only passive resistive elements and is symmetric:

$S_{12}=S_{21}=0.707$

Final Scattering Matrix

Both methods produce the same scattering matrix:

$ [S] = \begin{bmatrix} 0 & 0.707\\ 0.707 & 0 \end{bmatrix} $

Exam Takeaway

• A matched attenuator has: $S_{11}=S_{22}=0$
• A reciprocal attenuator satisfies: $S_{12}=S_{21}$
• ABCD parameters can be converted directly into S-parameters using standard conversion formulas.
• Always verify matched conditions before calculating transmission coefficients.
• For this T-attenuator: $ [S] = \begin{bmatrix} 0 & 0.707\\ 0.707 & 0 \end{bmatrix} $