Numerical 1

Problem Statement: There is a source
\[
Z_s = 50~\Omega
\]
that needs to be connected to a load with an impedance of 
\[
Z_L = 100~\Omega
\]
Design a lossless quarter-wave transmission line matching network to ensure maximum power transfer (i.e., a perfect match).

let there be a network Mismatch without a Matching Network

First, let's analyze what happens if we connect the 50 Ω source directly to the 100 Ω load using a standard 50 Ω transmission line 
\[
Z_0 = 50~\Omega
\]

1st Calculate the Reflection Coefficient (\(\Gamma\)):

The reflection coefficient at the load (\(Z_L\)) is calculated using the formula:
\[
\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}
\]

Substitute the values:
\[
\Gamma = \frac{100 - 50}{100 + 50}
\]
\[
\Gamma = \frac{50}{150}
\]
\[
\Gamma = \frac{1}{3}
\]

fig: Network

here,  reflection coefficient of \( \tfrac{1}{3} \) is significant. It means that the amplitude of the reflected voltage wave is one-third that of the incident wave. In terms of power, the reflected power is
\[
|\Gamma|^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \approx 11.1\%
\]
of the incident power. This reflected power returns to the source, creating standing waves and resulting in an inefficient power transfer. Our goal is to make 
\[
\Gamma = 0
\]

Lets deign the Quarter-Wave Transformerto solve this mismatch, we need to insert a special section of transmission line between our main 50 Ω line and the 100 Ω load.

Now,
\[
Z_{\text{in}} = 50~\Omega
\]
This way, the source and its main 50 Ω line see a perfectly matched load.

The Tool: We use a transmission line segment that is exactly one-quarter wavelength long 
\[
\ell = \frac{\lambda}{4}
\]
We previously derived the input impedance for this special case:
\[
Z_{\text{in}} = \frac{Z_0^2}{Z_L}
\]

Here, \(Z_{\text{in}}\) is the desired input impedance (50 Ω), \(Z_L\) is the load we are trying to match (100 Ω), and \(Z_0\) is the characteristic impedance of the quarter-wave line section that we need to find.

Set up the Matching Equation:  
We need
\[
Z_{\text{in}} = Z_s = 50~\Omega
\]

Substitute known values into the transformer formula:
\[
50 = \frac{Z_0^2}{100}
\]

Solve for the required Characteristic Impedance (\(Z_0\)):
\[
Z_0^2 = 50 \cdot 100
\]
\[
Z_0^2 = 5000
\]
\[
Z_0 = \sqrt{5000} \approx 70.71~\Omega
\]

Now to perfectly match the 50 Ω source to the 100 Ω load, we must insert a section of lossless transmission line with the following properties:

\[
\ell = \frac{\lambda}{4}, 
\quad Z_0 \approx 70.71~\Omega,
\quad \Gamma = 0
\]

When this matching section is in place, the 100 Ω load impedance is "transformed" to appear as a 50 Ω impedance at the input of the quarter-wave section. The 50 Ω source is therefore perfectly matched, the reflection coefficient becomes zero, and maximum power is delivered to the load.