A two-port network has scattering parameters given \(\left[ s \ri

SOLUTION

Concept:

Consider a two-port network as shown:

Where,

a₁ = Incident wave at port 1

b₁ = reflected wave at port 1

a₂ = Incident wave at port 2

b₂ = reflected wave at port 2

Scattering parameters are defined as:

\({s_{11}} = \frac{{{b_1}}}{{{a_1}}},\;\;{s_{12}} = \frac{{{b_1}}}{{{a_2}}}\) 

\({s_{21}} = \frac{{{b_2}}}{{{a_1}}}\;and\;{s_{22}} = \frac{{{b_2}}}{{{a_2}}}\) 

\({s_{11}} = \frac{{{b_1}}}{{{a_1}}} = \frac{{V_1^ - }}{{V_1^ + }}\) 

\({s_{12}} = \frac{{{b_1}}}{{{a_2}}} = \frac{{V_1^ - }}{{V_2^ + }}\) 

\({s_{21}} = \frac{{{b_2}}}{{{a_1}}} = \frac{{V_2^ - }}{{V_1^ + }}\) 

\({s_{22}} = \frac{{{b_2}}}{{{a_2}}} = \frac{{V_2^ - }}{{V_2^ + }}\) 

V⁺ = Incident voltage wave

V^- = Reflected voltage wave

Boundary conditions in a Transmission line:

At port 2:

\(V_2^ - = - V_2^ + \) 

or b₂ = -a₂   ---(1)

Now, b₁ = s₁₁a₁ + s₁₂ a₂   ---(2)

b₂ = s₂₁ a₁ + s₂₂ a₂     ---(3)

From (1) and (2), we get:

-a₂ = s₂₁ a₁ + s₂₂ a₂

- (1 + s₂₂) a₂ = s₂₁ a₁

\({a_2} = \frac{{ - {s_{21}}\;{a_1}}}{{1 + {s_{22}}}}\)     ---(4)

From (1) and (4), we get:

\({b_1} = {s_{11}}\;{a_1} + {s_{12}}\left( {\frac{{ - {s_{21}}{a_1}}}{{1 + {s_{22}}}}} \right)\)

\({b_1} = \frac{{\left( {{s_{11}}\left( {1 + {s_{22}}} \right) - {s_{12}}{s_{21}}} \right){a_1}}}{{1 + {s_{22}}}}\)  

\({s_{11}} = \;\frac{{{b_1}}}{{{a_1}}} = \frac{{{s_{11}} + {s_{11}}{s_{22}} - {s_{12}}{s_{21}}}}{{1 + {s_{22}}}}\) 

Option B is correct.

Important:

1) If the port is short-circuited:

Reflected voltage wave = - Incident voltage wave

2) If the port is terminated with load, such that Z_L = Z₀, with Z₀ the characteristic impedance of that port, then the reflected wave = 0