Question

Match List - I with List - II (Value of reflection coefficient).

List - I	List - II
A. Short circuit B. Open circuit C. Line characteristic impedance D. 2 times of line characteristic impedance	I. 0 II. -1 III. +1 IV. +\dfrac{1}{3}

• A. A-II, B-III, C-III, D-IV
• B. A-IV, B-III, C-I, D-II
• C. A-II, B-III, C-I, D-IV
• D. A-IV, B-I, C-III, D-II

Hint:

Reflection coefficient: \[ \Gamma=\frac{Z_L-Z_0}{Z_L+Z_0} \] Important special cases: \[ Z_L=0 \Rightarrow \Gamma=-1 \] \[ Z_L=\infty \Rightarrow \Gamma=+1 \] \[ Z_L=Z_0 \Rightarrow \Gamma=0 \]

Answer 1

The Correct Option is C

Concept: The reflection coefficient for a transmission line is given by: \[ \Gamma=\frac{Z_L-Z_0}{Z_L+Z_0} \] where:
• \(Z_L\) = Load impedance
• \(Z_0\) = Characteristic impedance of transmission line The value of reflection coefficient depends upon the load condition.

Step 1: Find reflection coefficient for short circuit. For short circuit: \[ Z_L=0 \] Substituting into reflection coefficient equation: \[ \Gamma=\frac{0-Z_0}{0+Z_0} \] \[ \Gamma=\frac{-Z_0}{Z_0} \] \[ \Gamma=-1 \] Therefore: \[ A \rightarrow II \]

Step 2: Find reflection coefficient for open circuit. For open circuit: \[ Z_L\rightarrow \infty \] Thus: \[ \Gamma=\frac{\infty-Z_0}{\infty+Z_0} \] \[ \Gamma=+1 \] Hence: \[ B \rightarrow III \]

Step 3: Find reflection coefficient when load equals characteristic impedance. If: \[ Z_L=Z_0 \] then: \[ \Gamma=\frac{Z_0-Z_0}{Z_0+Z_0} \] \[ \Gamma=0 \] Hence: \[ C \rightarrow I \] This represents perfect impedance matching.

Step 4: Find reflection coefficient for \(Z_L=2Z_0\). Using: \[ Z_L=2Z_0 \] Substituting: \[ \Gamma=\frac{2Z_0-Z_0}{2Z_0+Z_0} \] \[ \Gamma=\frac{Z_0}{3Z_0} \] \[ \Gamma=\frac{1}{3} \] Thus: \[ D \rightarrow IV \]

Step 5: Write the final matching. Therefore: \[ A-II,\ B-III,\ C-I,\ D-IV \]

Step 6: Write the final answer. Hence the correct option is: \[ \boxed{(3)\ A-II,\ B-III,\ C-I,\ D-IV} \]