Transmission Line Input Impedance Matching

This question involves matching the input impedance ($Z_{in}$) expressions of a transmission line section with specific load conditions.

General Transmission Line Input Impedance ($Z_{in}$)

The general formula for the input impedance ($Z_{in}$) of a transmission line section of length $l$, with characteristic impedance $Z_0$ and terminated by a load impedance $Z_L$, is:

$Z_{in} = Z_0 \left[ \frac{Z_L + jZ_0 \tan(\beta l)}{Z_0 + jZ_L \tan(\beta l)} \right]$

This corresponds to match A with IV.

Input Impedance ($Z_{in}$) with Short-Circuited Load ($Z_L=0$)

For a short-circuited load ($Z_L = 0$), the general formula simplifies:

$Z_{in} = Z_0 \left[ \frac{0 + jZ_0 \tan(\beta l)}{Z_0 + j(0) \tan(\beta l)} \right] = Z_0 \left[ \frac{jZ_0 \tan(\beta l)}{Z_0} \right] = jZ_0 \tan(\beta l)$

This corresponds to match B with III.

Input Impedance ($Z_{in}$) with Open-Circuited Load ($Z_L=\infty$)

For an open-circuited load ($Z_L = \infty$), we analyze the general formula. Dividing the numerator and denominator by $Z_L$ yields:

$Z_{in} = Z_0 \left[ \frac{1 + j\frac{Z_0}{Z_L} \tan(\beta l)}{\frac{Z_0}{Z_L} + j \tan(\beta l)} \right]$

As $Z_L \to \infty$, $\frac{1}{Z_L} \to 0$. The expression becomes:

$Z_{in} = Z_0 \left[ \frac{1 + j(0) \tan(\beta l)}{0 + j \tan(\beta l)} \right] = Z_0 \left[ \frac{1}{j \tan(\beta l)} \right] = \frac{Z_0}{j \tan(\beta l)} = -jZ_0 \cot(\beta l)$

This corresponds to match C with I.

Input Impedance ($Z_{in}$) with Matched Load ($Z_L=Z_0$)

For a matched load ($Z_L = Z_0$), the general formula simplifies:

$Z_{in} = Z_0 \left[ \frac{Z_0 + jZ_0 \tan(\beta l)}{Z_0 + jZ_0 \tan(\beta l)} \right] = Z_0 \left[ \frac{Z_0 (1 + j \tan(\beta l))}{Z_0 (1 + j \tan(\beta l))} \right] = Z_0$

This corresponds to match D with II.

Final Matching Summary

The correct matches are derived as:

• A matches with IV
• B matches with III
• C matches with I
• D matches with II

The complete matching is A-IV, B-III, C-I, D-II.