Question E7C04:
How does an impedance-matching circuit transform a complex impedance to a resistive impedance?  

Choices:  
A: It introduces negative resistance to cancel the resistive part of impedance  
B: It introduces transconductance to cancel the reactive part of impedance  
C: It cancels the reactive part of the impedance and changes the resistive part to the desired value  
D: Reactive currents are dissipated in matched resistances    

Correct Answer is  C: It cancels the reactive part of the impedance and changes the resistive part to the desired value  

Correct answer: C — It cancels the reactive part of the impedance and changes the resistive part to the desired value.

1) Why C is correct
- A load has impedance $Z_L = R_L + jX_L$. An ideal matching network (made of L/C or transmission line sections) uses reactances to:
  - Cancel the load’s reactance so the input looks purely resistive ($X_{\text{in}} = 0$).
  - Transform the remaining resistance to the target value (e.g., 50 Ω).
- After matching, the source sees $Z_{\text{in}} = R_0 + j0$, so power transfer is maximized and SWR is 1:1.

2) Why the others are incorrect
- A: “Negative resistance” is not used in passive matching; you don’t cancel resistance. You transform it with reactances.
- B: “Transconductance” is a device parameter (A/V), not how passive matching networks work. LC networks supply reactance, not transconductance.
- D: Reactive currents are not “dissipated” in reactances; ideal L and C store and return energy with zero average power. Only resistors dissipate power.

3) Useful formulas and rules of thumb
- Impedance form:
  $$Z = R + jX$$
- Conjugate match condition (seen by the source):
  $$Z_{\text{in}} = R_0 + j0 \quad\text{(purely resistive and equal to the source or line impedance)}$$
- Reflection coefficient (perfect match when $Z_L = Z_0$):
  $$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}$$
- L-network (lossless) transformation between resistances $R_{\text{low}}$ and $R_{\text{high}}$:
  $$Q = \sqrt{\frac{R_{\text{high}}}{R_{\text{low}}} - 1}$$
  $$X_{\text{series}} = Q\,R_{\text{low}} \quad\text{and}\quad X_{\text{shunt}} = \frac{R_{\text{high}}}{Q}$$
  The choice of L vs C element (inductive or capacitive) and which goes in series or shunt depends on whether you must add $+jX$ or $-jX$ to cancel the load’s reactance and on whether you are stepping resistance up or down.

Key idea: Matching networks use reactive elements to null out $jX$ and to transform $R$ to the desired value without dissipating power.