Question E5A05:
What is the result of increasing the Q of an impedance-matching circuit?  

Choices:  
A: Matching bandwidth is decreased  
B: Matching bandwidth is increased  
C: Losses increase  
D: Harmonics increase    

Correct Answer is  A: Matching bandwidth is decreased  

Correct answer: A — Matching bandwidth is decreased.

Why A is correct:
- The quality factor relates bandwidth and center frequency as:
  $$Q = \frac{f_0}{BW} \quad \Rightarrow \quad BW \approx \frac{f_0}{Q}$$
- Increasing $Q$ makes the network more selective, so the bandwidth over which a good match is maintained becomes narrower.

Why the other choices are incorrect:
- B: Matching bandwidth is increased — Incorrect. From $BW \approx f_0/Q$, increasing $Q$ decreases bandwidth.
- C: Losses increase — Generally incorrect. By definition,
  $$Q = 2\pi \frac{\text{energy stored}}{\text{energy dissipated per cycle}}$$
  Higher $Q$ means less relative loss (for a given amount of stored energy). In practical matching networks, raising “loaded $Q$” can increase circulating currents and thus component loss if component $Q$ is finite, but as a rule, higher $Q$ itself does not imply higher loss.
- D: Harmonics increase — Incorrect. A higher-$Q$ network is more selective and tends to better suppress out-of-band signals, including harmonics.

Useful formulas and rules of thumb:
- Bandwidth relation:
  $$BW \approx \frac{f_0}{Q}, \quad \text{fractional BW} \approx \frac{1}{Q}$$
- Component-level $Q$ (series form at resonance):
  $$Q = \frac{|X|}{R} = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}$$
- Minimum loaded $Q$ for an L-network matching real resistances ($R_H > R_L$):
  $$Q_{\min} = \sqrt{\frac{R_H}{R_L} - 1}$$
  Larger transformation ratios force higher loaded $Q$, hence narrower matching bandwidth.