Question E9D08:
What happens as the Q of an antenna increases?  

Choices:  
A: SWR bandwidth increases  
B: SWR bandwidth decreases  
C: Gain is reduced  
D: More common-mode current is present on the feed line    

Correct Answer is  B: SWR bandwidth decreases  

Correct answer: B — SWR bandwidth decreases

Why B is correct:
- A higher Q means a sharper resonance. The antenna’s impedance changes more rapidly with frequency around resonance, so the frequency span over which the SWR stays below a given value (e.g., 2:1) becomes smaller. In other words, higher Q → narrower usable (SWR) bandwidth.

Why the others are incorrect:
- A: SWR bandwidth increases — Opposite of what happens; higher Q narrows bandwidth.
- C: Gain is reduced — Q primarily describes energy storage vs. loss and relates to bandwidth and selectivity, not directly to far-field gain. An antenna can have high Q yet similar peak gain at resonance; any gain change is not a general consequence of higher Q.
- D: More common-mode current is present on the feed line — Common-mode current is mainly about feedline/antenna imbalance and lack of choking, not the antenna’s Q.

Useful formulas and rules of thumb:
- Definition:  
  $$Q = 2\pi \frac{\text{energy stored}}{\text{energy dissipated per cycle}}$$
- For a resonant system (3 dB bandwidth):  
  $$Q \approx \frac{f_0}{BW_{3\,\text{dB}}}$$
- For SWR bandwidth around resonance (assuming a good match at $f_0$):  
  $$\frac{BW_S}{f_0} \approx \frac{S - 1}{Q\sqrt{S}}$$
  So for 2:1 SWR:  
  $$BW_{2:1} \approx 0.707 \frac{f_0}{Q} \quad\text{or}\quad Q \approx 0.707 \frac{f_0}{BW_{2:1}}$$

Key takeaway: Increase Q → decrease SWR bandwidth.