Question E9B01:
What is the 3 dB beamwidth of the antenna radiation pattern shown in Figure E9-1?  

Choices:  
A: 75 degrees  
B: 50 degrees  
C: 25 degrees  
D: 30 degrees    

Correct Answer is  B: 50 degrees  

To find the 3 dB beamwidth, locate the main lobe’s peak on the polar plot, then find the two angles on either side where the pattern drops to −3 dB (half power). The 3 dB beamwidth is the angular separation between those two points.

Why 50 degrees is correct:
- From the figure, the −3 dB points occur about ±25° from the peak (boresight).
- Therefore, the full width at half maximum (FWHM) is:
  $$\text{BW}_{3\text{dB}} = (+25^\circ) - (-25^\circ) = 50^\circ$$

Why the other choices are incorrect:
- 75 degrees: Likely includes a side lobe or uses a deeper level (e.g., −10 dB) instead of −3 dB.
- 25 degrees: That’s the half-beamwidth (one side only), not the full width.
- 30 degrees: Likely a misread of the scale or using a point near −6 dB instead of −3 dB.

Useful formulas and rules of thumb:
- 3 dB beamwidth is the FWHM of the main lobe.
- −3 dB corresponds to half power:
  $$10\log_{10}\!\left(\frac{P}{P_{\max}}\right) = -3 \;\Rightarrow\; \frac{P}{P_{\max}} \approx 0.5$$
- In terms of field (voltage) magnitude:
  $$20\log_{10}\!\left(\frac{E}{E_{\max}}\right) = -3 \;\Rightarrow\; \frac{E}{E_{\max}} \approx 0.707$$
- Beamwidth definition:
  $$\text{BW}_{3\text{dB}} = \theta_2 - \theta_1$$
  where $\theta_1$ and $\theta_2$ are the angles at the −3 dB points around the main lobe.