Beamwidth of reflector antenna problem

Question

The beamwidth of a reflector antenna of diameter $=70 $ cm at $30$ GHz is:
a) $3.7$ degree
b) $0.5$ degree
c) $1.1$ degree
d) $1$ degree

My Approach:
We Know Directivity : $D= \frac{4 \pi}{\lambda^2}A_e$
Here, $A_e \approx \pi (\frac{d}{2})^2$
so, $D= (\frac{\pi d}{\lambda})^2=48361.06$ , as $d=70$ cm & $f= 30$ GHz (Given)

Now if i take the relation between beamwidth and directivity as $D=\frac{41253}{\theta}$ ;
then, beamwidth ($\theta$)= $0.853 $ degree $\approx $ (d) $1$ degree

And if i take $D=\frac{32400}{\theta}$ ;
then, $\theta= 0.67$ degree $\approx $ (b) $0.5$ degree

so which formula is applicable here? or if any other process is valid for this question then please explain it...

For a typical parabolic reflector the half power beam width in degrees is approximately $\frac{70,\rm wavelength}{\rm diameter}$ https://en.m.wikipedia.org/wiki/Parabolic_antenna#Beamwidth — Farcher, Mar 21 '18 at 06:40

score 2 · Answer 1 · edited Sep 14 '18 at 18:57

I don't know where those constants in your formula came from. $D=41253/\theta$ and $D=32400/\theta$. Maybe you forgot the gain efficiency.

Using the formula as commented and as referenced to Wikipedia, $$ BW = \frac{70\cdot \text{wavelength}}{\text{diameter}} $$ $$ BW = \frac{70\cdot 3\times10^8/30\times10^9}{0.7}=1^\circ $$

And using the Gain as you have pointed out. $$ G = e_a\frac{\pi d}{\lambda}=\frac{\pi k}{\theta}e_a $$ $$ 0.55\cdot\frac{\pi \cdot 0.7}{3\times10^8/30\times10^9}=\frac{\pi \cdot 70}{\theta}\cdot 0.55 $$ Solving for beam with $\theta$ gives $$\theta=1^\circ$$

Beamwidth of reflector antenna problem

1 Answers1