Question

a radar for tracking aircraft broadcasts at 12 GHz microwave beam from a 2.0 m-diameter circular radar antenna. From a wave perspective, the antenna is a circular aperture through which the microwaves diffract.

A. what is the diameter of the radar beam at a distance of 30 km?

B. if the antenna emits 100 kW of power what is the average microwave intensity at 30 km?

Answer 1

(A)What is the diameter of the radar beam at a distance of 30 km?

The wavelength of the wave is,

\(\begin{aligned} \lambda &=\frac{c}{f} \\ &=\frac{3 \times 10^{8} \mathrm{~m} / \mathrm{s}}{12 \times 10^{9} \mathrm{~Hz}} \\ &=0.025 \mathrm{~m} \\ &=25 \mathrm{~mm} \end{aligned}\)

The angular resolution of the system is,

\(\sin \theta=1.22\left(\frac{\lambda}{D}\right)\)

\( \begin{aligned} =1.22\left(\frac{25 \mathrm{~mm}}{2 \mathrm{~m}}\right) \\ \begin{aligned} \theta &=\sin ^{-1} (0.01525) \\ &=15.25 \mathrm{mrad} \end{aligned} \end{aligned} \)

The diameter of the radar beam from the half angle is,

\(d=2(15.25 \mathrm{mrad})\left(30 \mathrm{~km}\left(\frac{1000 \mathrm{~m}}{1 \mathrm{~km}}\right)\right)\)

\(=915 \mathrm{~m}\)

(B) if the antenna emits 100 kW of power what is the average microwave intensity at 30 km?

The average microwave intensity is,

\(\begin{aligned} I &=\frac{P}{A} \\ &=\frac{P}{\pi r^{2}} \\ &=\frac{4 P}{\pi d^{2}} \\ &=\frac{4\left(100 \times 10^{3} \mathrm{~W}\right)}{\pi(915 \mathrm{~m})^{2}} \\ &=0.152 \mathrm{~W} / \mathrm{m}^{2} \end{aligned}\)