Practice Problems (KEY)

Practice Problems (KEY)#

  1. A CIA operative with SOF support is trying to upload data to a Cessna flying at 1000 ft AGL in the People’s Republic of Pineland. He is using a VSAT (very small aperture antenna), a parabolic dish, to upload the data. The CIA operator’s antenna is 1 m in diameter, has an efficiency of 80%, and transmits 120 W at 3.7 GHz. The Cessna has a parabolic dish that is 500 mm in diameter with an efficiency of 75%. Assuming the Cessna is 80 km away and requires at least 200 nW, does the CIA operative have valid communications? If so, what is the received power at the Cessna? (Assume the CIA terminal is at 30 ft AGL)

\[ r_{\max} = \sqrt{2h_{\text{CIA}}} + \sqrt{2h_{\text{Cessna}}} = \sqrt{2 \cdot 30} + \sqrt{2 \cdot 1000} = 52.47\ \text{mi} \]
\[ 52.47\ \text{mi} \times 1.61 = 84.47\ \text{km} \]

84.5 km > 80 km → Yes, valid channel

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{3.7 \times 10^{9}} = 0.08108\ \text{m} \]
\[ G_T = \frac{(2\pi r)^2}{\lambda^2}\eta = \frac{(2\pi \cdot 0.5)^2}{(0.08108)^2}(0.8) = 1201 \]
\[ G_R = \frac{(2\pi r)^2}{\lambda^2}\eta = \frac{(2\pi \cdot 0.25)^2}{(0.08108)^2}(0.75) = 281.5 \]
\[ P_R = P_T G_T G_R \left(\frac{\lambda}{4\pi R}\right)^2 = (120)(1201)(281.5)\left(\frac{0.08108}{4\pi \cdot 80000}\right)^2 = 264\ \text{nW} \]

Yes, received

  1. A search and rescue helicopter is at 450 ft AGL, while the airman is on a 50 ft hill. The radio transmits 6 W at 2 MHz with antenna gain 3. The helicopter has gain 3.2 and requires 3 µW.

\[ r_{\text{mi}} = \sqrt{2 \cdot 450} + \sqrt{2 \cdot 50} = 40\ \text{mi} \]
\[ r_{\text{km}} = 1.61 \cdot 40 = 64.4\ \text{km} > 50\ \text{km} \]
\[ \lambda = \frac{3 \times 10^{8}}{2 \times 10^{6}} = 150\ \text{m} \]
\[ R = \frac{\lambda}{4\pi} \sqrt{\frac{P_T G_T G_R}{P_R}} = \frac{150}{4\pi} \sqrt{\frac{(6)(3)(3.2)}{3 \times 10^{-6}}} = 52.3\ \text{km} \]

Max range = 52.3 km (limiting factor)

  1. T / F: Decibels are fundamentally used to depict a voltage ratio.

Decibels fundamentally represent a power ratio.

  1. Two communication towers are 40 miles apart.

a. Required height:

\[ 40 = 2\sqrt{2h} \]
\[ 20^2 = 2h \Rightarrow h = 200\ \text{ft} \]

b. Dish radius:

\[ \lambda = \frac{3 \times 10^{8}}{3.5 \times 10^{9}} = 0.08571\ \text{m} = 85.71 {mm} \]
\[ r = \frac{\lambda}{2\pi}\sqrt{\frac{G}{\eta}} = \frac{0.08571}{2\pi}\sqrt{\frac{1986}{0.91}} = 0.637\ \text{m} = 637 {mm} \]

  1. UAS detection problem

\[ R_{\text{LOS}} = \sqrt{2 \cdot 800} + \sqrt{2 \cdot 25} = 47.07\ \text{mi} = 75.78\ \text{km} \]

80 km > 75.78 km → No LOS → cannot detect

(If LOS existed:)

\[ \lambda = \frac{3 \times 10^{8}}{24 \times 10^{6}} = 12.5\ \text{m} \]
\[ P_R = (120)(1.5)(3)\left(\frac{12.5}{4\pi \cdot 80000}\right)^2 = 83.5\ \text{nW} \]

Would be detectable if LOS existed.

  1. HC-130 problem

a. LOS limit:

\[ R_{\text{LOS}} = \sqrt{2 \cdot 10000} = 141.4\ \text{mi} = 227.7\ \text{km} \]
\[ \lambda = \frac{3 \times 10^{8}}{243 \times 10^{6}} = 1.235\ \text{m} \]
\[ R = \frac{\lambda}{4\pi} \sqrt{\frac{P_T G_T G_R}{P_R}} = 25550\ \text{km} \]

System is LOS-limited → max = 227.7 km

b. SNR at LOS distance:

\[ P_R = (4)(2.8)(3)\left(\frac{1.235}{4\pi \cdot 227700}\right)^2 = 6.26\ \text{pW} \]
\[ SNR = \frac{6.26 \times 10^{-12}}{66.24 \times 10^{-18}} = 94500 \]
\[ SNR_{dB} = 10\log_{10}(94500) = 49.75\ \text{dB} \]