Practice Problems (KEY)

Practice Problems (KEY)#

Problem 1

A RADAR transmitting \(100\ \text{kW}\) at \(350\ \text{MHz}\) receives a return from a plane with an RCS of \(100\ \text{m}^2\) after \(250\ \mu s\).

\[ R = \frac{c \Delta t}{2} = \frac{(3 \times 10^8\ \text{m/s})(250 \times 10^{-6}\ \text{s})}{2} = 37.5\ \text{km} \]
Problem 2

An F-15 receives a TACAN signal after \(500\ \mu s\).

\[ d = c t = (3 \times 10^8\ \text{m/s})(500 \times 10^{-6}\ \text{s}) = 150{,}000\ \text{m} = 150\ \text{km} \]

The aircraft is \(150\ \text{km}\) away when it receives the signal.

Since the aircraft only has enough fuel for \(100\ \text{km}\), it must refuel.

Problem 3

Given:

  • \(\sigma = 20\ \text{m}^2\)

  • \(f = 1\ \text{GHz}\)

  • \(P_T = 50{,}000\ \text{W}\)

  • \(G = 50\)

  • \(R = 50\ \text{km} = 50{,}000\ \text{m}\)

  • \(P_{R,\min} = 10\ \text{pW}\)

Wavelength:

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8}{1 \times 10^9} = 0.3\ \text{m} \]

RADAR Equation:

\[ P_R = P_T G^2 \sigma \frac{\lambda^2}{(4\pi)^3 R^4} \]
\[ P_R = (50{,}000)(50)^2 (20) \frac{(0.3)^2}{(4\pi)^3 (50{,}000)^4} = 18.14\ \text{fW} \]

Since \(18.14\ \text{fW} \ll 10\ \text{pW}\):

The RADAR cannot lock on.

Problem 4
\[ R \propto \sigma^{1/4} \]
\[ \frac{R_{\text{fighter}}}{R_{\text{missile}}} = \left( \frac{20}{0.2} \right)^{1/4} = (100)^{1/4} = 3.16 \]

The fighter can be detected 3.16 times farther away than the missile.

Problem 5

Mobile RADAR Unit

UAS

\(f = 450\ \text{MHz}\)

\(\sigma = 0.4\ \text{m}^2\)

\(P_T = 1500\ \text{W}\)

\(G_R = 3\)

\(G = 200\)

\(P_{R,\min} = 1.25\ \mu\text{W}\)

\(P_{R,\min} = 1\ \text{fW}\)

(a) Range from Time Delay:

\[ R = \frac{c t}{2} = \frac{(3 \times 10^8)(133.3 \times 10^{-6})}{2} = 20.0\ \text{km} \]

(b) Received Power at \(30\ \text{km}\):

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8}{450 \times 10^6} = 0.6667\ \text{m} \]
\[ P_R = P_T G^2 \sigma \frac{\lambda^2}{(4\pi)^3 R^4} \]
\[ P_R = (1500)(200)^2 (0.4) \frac{(0.6667)^2}{(4\pi)^3 (30{,}000)^4} = 6.64\ \text{fW} \]

Since \(6.64\ \text{fW} > 1\ \text{fW}\):

RADAR detection is possible (if LOS is established).

(c) Who Sees Who First?

LOS Range:

\[ R_{\text{LOS}} = \sqrt{2h_{\text{UAS}}} + \sqrt{2h_{\text{RADAR}}} = \sqrt{2 \cdot 200'} + \sqrt{2 \cdot 45'} = 47.47\ \text{km} \]

RWR Range:

\[ R_{\text{RWR}} = \sqrt{\frac{P_T G_T G_R \lambda^2}{(4\pi)^2 P_{R,\min}}} \]
\[ R_{\text{RWR}} = \sqrt{\frac{(1500)(200)(3)(0.6667)^2}{(4\pi)^2 (1.25 \times 10^{-6})}} = 45.02\ \text{km} \]

RADAR Range:

\[ R_{\text{RADAR}} = \sqrt[4]{\frac{P_T G^2 \sigma \lambda^2}{(4\pi)^3 P_{R,\min}}} \]
\[ R_{\text{RADAR}} = \sqrt[4]{\frac{(1500)(200)^2 (0.4)(0.6667)^2}{(4\pi)^3 (1 \times 10^{-15})}} = 48.15\ \text{km} \]

Final Comparison:

  • \(R_{\text{LOS}} = 47.47\ \text{km}\)

  • \(R_{\text{RWR}} = 45.02\ \text{km}\)

  • \(R_{\text{RADAR}} = 48.15\ \text{km}\)

The system is LOS-limited, so detection occurs at \(R = 47.47\ \text{km}\).

Answer: The RADAR detects the UAS first.