# 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.**