An airplane is flying overhead with an approach angle of 60°. If a RADAR transmits at 300 MHz and receives a return at 300.000125 MHz, how fast is the airplane traveling? **Understand:** The Doppler equation can be rearranged to solve for velocity given the transmitted and received frequencies. **Identify Key Information:** - **Knowns:** $f_o = 300\ \text{MHz}$, $f = 300.000125\ \text{MHz}$, $\theta = 60°$ - **Unknowns:** Target velocity, $v$ - **Assumptions:** None. **Plan:** Rearrange the closing-target Doppler equation to solve for $v$. **Solve:** $$ v = \frac{c(f - f_{0})}{2f_{0}\cos\theta} = \frac{(3 \times 10^{8}\ m/s)(300.000125\ \text{MHz} - 300\ \text{MHz})}{2(300\ \text{MHz})\cos(60^\circ)} = 125\ m/s $$ **Answer:** The airplane is traveling toward the RADAR at 125 m/s.

A RADAR with a PRI of 200 $\mu$s is trying to detect an aircraft that is 50 km away. The RADAR operator does not realize the power level is too low to be a viable signal. How far away will the RADAR think the aircraft is? **Understand:** When a target is beyond the maximum unambiguous range, the RADAR mistakes the return for a closer target from the next pulse interval. **Identify Key Information:** - **Knowns:** $PRI = 200\ \mu\text{s}$, actual range $= 50\ \text{km}$ - **Unknowns:** Maximum unambiguous range; apparent range reported by the RADAR - **Assumptions:** The RADAR attributes all returns to the most recent pulse. **Plan:** First calculate the maximum unambiguous range. Since the target is beyond it, determine the apparent range by subtracting the unambiguous range from the true range. **Solve:** $$ R_{unamb} = \frac{c(PRI)}{2} = \frac{(3 \times 10^{8}\ m/s)(200 \times 10^{-6}\ s)}{2} = 30\ \text{km} $$ Since the aircraft at 50 km is outside this range, the correct round-trip time would be $$ t = \frac{2R}{c} = \frac{2(50{,}000\ m)}{3 \times 10^{8}\ m/s} = 333.3\ \mu s $$ The RADAR assumes the echo belongs to the next pulse (emitted 200 $\mu$s later), so it computes the range using only the remaining time $$ R_{apparent} = \frac{c(t - PRI)}{2} = \frac{(3 \times 10^{8}\ m/s)(333.3 - 200) \times 10^{-6}\ s}{2} = 20\ \text{km} $$ **Answer:** The RADAR will report the aircraft as being 20 km away.

A RADAR altimeter uses a pulse width of 1 $\mu$s. Would you feel safe flying in this airplane? Why or why not? **Understand:** A RADAR altimeter measures altitude above the ground. Its range resolution determines how close to the ground the aircraft can be before the RADAR cannot distinguish the aircraft from the terrain. **Identify Key Information:** - **Knowns:** $\tau = 1\ \mu\text{s}$ - **Unknowns:** Range resolution $\Delta R$ - **Assumptions:** None. **Plan:** Calculate $\Delta R$ and evaluate whether it is suitable for an altimeter application. **Solve:** $$ \Delta R = \frac{c\tau}{2} = \frac{(3 \times 10^{8}\ m/s)(1 \times 10^{-6}\ s)}{2} = 150\ \text{m} $$ **Answer:** No — the RADAR cannot differentiate between objects closer than 150 m apart. An altimeter with 150 m of resolution cannot reliably distinguish the aircraft from the ground at low altitudes, making this design unsafe.

A SAM RADAR uses a PRF of 6 kHz and a pulse width of 300 ns. **a.** PRI is the reciprocal of PRF: $$ PRI = \frac{1}{PRF} = \frac{1}{6000\ \text{Hz}} = 166.7\ \mu s $$ **b.** The maximum unambiguous range is $$ R_{unamb} = \frac{c(PRI)}{2} = \frac{(3 \times 10^{8}\ m/s)(166.7 \times 10^{-6}\ s)}{2} = 25\ \text{km} $$ Since you are at 28 km — outside the maximum unambiguous range — the RADAR does not know your exact location without additional processing. **c.** The range resolution is $$ \Delta R = \frac{c\tau}{2} = \frac{(3 \times 10^{8}\ m/s)(300 \times 10^{-9}\ s)}{2} = 45\ \text{m} $$ The two-ship is spaced only 10 m apart, which is less than $\Delta R = 45\ \text{m}$. The SAM RADAR will see both aircraft as a single return. **Answer:** PRI = 166.7 μs; the RADAR does not know your exact position; the two aircraft appear as one.

Design a RADAR that resolves objects separated by 25 m and unambiguously detects targets out to 75 km. **Understand:** Range resolution is set by pulse width; maximum unambiguous range is set by PRI (and therefore PRF). Each requirement drives one design parameter independently. **Identify Key Information:** - **Knowns:** Required $\Delta R = 25\ \text{m}$; required $R_{unamb} = 75\ \text{km}$ - **Unknowns:** Pulse width $\tau$; PRF - **Assumptions:** None. **Plan:** Solve the range resolution equation for $\tau$, then solve the unambiguous range equation for PRI and convert to PRF. **Solve:** **a.** Solving the range resolution equation for $\tau$: $$ \tau = \frac{2\Delta R}{c} = \frac{2(25\ m)}{3 \times 10^{8}\ m/s} = 166.7\ \text{ns} $$ **b.** Solving the unambiguous range equation for PRI: $$ PRI = \frac{2R_{unamb}}{c} = \frac{2(75{,}000\ m)}{3 \times 10^{8}\ m/s} = 500\ \mu s $$ $$ PRF = \frac{1}{PRI} = \frac{1}{500 \times 10^{-6}\ s} = 2\ \text{kHz} $$ **Answer:** Use a pulse width of 166.7 ns and a PRF of 2 kHz.

A RADAR ground station wants its maximum unambiguous range to equal its LOS range of 20 km. Find the required PRF and the required antenna height. **Understand:** Setting the unambiguous range equal to the LOS range ensures any detected target has an unambiguous position. The antenna height determines LOS range using the $R_{LOS}$ equation (assuming ground-level targets). **Identify Key Information:** - **Knowns:** $R_{unamb} = R_{LOS} = 20\ \text{km}$ - **Unknowns:** PRF; antenna height $h$ - **Assumptions:** Targets are at ground level ($h_{target} \approx 0$). **Plan:** Solve the unambiguous range equation for PRF. Then solve the LOS equation for antenna height, converting units carefully. **Solve:** **a.** $$ PRF = \frac{c}{2R_{unamb}} = \frac{3 \times 10^{8}\ m/s}{2(20{,}000\ m)} = 7.5\ \text{kHz} $$ **b.** Convert 20 km to miles: $20\ \text{km} \div 1.61 = 12.42\ \text{mi}$ $$ R_{LOS} = \sqrt{2h} \rightarrow h = \frac{R_{LOS}^{2}}{2} = \frac{(12.42\ \text{mi})^{2}}{2} = 77.1\ \text{ft} $$ **Answer:** Use a PRF of 7.5 kHz and mount the antenna at least 77.1 ft AGL.

A UAS is moving directly away from you (approach angle = 0°) at 90 m/s. If your RADAR uses a frequency of 2.4 GHz, at what frequency will the signal return? **Understand:** Because the UAS is moving away, the return frequency will be lower than the transmitted frequency (negative Doppler shift). **Identify Key Information:** - **Knowns:** $f_o = 2.4\ \text{GHz}$, $v = 90\ \text{m/s}$, $\theta = 0°$ - **Unknowns:** Return frequency $f$ - **Assumptions:** None. **Plan:** Apply the retreating-target Doppler equation directly. **Solve:** $$ f = f_{o}\left[1 - \frac{2v\cos\theta}{c}\right] = (2.4\ \text{GHz})\left[1 - \frac{2(90\ m/s)\cos(0^\circ)}{3 \times 10^{8}\ m/s}\right] = 2.39999856\ \text{GHz} $$ **Answer:** The RADAR signal will return at 2.39999856 GHz. Do not round Doppler results — the useful information is in the small frequency shift.

A RADAR's operating frequency is 4 GHz. An enemy aircraft is approaching at 800 km/h directly toward the RADAR (approach angle = 0°). **Understand:** Convert the velocity to m/s first, then apply the Doppler equation with the appropriate sign for each case. **Identify Key Information:** - **Knowns:** $f_o = 4\ \text{GHz}$, $v = 800\ \text{km/h}$, $\theta = 0°$ - **Unknowns:** Return frequency for each case - **Assumptions:** None. **Plan:** Convert velocity, then apply closing (+) and retreating (−) forms. **Solve:** $$ v = \frac{800\ \text{km}}{1\ \text{hr}} \times \frac{1000\ m}{1\ \text{km}} \times \frac{1\ \text{hr}}{3600\ s} = 222.22\ m/s $$ **a.** Aircraft approaching: $$ f = f_{o}\left[1 + \frac{2v\cos\theta}{c}\right] = (4\ \text{GHz})\left[1 + \frac{2(222.22\ m/s)\cos(0^\circ)}{3 \times 10^{8}\ m/s}\right] = 4.000005926\ \text{GHz} $$ **b.** Aircraft retreating: $$ f = f_{o}\left[1 - \frac{2v\cos\theta}{c}\right] = (4\ \text{GHz})\left[1 - \frac{2(222.22\ m/s)\cos(0^\circ)}{3 \times 10^{8}\ m/s}\right] = 3.999994074\ \text{GHz} $$ **Answer:** Approaching: 4.000005926 GHz. Retreating: 3.999994074 GHz.

An unidentified aircraft is traveling toward you with an approach angle of 40°. Your RADAR transmits at 1.2 GHz and receives a return at 1.2000008465 GHz. How fast is the aircraft traveling? **Understand:** Since $f > f_o$, the aircraft is closing. Rearrange the closing-target Doppler equation to solve for velocity. **Identify Key Information:** - **Knowns:** $f_o = 1.2\ \text{GHz}$, $f = 1.2000008465\ \text{GHz}$, $\theta = 40°$ - **Unknowns:** Target velocity $v$ - **Assumptions:** None. **Plan:** Solve the closing-target Doppler equation for $v$. **Solve:** $$ f = f_{o}\left[1 + \frac{2v\cos\theta}{c}\right] \quad \Rightarrow \quad v = \frac{c(f - f_{0})}{2f_{0}\cos\theta} $$ $$ v = \frac{(3 \times 10^{8}\ m/s)(1.2000008465\ \text{GHz} - 1.2\ \text{GHz})}{2(1.2\ \text{GHz})\cos(40^\circ)} = 138.1\ m/s $$ **Answer:** The aircraft is traveling toward you at 138.1 m/s.

An F-16 ($RCS = 5\ m^2$) is flying at 1200 ft AGL approaching a RADAR station. The F-16's RWR has a gain of 3 and requires 3 $\mu$W to detect a signal. The RADAR station transmits from 50 ft AGL at 200 MHz with 4 kW. The RADAR antenna has a gain of 250 and requires 14 fW to identify a target. The RADAR has a PRF of 2 kHz and a pulse width of 200 ns. **Understand:** This problem requires five separate calculations using the RADAR equation, LOS equation, unambiguous range equation, Friis equation (for the RWR), and range resolution equation. **Identify Key Information:** - **Knowns:** All RADAR and RWR parameters as stated above. - **Unknowns:** $R_{RADAR}$, $R_{LOS}$, $R_{unamb}$, $R_{RWR}$, $\Delta R$ - **Assumptions:** Same antenna used for transmit and receive on the RADAR. **Plan:** Solve each quantity in turn using its respective equation. Start with $\lambda$ since it is needed for the RADAR and RWR range calculations. **Solve:** $$ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}\ m/s}{200 \times 10^{6}\ Hz} = 1.5\ \text{m} $$ **a.** Maximum RADAR detection range: $$ R_{RADAR} = \sqrt[4]{\frac{P_{T}G^{2}(RCS)\lambda^{2}}{(4\pi)^{3}P_{R,min}}} = \sqrt[4]{\frac{(4000\ W)(250)^{2}(5\ m^{2})(1.5\ m)^{2}}{(4\pi)^{3}(14 \times 10^{-15}\ W)}} = 100.3\ \text{km} $$ **b.** LOS range: $$ R_{LOS} = \sqrt{2(50\ \text{ft})} + \sqrt{2(1200\ \text{ft})} = 58.99\ \text{mi} \times 1.61\ \frac{\text{km}}{\text{mi}} = 95.0\ \text{km} $$ **c.** Maximum unambiguous range: $$ R_{unamb} = \frac{c}{2(PRF)} = \frac{3 \times 10^{8}\ m/s}{2(2 \times 10^{3}\ \text{Hz})} = 75\ \text{km} $$ **d.** Maximum RWR detection range (Friis equation): $$ R_{RWR} = \frac{\lambda}{4\pi}\sqrt{\frac{P_{T}G_{T}G_{R}}{P_{R,min}}} = \frac{1.5\ m}{4\pi}\sqrt{\frac{(4000\ W)(250)(3)}{3 \times 10^{-6}\ W}} = 119.4\ \text{km} $$ **e.** Range resolution: $$ \Delta R = \frac{c\tau}{2} = \frac{(3 \times 10^{8}\ m/s)(200 \times 10^{-9}\ s)}{2} = 30\ \text{m} $$ **Answer:** $R_{RADAR} = 100.3\ \text{km}$; $R_{LOS} = 95.0\ \text{km}$; $R_{unamb} = 75\ \text{km}$; $R_{RWR} = 119.4\ \text{km}$; $\Delta R = 30\ \text{m}$. The LOS range (95 km) is the binding constraint for detection — the RADAR has the power to detect at 100.3 km but can only achieve LOS out to 95 km. The F-16's RWR detects the RADAR at 119.4 km, before the RADAR can detect the F-16, so the F-16 sees the RADAR first.