Lesson 39 Practice Problems (KEY)

Lesson 39 Practice Problems (KEY)#

Problem 1

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.

Problem 2

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.

Problem 3

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.

Problem 4 — Multi-Part

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.

Problem 5 — Multi-Part

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.

Problem 6 — Multi-Part

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.

Problem 7

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.

Problem 8 — Multi-Part

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.

Problem 9

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.

Problem 10 — Multi-Part

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.