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Radar Systems

Solved exercises on radar systems: radar equation in dB, Doppler, compression, blind speeds, arrays, clutter, FMCW and SAR

Ten exercises at three levels with step-by-step solutions and graphs: range and second trace, budget in decibels, Doppler, pulse compression, staggered PRFs, fluctuating targets, array design, MTI and phase noise, an FMCW car radar, a real satellite SAR.

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Exercises on radar systems of the kind found in exams, at three levels: basic, with the fundamental formulas (range, radar equation in decibels, Doppler, pulse compression); intermediate, with design choices (blind speeds, detection, arrays, clutter); exam, with the design of an automotive radar and the analysis of a real satellite SAR. Constants: c = 3\cdot 10^8 m/s, kT_0 = -174 dBm/Hz. The result is given at once; the full solution, step by step and with the graph, opens with a click. The proofs of the formulas are in the lessons of the course.

Basic level

Exercise 1. Range and ambiguity

A radar receives an echo 400 μs after transmitting a pulse. How far is the target? If the radar transmits with a PRF of 2 kHz, what is the unambiguous range, and where does a target located at 90 km appear?

Result: 60 km; 75 km; the target at 90 km appears at 15 km.

Solution

1. The wave makes a round trip: R = \frac{c\tau}{2} = \frac{3\cdot 10^8\cdot 400\cdot 10^{-6}}{2} = 60 km. A handy rule to remember: 150 m per microsecond, so 400 μs give 400\cdot 150 = 60\,000 m.

2. Between two pulses 1/PRF = 500 μs elapse. An echo arriving before the next pulse is attributed correctly; the maximum unambiguous range is R_{amb} = \frac{c}{2\,PRF} = \frac{3\cdot 10^8}{4000} = 75 km.

3. The target at 90 km answers after 2\cdot 90\,000/c = 600 μs, that is 100 μs after the next pulse. The radar attributes it to that pulse and places it at 100\cdot 0.15 = 15 km: 90 - 75 = 15 km. It is a “second-trace” target. To unmask it, the PRF is changed from one group of pulses to the next: a true target stays at the same range, the second-trace one jumps.

A second-trace echo
On the time axis two transmitted pulses 500 microseconds apart; the echo at 600 microseconds from the first pulse falls 100 microseconds after the second, and the radar places it at 15 kilometres instead of 90.

Exercise 2. The radar equation in decibels

An S-band radar (3 GHz) transmits 100 kW peak with a 30 dB antenna. Compute the power received from a 2 m² aircraft at 50 km, and the signal-to-noise ratio with 1 MHz of bandwidth and a 3 dB noise figure.

Result: P_r = -97.9 dBm; noise −111.0 dBm; SNR = 13.0 dB.

Solution

1. In decibels the radar equation P_r = \frac{P_tG^2\lambda^2\sigma}{(4\pi)^3R^4} becomes a sum. With \lambda = c/f = 0.1 m:

termvaluedB
P_t100 kW+80.0 dBm
G^230 dB twice+60.0
\lambda^20.01 m²−20.0
\sigma2 m²+3.0
1/(4\pi)^31/1984−33.0
1/R^4R = 5\cdot 10^4 m−188.0

Sum: P_r = -97.9 dBm, that is 1.6\cdot 10^{-13} W.

2. The noise: N = kT_0 + 10\log_{10}B + F = -174 + 60 + 3 = -111.0 dBm.

3. SNR = -97.9 - (-111.0) = 13.0 dB: exactly the value needed for P_d = 0.9 and P_{fa} = 10^{-6} with one pulse (lesson on detection). 50 km is therefore the range of this radar for this aircraft, without integration and without losses. With 6 dB of losses the range would drop by 10^{-6/40} = 0.71 times, to 35 km; integrating 10 pulses (about 8 dB) would bring it back up.

Exercise 3. The Doppler of a car

A 24 GHz police radar measures a car approaching at 120 km/h. What is the Doppler shift? And if the radar is at the side of the road and sees the car at an angle of 30° to the direction of travel?

Result: 5.34 kHz; 4.62 kHz (the measured speed would be 104 km/h).

Solution

1. \lambda = c/f = 1.25 cm, v = 120/3.6 = 33.3 m/s: f_D = \frac{2v}{\lambda} = \frac{66.7}{0.0125} = 5.34 kHz.

2. Doppler sees only the radial component: v_r = v\cos 30° = 28.9 m/s, f_D = 4.62 kHz. The radar, if it does not know the angle, measures 120\cos 30° = 104 km/h: the cosine error always favours the driver (the measured speed is less than or equal to the true one). This is why Doppler speed cameras are installed at small, known angles, and the correction is applied by design.

Exercise 4. Pulse compression

A radar transmits a 50 μs chirp with 5 MHz of bandwidth. Compute the compression ratio, the range resolution and the blind zone in front of the radar. How long would a simple pulse with the same resolution be, and how much more peak power would be needed for the same energy?

Result: BT = 250; 30 m; 7.5 km; a 0.2 μs pulse with 250 times the peak power (+24 dB).

Solution

1. BT = 5\cdot 10^6\cdot 50\cdot 10^{-6} = 250.

2. After the matched filter the resolution depends only on bandwidth: \Delta R = \frac{c}{2B} = 30 m.

3. While transmitting, the radar does not receive: the blind zone is \frac{cT}{2} = 7.5 km. For close targets a short pulse is added.

4. A simple pulse with the same resolution lasts 1/B = 0.2 μs; for the same energy P\cdot T a peak power 50/0.2 = 250 times higher would be needed, that is 24 dB. This is exactly the advantage of compression.

Intermediate level

Exercise 5. Blind speeds and staggered PRF

An S-band surveillance radar (3 GHz) uses a two-pulse MTI canceller. Compute the first blind speed with a PRF of 1000 Hz and of 1250 Hz. Alternating the two PRFs, what is the first common blind speed? In the range from 20 to 240 m/s, what is the worst response when the better of the two PRFs is chosen for each speed?

Result: 50 m/s and 62.5 m/s; 250 m/s; 0.68, that is −3.3 dB relative to unit gain.

Solution

1. The 1 - z^{-1} canceller has response 2|\sin(\pi f_D/PRF)|, zero for f_D = k\cdot PRF. With f_D = 2v/\lambda and \lambda = 0.1 m: v_b = k\frac{\lambda\,PRF}{2}, that is 50 m/s (and multiples) with 1000 Hz, 62.5 m/s with 1250 Hz.

2. A speed is blind for both if it is a multiple of both 50 and 62.5: the least common multiple is 250 m/s. The 4:5 ratio between the PRFs multiplies the first blind speed by 5 compared with 1000 Hz. With three well-chosen PRFs one gets beyond the speeds of the targets of interest.

3. Computing the maximum of the two responses on a fine grid, the minimum value between 20 and 240 m/s is 0.68, that is −3.3 dB, at the points where the two responses cross (at 55.6 and 194.4 m/s): no speed of interest is lost, at the cost of a few decibels in some zones. Below 20 m/s the response falls towards zero anyway: it is the notch that removes stationary clutter.

MTI response with two PRFs
In blue the response with a PRF of 1000 hertz, with nulls every 50 metres per second; in orange that with 1250 hertz, with nulls every 62.5; the black line is the better of the two, which never goes to zero before 250 metres per second.

Exercise 6. Constant and fluctuating target

With P_{fa} = 10^{-6} and an SNR of 15 dB on a single pulse, compute P_d for a constant target and for a Swerling 1 target. With how many dB per pulse is P_d = 0.9 obtained (constant target) integrating 10 pulses non-coherently?

Result: 0.997 (constant); 0.655 (Swerling 1); 5.0 dB per pulse.

Solution

1. Constant target: the threshold is \eta = \sigma\sqrt{-2\ln P_{fa}} = 5.26\,\sigma; with SNR = 31.6 the echo amplitude is \sqrt{2\cdot 31.6} = 7.95\,\sigma, well above the threshold. Marcum’s function gives P_d = 0.997.

2. Swerling 1: P_d = P_{fa}^{1/(1 + SNR)} = (10^{-6})^{1/32.6} = e^{-13.82/32.6} = 0.655. With the same mean SNR, the fluctuating target is lost one time in three: in the looks where its radar cross section is small, the echo stays below the threshold.

3. Albersheim with N = 10: SNR = -5\log_{10}10 + \big(6.2 + \frac{4.54}{\sqrt{10.44}}\big)\log_{10}(A + 0.12AB + 1.7B) = 4.99 dB, with A = \ln(0.62/10^{-6}) = 13.34, B = \ln 9 = 2.20. Against the 13.1 dB of a single pulse, integration gains 8.1 dB.

Probability of detection versus signal-to-noise ratio
With a false alarm probability of one in a million: the blue curve, constant target, rises steeply from 10 to 15 decibels; the red curve, Swerling 1, rises slowly and reaches 0.9 only at 21 decibels.

Exercise 7. Designing an array

A linear array of 64 elements is spaced 0.6\lambda. Up to what angle can it be steered without grating lobes? How wide is the beam at broadside and at 40°? What phase shift between adjacent elements is needed to steer to 40°?

Result: 41.8°; 1.32° and 1.73°; 138.8°.

Solution

1. Condition: d \le \frac{\lambda}{1 + |\sin\theta_{max}|}, hence \sin\theta_{max} \le \frac{1}{0.6} - 1 = 0.667, \theta_{max} = 41.8°. Spacing the elements more than half a wavelength saves (expensive) modules but limits the scan range.

2. \theta_{3dB} \approx \frac{0.886\,\lambda}{Nd\cos\theta_0}: at broadside \frac{0.886}{38.4} = 0.0231 rad, that is 1.32°; at 40° divide by \cos 40° = 0.766: 1.73°.

3. \Delta\varphi = 360°\cdot\frac{d}{\lambda}\sin\theta_0 = 360\cdot 0.6\cdot 0.643 = 138.8°. With a 4-bit phase shifter (22.5° steps) 135° would be used, with a residual error that, repeating along the array, is spread with a known random phase.

Array factor of 64 elements spaced 0.6 wavelengths
In blue the beam steered to 40 degrees, without grating lobes; in orange steered to 55 degrees, beyond the limit: a grating lobe appears around minus 60 degrees.

Exercise 8. How much the MTI must cancel

A radar with a 1.2° beam and a 0.5 μs compressed pulse looks at a small 0.5 m² light aircraft at 30 km, over terrain with \sigma^0 = -25 dB. Compute the clutter radar cross section, the signal-to-clutter ratio, the improvement factor needed to reach 13 dB, and the maximum compatible transmitter phase noise.

Result: \sigma_c = 149 m² (21.7 dBsm); SCR = −24.7 dB; 37.7 dB needed; phase noise below 9.2 mrad (0.53°).

Solution

1. Cell area: A_c = R\,\theta_B\,\frac{c\tau}{2} = 30\,000\cdot 0.0209\cdot 75 = 47\,100 m². \sigma_c = \sigma^0A_c = 0.00316\cdot 47\,100 = 149 m².

2. SCR = 10\log_{10}\frac{0.5}{149} = -24.7 dB: the aircraft is 300 times weaker than the clutter of its cell.

3. To reach +13 dB an improvement of 13 + 24.7 = 37.7 dB is needed.

4. The two-pulse canceller with phase noise \sigma_\varphi leaves a residue 2\sigma_\varphi^2 (lesson on transmitters and receivers): \frac{1}{2\sigma_\varphi^2} \ge 10^{3.77} gives \sigma_\varphi \le 9.2 mrad. A magnetron, with random phase, would not make it even with coherent-on-receive; a coherent transmitter with a good oscillator is needed. Note that shortening the pulse (more bandwidth) shrinks the cell and directly improves the signal-to-clutter ratio.

Exam level

Exercise 9. Designing an FMCW car radar

A 77 GHz radar is required with a range resolution of 15 cm, an unambiguous velocity of ±40 m/s, a velocity resolution of at least 0.1 m/s and a range of 100 m. Compute bandwidth, maximum chirp duration, number of chirps (a power of 2), frame time, chirp slope and beat frequency at 100 m.

Result: B = 1 GHz; T_c \le 24.3 μs; 800 chirps, hence 1024; 24.9 ms; 41 MHz/μs; 27.4 MHz.

Solution

1. \Delta R = \frac{c}{2B}, hence B = \frac{c}{2\cdot 0.15} = 1 GHz (within the 77-81 GHz band).

2. The unambiguous velocity is \frac{\lambda}{4T_c} with \lambda = 3.9 mm: T_c \le \frac{\lambda}{4\cdot 40} = 24.3 μs.

3. \Delta v = \frac{\lambda}{2NT_c}, hence N \ge \frac{\lambda}{2\cdot 0.1\cdot T_c} = 800; with the FFT 1024 are used, giving \Delta v = 0.078 m/s.

4. Frame: 1024\cdot 24.3 μs = 24.9 ms, compatible with an update every 50 ms.

5. Slope S = B/T_c = 41 MHz/μs; at 100 m the beat is \frac{2SR}{c} = 27.4 MHz: the quadrature ADC must sample at least at 27.4 MS/s (more in practice, for the anti-aliasing filter). The design shows the trade-offs: short chirps for velocity, many chirps for velocity resolution, bandwidth for range, and a fast ADC for maximum range.

The choices of the FMCW radar
Top, the sequence of 1 GHz chirps 24.3 microseconds long, repeated 1024 times in 24.9 milliseconds; bottom, the beat frequency growing linearly with range up to 27.4 megahertz at 100 metres.

Exercise 10. A real satellite SAR

A C-band satellite (5.405 GHz) with a 12.3 m antenna flies at 7.6 km/s and observes at a range of 850 km, with 56.5 MHz of bandwidth and an incidence angle of 35° (values similar to those of Sentinel-1). Compute the azimuth resolution with the real antenna, the length of the synthetic aperture, the SAR azimuth resolution, the minimum PRF, the illumination time of a point and the slant and ground range resolutions.

Result: 3.8 km; 3.8 km; 6.15 m; 1236 Hz; 0.50 s; 2.65 m slant, 4.6 m on the ground.

Solution

1. \lambda = 5.55 cm. Real antenna: \frac{\lambda R}{D} = \frac{0.0555\cdot 850\,000}{12.3} = 3.8 km.

2. Each point stays in the beam for a stretch of orbit as long as the footprint: L = 3.8 km, travelled in 3833/7600 = 0.50 s.

3. SAR resolution: \frac{D}{2} = 6.15 m, independent of range and wavelength.

4. PRF \ge \frac{2v}{D} = \frac{15\,200}{12.3} = 1236 Hz (Sentinel-1 uses PRFs between about 1 and 3 kHz depending on the mode).

5. In slant range \frac{c}{2B} = 2.65 m; projected on the ground, divide by \sin 35°: 4.6 m. With small incidence angles the ground resolution gets much worse (with 20° it would be 7.8 m). In wide-swath modes, to cover 250 km, the satellite divides its time among sub-swaths and the real azimuth resolution worsens to about 20 m: a trade-off between coverage and detail.

The geometry of the satellite SAR
The satellite at 850 kilometres illuminates the ground with a beam 3.8 kilometres wide in azimuth; each point is seen along a stretch of orbit of the same length, the synthetic aperture, in half a second. In range the 2.65-metre slant cell projects onto 4.6 metres on the ground.

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