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Knowledge Hubintegrated sensing and communication2. Radar Basics for Communication Engineers

integrated sensing and communication learning note

2. Radar Basics for Communication Engineers

How a radio measures range, speed and direction from an echo, why echoes are so much weaker than communication signals, and how a receiver decides that a target is really there.

At a glance

  • Three numbers from one echoDelay gives range, Doppler gives radial speed, and the phase across an array gives direction.
  • Echoes fade as R⁴A communication signal travels one way and fades as R²; an echo travels out and back, so it fades as R⁴.
  • Bandwidth separates, SNR sharpensResolution (telling two targets apart) depends on bandwidth; accuracy (pinning one target down) also improves with SNR.

Why Start With Radar

ISAC asks one radio to do two jobs. The communication job is familiar: send bits, decode bits. The sensing job comes from radar, and it uses a different vocabulary. This note builds that vocabulary from scratch, so the rest of the track reads easily.

A radar does three things with an echo:

  • Range: how long the echo took to come back tells how far the object is.
  • Velocity: how much the echo's frequency shifted tells how fast the object moves toward or away from the radio.
  • Angle: how the echo's phase changes across an antenna array tells which direction it came from.

Then it has to answer a fourth question first: is there an echo at all, or just noise?

References for this section3

Range: Timing the Echo

A wave travels at the speed of light c≈3×108c \approx 3\times10^8 m/s. If the echo arrives τ\tau seconds after transmission, the wave has covered the distance twice, out and back:

R=c τ2R = \frac{c\,\tau}{2}
  • RR: range to the target in metres.
  • τ\tau: round-trip delay in seconds.
  • Scale: 1 µs of delay is 150 m of range; 1 ns is 15 cm.

This is the monostatic case: transmitter and echo receiver at the same place. When they are apart (bistatic), the delay measures the total path transmitter → target → receiver, and the target lies somewhere on an ellipse with the two radios as foci.

References for this section3

Velocity: The Doppler Shift

A target moving toward the radio compresses the wave; moving away stretches it. The echo comes back at a slightly different frequency:

fD=2vrλ=2vrfccf_D = \frac{2 v_r}{\lambda} = \frac{2 v_r f_c}{c}
  • fDf_D: Doppler shift in hertz, positive for an approaching target.
  • vrv_r: radial speed, the part of the velocity along the line of sight.
  • λ\lambda, fcf_c: wavelength and carrier frequency.
  • Why the 2: the target sees a shifted wave and then re-radiates it while still moving, so the shift happens twice.

At 28 GHz, a pedestrian walking at 1.5 m/s gives about 280 Hz; a car at 30 m/s gives about 5.6 kHz. Only the radial part counts: a car crossing the beam at right angles produces almost no Doppler.

References for this section3

Angle: Where the Echo Came From

With several antennas spaced half a wavelength apart, an echo from angle θ\theta reaches neighbouring elements with a phase difference

Δϕ=πsin⁡θ.\Delta\phi = \pi \sin\theta .

Measuring that phase step across the array gives the direction. The array can tell two directions apart when they differ by roughly one beamwidth, about 2/N2/N radians for NN elements. This is the same physics as beam steering, used in reverse.

References for this section3

Why Echoes Are So Weak

A communication signal travels one way, and its power spreads over a sphere: it falls as 1/R21/R^2. An echo does that twice: once on the way to the target, and again after the target scatters it back. The radar equation captures this:

Pr=Pt Gt Gr λ2 σ(4π)3 R4P_r = \frac{P_t\, G_t\, G_r\, \lambda^2\, \sigma}{(4\pi)^3\, R^4}
  • PtP_t, GtG_t, GrG_r: transmit power and antenna gains.
  • σ\sigma: radar cross section (RCS) in m², how strongly the target scatters. A person is roughly 1 m², a car roughly 10–100 m², a small drone roughly 0.01 m².
  • R4R^4: the price of the round trip.

Compare a one-way link to a user at the same distance: its received power falls only as R2R^2. Double the distance and the user loses 6 dB, while the echo loses 12 dB. That is why sensing range is usually much shorter than communication range from the same base station, and why ISAC systems care so much about processing gain: integrating many symbols adds the energy back.

Why the echo fades so much faster

-160-140-120-100-80-60-40-20010 m100 m1 km10 kmreceived power (dBm)noise flooruser (∝ 1/R²)echo (∝ 1/R⁴)distance (log scale)
Target
Carrier
Communication range6032 muser SNR ≥ 0 dB
Sensing range310 mecho ≥ 13 dB after integration
Processing gain18.1 dB64 symbols
Ratio19×comm reach / sensing reach
Prcomm=PtGtGrλ2(4π)2R2,Precho=PtGtGrλ2σ(4π)3R4,SNRecho=PrechoN⋅MP_r^{\text{comm}} = \frac{P_t G_t G_r \lambda^{2}}{(4\pi)^{2} R^{2}}, \qquad P_r^{\text{echo}} = \frac{P_t G_t G_r \lambda^{2} \sigma}{(4\pi)^{3} R^{4}}, \qquad \mathrm{SNR}_{\text{echo}} = \frac{P_r^{\text{echo}}}{N}\cdot M
Rsense=(PtG2λ2σM(4π)3N SNRmin⁡)1/4=310 m(σ=10 m2, M=64)R_{\text{sense}} = \left(\frac{P_t G^{2} \lambda^{2} \sigma M}{(4\pi)^{3} N\, \mathrm{SNR}_{\min}}\right)^{1/4} = 310\ \text{m} \quad (\sigma = 10\ \text{m}^2,\ M = 64)
Received power against distance for a one-way link to a user (teal, falls 20 dB per decade) and for the echo from a car (orange, falls 40 dB per decade). The dashed orange line adds the processing gain from coherently combining 64 symbols. The communication range is where the user's SNR reaches 0 dB; the sensing range is where the integrated echo reaches 13 dB above noise, a typical detection threshold. 100 MHz bandwidth, 7 dB noise figure.
References for this section3

Resolution Versus Accuracy

Two different questions are easy to mix up:

  • Resolution: can I tell two nearby targets apart?
  • Accuracy: how precisely can I locate one target?

Range resolution depends only on bandwidth:

ΔR=c2B\Delta R = \frac{c}{2B}

100 MHz resolves targets 1.5 m apart; 1 GHz resolves 15 cm. Accuracy for a single target is much finer than that and improves with SNR, roughly as ΔR/SNR\Delta R / \sqrt{\mathrm{SNR}}. A high-SNR radar can locate one car to a few centimetres while still being unable to separate two cars parked half a metre apart.

Resolution: can we see two targets or just one blob?

target 1target 226.0 m30.5 m35.0 mrangetwo peaks: resolved
Resolution cell c/2B99.9 cm
Target gap1.00 m
VerdictTwo targetsgap ≥ resolution cell
ΔR=c2B=3×1082×150×106=99.9 cm,resolved if  ∣R2−R1∣≥ΔR\Delta R = \frac{c}{2B} = \frac{3\times10^{8}}{2\times 150\times10^{6}} = 99.9\ \text{cm}, \qquad \text{resolved if } \ |R_2 - R_1| \ge \Delta R
Two equal targets sit close together in range. Each one produces a peak whose width is set by the bandwidth, about c/2B. When the gap is smaller than that, the two peaks merge into one and the receiver sees a single, wider target. Wider bandwidth narrows the peaks and splits them apart.
References for this section3

Detection: Is Anything There?

Before measuring anything, the receiver has to decide whether a peak is a target or just noise. It compares each range cell against a threshold:

  • Detection probability PDP_D: the chance a real target crosses the threshold.
  • False-alarm probability PFAP_{FA}: the chance noise alone crosses it.

Raise the threshold and false alarms drop, but weak targets are missed. Lower it and you catch more targets, along with more ghosts. For noise whose power is exponentially distributed (complex Gaussian noise), the false-alarm probability for a threshold TT times the noise power is

PFA=e−T.P_{FA} = e^{-T}.

Real receivers rarely know the noise power exactly, so they estimate it from neighbouring cells and scale the threshold from that estimate. This is CFAR (constant false-alarm rate) detection, the standard first step in almost every radar and ISAC receiver.

Detection: target or just noise?

thresholdstrongweak-1001020power vs mean noise (dB)range cell
Noise realisation
Targets detected1 of 2
False alarms here0expected 0.00
P_FA per cell3.4e-6
PFA=e−T=e−1011/10=3.41×10−6,E[false alarms]=(Ncells−2) PFAP_{FA} = e^{-T} = e^{-10^{11/10}} = 3.41\times10^{-6}, \qquad \mathbb{E}[\text{false alarms}] = (N_{\text{cells}} - 2)\,P_{FA}
A range profile of 128 cells: noise everywhere, a strong target (16 dB) and a weaker one you can adjust. Every cell above the threshold is declared a target. Lower the threshold to catch the weak target and watch false alarms appear; raise it and the weak target disappears. The false-alarm probability per cell follows e^(−T) for noise power normalised to 1. Press “New noise” for another realisation.
References for this section3

Takeaway

  • Echo to numbers: delay → range, Doppler → radial speed, phase across the array → angle.
  • Budget: echoes fade as R4R^4, so sensing range is shorter than communication range and needs processing gain.
  • Two different goals: bandwidth sets resolution; SNR sharpens accuracy; a threshold decides what counts as a target.

The next note shows how a normal OFDM communication signal can perform all three measurements without sending any extra radar pulse.

References for this section3

Check your understanding

0 of 4 answered

  1. An echo returns 400 ns after transmission. How far away is the target?

  2. A car approaches at 15 m/s and the carrier is 28 GHz (λ ≈ 10.7 mm). What Doppler shift does the echo carry?

  3. Doubling the distance to a target changes the echo power by how much?

  4. Two cars are 0.5 m apart in range. Which bandwidth is the smallest that can resolve them?

Complete references and further reading