Finding Object Velocity and Range Using FMCW Radar

FMCW Radar: The Mathematical Foundation

How one radar waveform can reveal an object's range, velocity, and direction using frequency, phase, Fourier transforms, and antenna geometry.

An FMCW radar does not directly measure range, velocity, and angle. Instead, it measures changes in frequency and phase. Those measurements are transformed into physical quantities.

Dimension	What changes?	Information
Samples within a chirp	Beat frequency	Range
Chirp to chirp	Doppler phase	Velocity
Receiver to receiver	Spatial phase	Angle

1. Start With an FMCW Chirp

FMCW stands for Frequency-Modulated Continuous Wave. Instead of transmitting a single frequency, the radar continuously sweeps its frequency.

A simple up-chirp can be written as:

[ f(t) = f_c + St ]

where:

• (f_c) = carrier frequency
• (S) = chirp slope in Hz/s
• (t) = time

The slope is:

[ S = \frac{B}{T_c} ]

where (B) is bandwidth and (T_c) is chirp duration.

Numerical Example

Suppose:

• Carrier frequency = 77 GHz
• Bandwidth = 1 GHz
• Chirp duration = 50 μs

Then:

[ S = \frac{1\times10^9}{50\times10^{-6}} = 2\times10^{13}\text{ Hz/s} ]

So the radar's frequency increases by approximately 20 THz every second.

2. Range: Why Does the Beat Frequency Give Distance?

Electromagnetic waves travel approximately at the speed of light:

[ c \approx 3\times10^8\text{ m/s} ]

If an object is at distance (R), the signal must travel to the object and back.

The round-trip delay is therefore:

[ \tau = \frac{2R}{c} ]

Because the transmitted chirp is changing frequency with time, the delayed signal has a slightly different instantaneous frequency from the signal currently being transmitted.

Mixing the transmitted signal with the received signal produces the beat frequency:

[ f_b = S\tau ]

Substituting the delay:

[ f_b = S\frac{2R}{c} ]

Therefore:

[ \boxed{R=\frac{c f_b}{2S}} ]

Numerical Range Example

Assume:

• (S=2\times10^{13}) Hz/s
• Object range (R=30) m
• (c=3\times10^8) m/s

The round-trip delay is:

[ \tau = \frac{2(30)}{3\times10^8} = 2\times10^{-7}\text{ s} = 200\text{ ns} ]

The beat frequency is:

[ f_b = (2\times10^{13})(2\times10^{-7}) = 4\times10^6 = \boxed{4\text{ MHz}} ]

Therefore, a target 30 m away produces approximately a 4 MHz beat frequency.

3. Why Do We Take an FFT for Range?

The ADC does not tell us directly:

'The target is 30 metres away.'

Instead, it gives us a sequence of voltage samples representing a signal that oscillates at approximately 4 MHz.

For example:

[ x[n] = A\cos(2\pi fb nTs) ]

The question becomes:

What frequencies are present in these samples?

The Fourier Transform answers exactly that question.

In practice, we use the Fast Fourier Transform (FFT), which is an efficient algorithm for calculating the Discrete Fourier Transform.

The DFT is:

[ X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N} ]

The FFT efficiently calculates these frequency components. A strong peak at frequency (f_b) means that the received signal contains a strong component at that beat frequency.

Important: The FFT is not magically calculating distance. It is finding frequency. We then use the relationship (R=cf_b/(2S)) to convert frequency into distance.

4. Range Resolution

An important property of FMCW radar is that range resolution is primarily determined by bandwidth.

[ \boxed{\Delta R = \frac{c}{2B}} ]

For (B=1) GHz:

[ \Delta R = \frac{3\times10^8}{2\times10^9} = \boxed{0.15\text{ m}} ]

So approximately 15 cm range resolution is possible under ideal assumptions.

More bandwidth → better range resolution.

5. Velocity: The Doppler Effect

Now imagine the target is moving. Motion changes the frequency of the returned wave. This is the Doppler effect.

The Doppler frequency for a monostatic radar is:

[ \boxed{f_D = \frac{2v}{\lambda}} ]

where:

• (v) = radial velocity
• (\lambda) = wavelength

The wavelength is:

[ \lambda = \frac{c}{f_c} ]

Numerical Velocity Example

For a 77 GHz radar:

[ \lambda = \frac{3\times10^8}{77\times10^9} \approx 3.896\text{ mm} ]

Suppose a car is approaching at (v=20) m/s.

[ f_D = \frac{2(20)}{0.003896} \approx \boxed{10.27\text{ kHz}} ]

6. Why Do We Take Another FFT for Velocity?

This is where FMCW radar becomes especially interesting.

We already used the samples within one chirp to determine range.

Now we look at the same range bin across many chirps.

Suppose the radar transmits chirps every (T_c) seconds.

The Doppler signal across chirps can be approximated as:

[ x[m] = Ae^{j2\pi fDmTc} ]

Notice what happened:

[ \text{Fast time} \rightarrow f_b \rightarrow \text{Range} ] [ \text{Slow time} \rightarrow f_D \rightarrow \text{Velocity} ]

Because the Doppler signal is also a frequency component, we again use an FFT.

First FFT: frequency across ADC samples → range.
Second FFT: frequency across chirps → velocity.

Numerical Slow-Time Example

Suppose chirps are separated by:

[ T_c=50\ \mu s ]

For (v=20) m/s we calculated:

[ f_D\approx10.27\text{ kHz} ]

The phase change between consecutive chirps is:

[ \Delta\phi = 2\pi fD Tc ]

Therefore:

[ \Delta\phi = 2\pi(10270)(50\times10^{-6}) \approx 3.23\text{ rad} ]

So the complex radar signal rotates by roughly 3.23 radians per chirp.

That phase rotation is the mathematical origin of Doppler estimation in the slow-time dimension.

7. Range-Doppler Map

After the first FFT, every chirp contains a set of range bins. Stack those chirps together:

[ \begin{bmatrix} \text{chirp 1}\ \text{chirp 2}\ \text{chirp 3}\ \vdots\ \text{chirp M} \end{bmatrix} ] [ \Downarrow ] [ X[\text{range},\text{chirp}] ]

Then perform another FFT along the chirp dimension.

[ X[n,m] \xrightarrow{\text{FFT in }n} X[kR,m] \xrightarrow{\text{FFT in }m} X[kR,k_D] ]

The result is a range-Doppler map.

Axis	Physical meaning
X-axis	Range
Y-axis	Velocity / Doppler
Brightness	Signal strength

8. Angle: Why Can Two Antennas Tell Direction?

Now consider two receiving antennas separated by distance (d).

A wave arriving at an angle does not reach both antennas at exactly the same time.

The path difference is approximately:

[ \Delta r = d\sin\theta ]

A path difference creates a phase difference:

[ \Delta\phi = \frac{2\pi}{\lambda}\Delta r ]

Substituting:

[ \boxed{ \Delta\phi = \frac{2\pi d\sin\theta}{\lambda} } ]

Solving for angle:

[ \boxed{ \theta = \sin^{-1} \left( \frac{\lambda\Delta\phi}{2\pi d} \right) } ]

Numerical Angle Example

Suppose:

• Radar frequency = 77 GHz
• (\lambda\approx3.896) mm
• Antenna spacing (d=\lambda/2)
• Target angle (\theta=30^\circ)

Then:

[ \Delta\phi = \frac{2\pi(\lambda/2)\sin(30^\circ)} {\lambda} ]

[ \Delta\phi = \pi(0.5) = \boxed{\frac{\pi}{2}\text{ rad}} ]

So a target at 30° produces approximately a 90° phase difference between the two receivers.

9. Why Only Two Receivers Are Limited

With two antennas, we essentially get one spatial phase measurement:

[ \Delta\phi = \phi2-\phi1 ]

That gives us an estimate of one angle.

But with more antennas, we obtain many spatial measurements:

[ \phi1,\phi2,\phi3,\ldots,\phiN ]

This allows us to perform spatial beamforming or a spatial FFT to obtain a more precise angular spectrum.

More antennas → more spatial samples → better angular discrimination.

10. Why MIMO Helps

MIMO stands for Multiple Input Multiple Output.

Multiple transmitters and receivers can create a larger virtual antenna array.

For example, with:

[ N{Tx}\times N{Rx} ]

appropriately arranged channels, the radar can obtain many virtual spatial samples.

This improves angular resolution and allows the radar to distinguish targets that are close together in angle.

11. Putting Everything Together

We can now see why the same raw radar data contains all three measurements.

Range

[ \boxed{f_b=S\frac{2R}{c}} ] [ \boxed{R=\frac{cf_b}{2S}} ]

Find the beat frequency using an FFT across samples within a chirp.

Velocity

[ \boxed{f_D=\frac{2v}{\lambda}} ] [ \boxed{v=\frac{\lambda f_D}{2}} ]

Find the Doppler frequency using an FFT across chirps.

Angle

[ \boxed{ \Delta\phi = \frac{2\pi d\sin\theta}{\lambda} } ] [ \boxed{ \theta = \sin^{-1} \left( \frac{\lambda\Delta\phi}{2\pi d} \right) } ]

Compare the phase received by spatially separated antennas.

12. The Mathematical Picture

The most useful way to think about FMCW radar is as a multidimensional signal-processing problem.

[ \boxed{ \text{Fast Time} \xrightarrow{\text{FFT}} \text{Range} } ] [ \boxed{ \text{Slow Time} \xrightarrow{\text{FFT}} \text{Velocity} } ] [ \boxed{ \text{Spatial Dimension} \xrightarrow{\text{Phase / FFT}} \text{Angle} } ]

Therefore, the radar data cube can conceptually be represented as:

[ \boxed{ X[\text{range},\text{velocity},\text{angle}] } ]

In real radar systems there are many additional complications: noise, clutter, multiple targets, acceleration, phase noise, leakage, range-Doppler coupling, antenna calibration, windowing, CFAR detection, and more.

13. Numerical Example

Consider a 77 GHz FMCW radar with:

• Bandwidth (B=1) GHz
• Chirp duration (T_c=50) μs
• Target range (R=30) m
• Target velocity (v=20) m/s
• Two receivers separated by (d=\lambda/2)
• Target angle (\theta=30^\circ)

Step 1 — Chirp slope

[ S= \frac{B}{T_c} = \frac{10^9}{50\times10^{-6}} = 2\times10^{13}\text{ Hz/s} ]

Step 2 — Range

[ f_b = S\frac{2R}{c} = (2\times10^{13}) \frac{60}{3\times10^8} = \boxed{4\text{ MHz}} ]

Step 3 — Wavelength

[ \lambda = \frac{3\times10^8}{77\times10^9} \approx \boxed{3.896\text{ mm}} ]

Step 4 — Doppler

[ f_D = \frac{2(20)}{0.003896} \approx \boxed{10.27\text{ kHz}} ]

Step 5 — Angle

[ d=\frac{\lambda}{2} ] [ \Delta\phi = \frac{2\pi d\sin30^\circ}{\lambda} = \boxed{\frac{\pi}{2}} ]

The radar therefore observes approximately:

Measurement	Observed quantity	Physical result
Fast-time FFT	4 MHz beat frequency	30 m range
Slow-time FFT	10.27 kHz Doppler	20 m/s velocity
Rx phase difference	π/2 rad	30° angle

14. Summary

FMCW radar turns distance into frequency, motion into Doppler frequency, and direction into spatial phase difference; FFTs are used because they efficiently reveal the frequency components hidden inside the sampled radar signal.

  %% Simple FMCW Radar: Fast-Time FFT + Slow-Time FFT
clear;
clc;
close all;

%% Radar parameters
c  = 3e8;              % Speed of light (m/s)
fc = 77e9;             % Carrier frequency (Hz)

B  = 1e9;              % Bandwidth = 1 GHz
Tc = 50e-6;            % Chirp duration = 50 us

S = B/Tc;              % Chirp slope (Hz/s)

%% Target
R = 30;                % Target range = 30 m
v = 20;                % Target velocity = 20 m/s

lambda = c/fc;

%% Sampling parameters
Fs = 20e6;             % ADC sampling frequency = 20 MHz
Ns = 1000;              % Samples per chirp
Nc = 128;               % Number of chirps

Ts = 1/Fs;

%% Create data matrix
% Rows    = fast time (samples within chirp)
% Columns = slow time (different chirps)

radarData = zeros(Ns, Nc);

%% Generate received beat signal
for m = 1:Nc

    % Time within this chirp
    t = (0:Ns-1).' * Ts;

    % Target range changes because target is moving
    Rm = R - v*(m-1)*Tc;

    % Round-trip delay
    tau = 2*Rm/c;

    % Beat frequency caused by range
    fb = S*tau;

    % Doppler frequency
    fd = 2*v/lambda;

    % Beat signal
    radarData(:,m) = exp(1j*2*pi*(fb + fd)*t);

end


%% ============================================================
%  FAST-TIME FFT
%  Samples within ONE chirp -> Range
% =============================================================

% Take FFT along rows
rangeFFT = fft(radarData, [], 1);

% Keep only positive frequencies
rangeFFT = rangeFFT(1:Ns/2, :);

% Frequency axis
fRange = (0:Ns/2-1).' * Fs/Ns;

% Convert beat frequency to range
rangeAxis = c*fRange/(2*S);


%% Plot Fast-Time FFT
figure;

plot(rangeAxis, abs(rangeFFT(:,1)));

xlabel('Range (m)');
ylabel('Magnitude');
title('Fast-Time FFT → Range');
grid on;

xlim([0 50]);


%% ============================================================
% SLOW-TIME FFT
% Same range bin across MANY chirps -> Velocity
% =============================================================

% Pick the range bin containing the target
[~, rangeBin] = max(abs(rangeFFT(:,1)));

% Extract this range bin across all chirps
slowTimeSignal = rangeFFT(rangeBin, :);

% FFT across chirps
dopplerFFT = fftshift(fft(slowTimeSignal, Nc));

% Doppler frequency axis
PRF = 1/Tc;

fDoppler = (-Nc/2:Nc/2-1) * PRF/Nc;

% Convert Doppler frequency to velocity
velocityAxis = lambda*fDoppler/2;


%% Plot Slow-Time FFT
figure;

plot(velocityAxis, abs(dopplerFFT));

xlabel('Velocity (m/s)');
ylabel('Magnitude');
title('Slow-Time FFT → Velocity');
grid on;

xlim([-30 30]);


%% ============================================================
% RANGE-DOPPLER MAP
% =============================================================

% FFT across slow-time/chirp dimension
RD = fftshift(fft(rangeFFT, Nc, 2), 2);

figure;

imagesc(velocityAxis, rangeAxis, abs(RD));

xlabel('Velocity (m/s)');
ylabel('Range (m)');
title('Range-Doppler Map');

colorbar;
axis xy;

ylim([0 50]);
xlim([-30 30]);