Ferkans — Interactive Telecom Tutor

The Range-Doppler Map as a 2D Image

A pulsed-Doppler radar transmits $N_p$ pulses with pulse repetition interval (PRI) $T_{\text{PRI}}$ . The received echoes form a 2D data matrix — fast time (within each pulse, encoding range) and slow time (across pulses, encoding velocity). Processing this matrix with a 2D FFT produces the range-Doppler map, which is the simplest "image" a radar can form.

For RF imaging, the range-Doppler map is the matched-filter image $\mathbf{A}^{H}\mathbf{y}$ restricted to the range and Doppler dimensions.

Definition:
Fast-Time / Slow-Time Data Matrix

The radar collects $N_p$ pulses, each containing $N_s$ range samples. The data matrix is:

$\mathbf{X} \in \mathbb{C}^{N_s \times N_p},$

where:

Column $n$ contains the $n$ -th pulse after demodulation.
Row $m$ corresponds to fast-time sample $m$ (range bin $m$ ).
Element $X_{m,n}$ contains the signal from range bin $m$ , pulse $n$ .

Fast-time processing (across columns) resolves range via matched filtering. Slow-time processing (across rows) resolves Doppler via coherent integration (DFT).

Definition:
Range-Doppler Processing via 2D FFT

The range-Doppler map is formed by:

Fast-time matched filtering: For each pulse $n$ , compute $y_n(\tau) = r_n(\tau) * s^*(-\tau)$ (or equivalently multiply in frequency and IFFT). This resolves targets in range.
Slow-time DFT: For each range bin $m$ , compute the DFT across the $N_p$ pulses: $Z_m[k] = \sum_{n=0}^{N_p - 1} y_n[m] \, e^{-j2\pi kn/N_p},$ resolving targets in Doppler.

The result $|Z_m[k]|^2$ is the range-Doppler map with:

Range resolution: $\Delta R = c/(2W)$ .
Doppler resolution: $\Delta f_d = 1/(N_p T_{\text{PRI}}) = 1/T_{\text{CPI}}$ .
Velocity resolution: $\Delta v = \lambda/(2T_{\text{CPI}})$ .

Theorem: Coherent Integration Gain

Coherent integration of $N_p$ pulses via the slow-time DFT increases the output $\text{SNR}$ by a factor of $N_p$ relative to the single-pulse $\text{SNR}$ :

$\text{SNR}_{\text{coh}} = N_p \cdot \text{SNR}_{1}.$

The integrated $\text{SNR}$ is:

$\text{SNR}_{\text{coh}} = \frac{N_p P_t G^2 \lambda^{2} \sigma}{(4\pi)^3 R^4 k_B T_s W L}.$

Each pulse adds signal coherently (amplitude adds) while noise adds incoherently (power adds). Signal power grows as $N_p^2$ while noise power grows as $N_p$ , giving $\text{SNR}$ gain $N_p$ .

Proof

Signal at DFT output

For a target at Doppler bin $k_0$ , the DFT output is $|Z_{m}[k_0]| = N_p \cdot |\alpha|$ , so signal power is $N_p^2 |\alpha|^2$ .

Noise at DFT output

Noise samples are i.i.d. across pulses. The DFT preserves noise power: $\mathbb{E}[|Z_m[k]|^2] = N_p \sigma^2$ .

SNR ratio

$\text{SNR}_{\text{coh}} = N_p^2 |\alpha|^2 / (N_p \sigma^2) = N_p \cdot |\alpha|^2/\sigma^2 = N_p \cdot \text{SNR}_{1}$ .

Definition:
PRF Design and Ambiguity Trade-Offs

The pulse repetition frequency (PRF) $f_{\text{PRF}} = 1/T_{\text{PRI}}$ determines both the unambiguous range and unambiguous velocity:

Unambiguous range: $R_{\text{ua}} = c/(2 f_{\text{PRF}})$ .
Unambiguous velocity: $v_{\text{ua}} = \lambda f_{\text{PRF}} / 4$ .

These are inversely related: $R_{\text{ua}} \cdot v_{\text{ua}} = c \lambda / 8$ .

This leads to three PRF regimes:

Low PRF: Unambiguous in range, ambiguous in velocity.
High PRF: Unambiguous in velocity, ambiguous in range.
Medium PRF: Ambiguous in both, but with manageable unfolding.

Example: Constructing a Range-Doppler Map

A pulsed radar operates at $f_0 = 10$ GHz ( $\lambda = 3$ cm), $W = 10$ MHz, PRF $= 1$ kHz, $N_p = 64$ pulses. Two targets exist: Target A at $R_A = 15$ km, $v_A = 30$ m/s; Target B at $R_B = 15.01$ km, $v_B = -50$ m/s. (a) Can the radar resolve these targets in range? (b) What are their Doppler frequencies? (c) Can the radar resolve them in Doppler?

Solution

Range resolution

$\Delta R = c/(2W) = 3 \times 10^8/(2 \times 10^7) = 15$ m. The range separation is $|R_B - R_A| = 10$ m $< \Delta R$ . The targets are not resolvable in range.

Doppler frequencies

$f_{d,A} = 2v_A/\lambda = 2 \times 30/0.03 = 2000$ Hz. $f_{d,B} = 2v_B/\lambda = 2 \times (-50)/0.03 = -3333$ Hz.

Doppler resolution and ambiguity

$\Delta f_d = 1/(N_p T_{\text{PRI}}) = 1/(64 \times 10^{-3}) = 15.6$ Hz. The Doppler separation $|f_{d,A} - f_{d,B}| = 5333$ Hz $\gg \Delta f_d$ . However, the unambiguous Doppler is $f_{\text{PRF}} = 1000$ Hz. $f_{d,A} = 2000$ Hz aliases to $2000 - 2 \times 1000 = 0$ Hz. $f_{d,B} = -3333$ Hz aliases to $-3333 + 4 \times 1000 = 667$ Hz. The targets are resolvable in Doppler (after aliasing).

Range-Doppler Map Visualization

Generate a range-Doppler map for configurable targets. Observe how bandwidth determines range resolution, CPI determines velocity resolution, and PRF determines ambiguity boundaries.

Parameters

B

(MHz)20

N_p

64

PRF (kHz)1

SNR (dB)15

Definition:
Range Migration

During the coherent processing interval (CPI) $T_{\text{CPI}} = N_p T_{\text{PRI}}$ , a target at radial velocity $v$ migrates through range bins:

$\Delta R_{\text{mig}} = v \cdot T_{\text{CPI}}.$

When $\Delta R_{\text{mig}} > \Delta R$ (i.e., migration exceeds one range bin), the target's energy smears across range bins, degrading coherent integration gain.

Range migration correction compensates this drift before Doppler processing. Common techniques include:

Keystone transform: resampling the data matrix to align range cells across pulses.
Range walk compensation: linear range shift per pulse based on estimated velocity.

Common Mistake: Forgetting Range-Doppler Coupling in LFM

Mistake:

Treating the range-Doppler map as having independent range and Doppler axes when using an LFM waveform.

Correction:

For LFM, a Doppler shift $f_d$ causes an apparent range shift $\Delta R = f_d c / (2\mu) = f_d cT / (2W)$ . A target at 100 m/s with $\lambda = 3$ cm and $W/T = 10^{12}$ Hz/s appears shifted by $\Delta R = f_d/(2\mu) \cdot c$ . For slow targets this is negligible; for fast targets or long CPIs, it must be corrected by the keystone transform or by adjusting the range-Doppler map axes.

Coherent Processing Interval (CPI)

The time duration $T_{\text{CPI}} = N_p T_{\text{PRI}}$ over which pulses are coherently integrated. Determines velocity resolution $\Delta v = \lambda/(2T_{\text{CPI}})$ and total integration gain $N_p$ .

Quick Check

A radar at $\lambda = 3$ cm needs to detect targets at up to $R = 150$ km with unambiguous velocity. What is the minimum PRF if the maximum target velocity is $v_{\max} = 300$ m/s?

20 kHz

40 kHz

1 kHz

10 kHz

Correction:

20 kHz

$v_{\text{ua}} = \lambda f_{\text{PRF}}/4$ , so $f_{\text{PRF}} = 4v_{\max}/\lambda = 4 \times 300/0.03 = 40$ kHz. But wait: the unambiguous range is $R_{\text{ua}} = c/(2f_{\text{PRF}}) = 3.75$ km, which is far less than 150 km. We need to use a medium-PRF or staggered PRF. $f_{\text{PRF}} = 20$ kHz gives $v_{\text{ua}} = 150$ m/s (not sufficient for 300 m/s, but with PRF staggering we can unfold). Actually, $f_{\text{PRF}} \geq 4 \times 300 / 0.03 = 40$ kHz for unambiguous velocity.

Key Takeaway

The 2D FFT (fast-time matched filter + slow-time DFT) produces the range-Doppler map with resolution $\Delta R = c/(2W)$ in range and $\Delta v = \lambda/(2T_{\text{CPI}})$ in velocity. Coherent integration provides $N_p$ gain in $\text{SNR}$ . Range migration must be corrected when $v \cdot T_{\text{CPI}} > \Delta R$ .

Range-Doppler Processing

The Range-Doppler Map as a 2D Image

Definition: Fast-Time / Slow-Time Data Matrix

Definition: Range-Doppler Processing via 2D FFT

Theorem: Coherent Integration Gain

Signal at DFT output

Noise at DFT output

SNR ratio

Definition: PRF Design and Ambiguity Trade-Offs

Example: Constructing a Range-Doppler Map

Range resolution

Doppler frequencies

Doppler resolution and ambiguity

Range-Doppler Map Visualization

Parameters

Definition: Range Migration

Common Mistake: Forgetting Range-Doppler Coupling in LFM

Coherent Processing Interval (CPI)

Quick Check

Key Takeaway

Definition:
Fast-Time / Slow-Time Data Matrix

Definition:
Range-Doppler Processing via 2D FFT

Definition:
PRF Design and Ambiguity Trade-Offs

Definition:
Range Migration