DDM-MIMO
How Doppler-Coded Transmit Separation Works
In a MIMO radar the number of independent angle measurements scales with the product of the transmit and receive element counts, because each transmit-receive pair forms one element of a synthetic virtual array. The challenge is making the Nt transmitted signals separable at every receiver so the processor can reconstruct which transmitter produced each echo. DDM solves this in the Doppler domain. All transmitters emit identical chirps simultaneously, but the phase of transmitter k is advanced by a fixed increment Δφk = 2πk/Nt from one chirp to the next. A constant phase progression across slow time is mathematically equivalent to a frequency offset, so after the Doppler FFT each transmitter's energy is parked at a different Doppler bin.
Because the Doppler axis is periodic, the unambiguous velocity span is partitioned into Nt equal sub-bands, one per transmitter. A real moving target also carries its own Doppler shift, which slides every channel copy by the same amount. The processor recovers the true velocity either by reserving one empty guard sub-band (the vacant slot reveals the alias offset) or by using a slightly irregular set of phase increments so the wrapping pattern is unique. This is the core trade against time division multiplexing MIMO, where transmitters are pulsed one at a time and the velocity ambiguity instead comes from the stretched effective chirp interval.
DDM is attractive for imaging radars with 12 or more transmitters because the alternative TDM approach would multiply the frame time by Nt, smearing fast-moving targets. Binary phase coding (BPSK-MIMO) and slow-time outer codes such as Hadamard sequences are related simultaneous-transmit methods; DDM is the special case where the codes are pure complex exponentials, which keeps the separation a single FFT bin shift rather than a correlation step.
Governing Relations for DDM Channel Spacing
Δφk = 2πk / Nt (k = 0, 1, …, Nt−1)
Unambiguous velocity (single Tx):
vmax = λ / (4 × Tc)
Per-channel velocity window in DDM:
vmax,DDM = vmax / Nt
Per-channel SNR penalty:
ΔSNR ≈ −10 log10(Nt) dB
Where λ = wavelength (≈ 3.9 mm at 77 GHz), Tc = chirp repetition interval, Nt = transmitter count. Example: 77 GHz, Tc = 50 μs → vmax ≈ 19.5 m/s; with Nt = 4, vmax,DDM ≈ 4.9 m/s and the per-channel SNR drops ≈ 6 dB.
DDM Versus Other MIMO Multiplexing Schemes
| Scheme | Transmit timing | Frame-time cost | Velocity ambiguity | Per-channel SNR | Best fit |
|---|---|---|---|---|---|
| DDM (Doppler division) | All Tx simultaneous | None (full CPI kept) | Split by Nt | −10log10(Nt) dB | Imaging / corner radar, many Tx |
| TDM (time division) | One Tx per chirp | ×Nt longer | Reduced by Nt | Full (one Tx at a time) | Low Tx count, slow scenes |
| BPSK / code division | All Tx simultaneous | None | Full | Code cross-talk limited | High dynamic range needs |
| FDM (frequency division) | All Tx simultaneous | None | Full | Full per band | Wide instantaneous bandwidth |
Frequently Asked Questions
How does DDM-MIMO separate the transmit channels without losing time on the chirp sequence?
Each transmitter gets a progressive phase shift that advances by Δφk = 2πk/Nt per chirp. After the slow-time Doppler FFT this ramp parks that transmitter at a distinct Doppler sub-band, so all Nt channels appear at evenly spaced velocity offsets. Because every transmitter fires on every chirp, no chirps are wasted cycling through transmitters one at a time, which is the penalty TDM-MIMO pays.
What is the maximum unambiguous velocity penalty in DDM-MIMO and how is it recovered?
With Nt transmitters sharing one Doppler window, usable unambiguous velocity per channel drops by Nt; for 4 transmitters that is about one quarter of the single-Tx value. It is recovered by leaving an empty guard sub-band so the vacant slot reveals the alias offset, or by using slightly non-uniform phase increments so the wrap pattern is unique. At 77 GHz with Tc = 50 μs, that takes vmax from ≈ 19.5 m/s to ≈ 4.9 m/s per channel.
Why does DDM-MIMO raise the noise floor compared with a single-transmitter measurement?
When Nt transmitters share the power budget and pulse repetition interval, each one's energy lands in only one Doppler sub-band while the others spill into the rest. Per-channel SNR therefore falls by about 10log10(Nt) dB, roughly 6 dB for 4 transmitters and 9 dB for 8. The larger virtual-array gain and the preserved coherent processing interval offset much of this loss.