# Reading — Doppler & MTI By the end of this lesson you should be able to: 1. Compute the **Doppler shift** from a target's radial velocity. 2. Explain how a **single delay-line MTI** rejects ground clutter. 3. Identify **blind speeds** and how PRF moves them around. 4. Describe **pulse-Doppler processing** as an FFT across pulses and read a **range-Doppler map**. ## The second observable Range tells you *where* a target is. Doppler tells you *what* it is doing. A target moving radially at velocity $v_r$ shifts the carrier frequency by $$ \boxed{\,f_d = \frac{2 v_r}{\lambda}\,} $$ The factor of 2 is, once again, the two-way path. At X-band ($\lambda = 3$ cm), a target closing at 300 m/s produces a 20 kHz shift; at 30 m/s, 2 kHz. Each pulse in the train carries a phase, and across pulses that phase advances in proportion to $f_d$ — so a coherent radar reads velocity straight off the pulse-to-pulse phase. ## Clutter versus targets A radar looking down sees **everything**: terrain, buildings, weather, sea, foliage. Ground clutter has an enormous RCS and sits at essentially **zero Doppler** because it is not moving. The aircraft return is far weaker and buried in clutter **on amplitude alone**. The escape is Doppler. The clutter sits at $f_d \approx 0$; the mover sits somewhere else. Separate them in frequency and the weak target pops out of the strong clutter. :::{admonition} Key Concept :class: key-concept Clutter is a confidently *stationary* signal. The radar cannot reject it by amplitude — clutter is far stronger than the target — but it can reject it by **Doppler**, because clutter lives at zero Doppler and movers do not. ::: ## Single delay-line MTI Moving-target indication (MTI) exploits exactly that. The simplest canceller subtracts each pulse return from the previous one: $$ y[n] = x[n] - x[n-1]. $$ Anything that did not change between pulses (stationary clutter) cancels to zero; anything that changed (a mover, whose phase advanced) survives. The frequency response of this one-tap filter, with $T = \text{PRI}$, is $$ |H(f)| = 2\,|\sin(\pi f T)|. $$ This has **notches at $f = 0, \text{PRF}, 2\,\text{PRF}, \dots$** — it kills zero-Doppler clutter and its PRF-spaced repeats, while passing Doppler shifts in between. It costs one PRI of memory and a subtractor. Higher-order cancellers and adaptive MTI deepen and widen the notch, but the idea is this. ## Blind speeds The notches are also a vulnerability. A target whose Doppler shift lands **on** a notch looks stationary and is cancelled along with the clutter. These **blind speeds** occur at $$ \boxed{\,v_{\text{blind},n} = \frac{n\,\lambda\,\text{PRF}}{2}\,} $$ At X-band with PRF = 1 kHz, the first blind speeds are 15, 30, 45 m/s, … — squarely in the range of real aircraft and ground vehicles. Raising the PRF to 10 kHz pushes the first blind speed to **150 m/s**, usually outside the common clutter band. This is a major reason pulse-Doppler radars favor medium-to-high PRF — and it ties straight back to Lesson 4's ambiguity tradeoff. :::{admonition} Key Concept :class: key-concept MTI notches reject clutter at $f = 0, \text{PRF}, 2\,\text{PRF}, \dots$, but a target on a notch is also rejected. Raising PRF moves the blind speeds higher and out of the way — at the cost of range ambiguity. The PRF choice is always a compromise. ::: ## Pulse-Doppler processing and the range-Doppler map Rather than a single subtraction, modern radars collect $N$ pulses in a coherent dwell and take an **$N$-point FFT across the pulses** at each range bin. The output is a **Doppler spectrum per range bin**, with bin width $$ \Delta f_d = \frac{\text{PRF}}{N}. $$ More pulses give finer Doppler resolution and, because the integration is coherent, an SNR gain proportional to $N$. Each target lands in its own **range-Doppler (R-D) cell**. Stacking the range bins against the Doppler bins gives the **range-Doppler map** — a 2D image where: - **Clutter forms a vertical ridge** at zero Doppler (all ranges, one velocity). - **Movers land off the ridge**, easy to threshold and track. The EW implication is sobering for the B-21: once you are moving relative to the radar, your Doppler signature is hard to hide. Stealth aims to push the *amplitude* return down into the clutter band; standoff jammers raise the noise floor across all Doppler bins. Both are attacks on this very picture, and both are Block 2 material. ::::{admonition} Quick Exercise :class: quick-exercise X-band radar, $\lambda = 3$ cm. 1. Compute $f_d$ for a B-21 closing at 250 m/s. 2. At PRF = 1 kHz, list the first two blind speeds. Are these realistic aircraft speeds? 3. The radar switches to PRF = 10 kHz. Where do the first two blind speeds move? 4. A jammer wants the B-21 to appear stationary on the R-D map. What property would it try to spoof? :::{admonition} Solution :class: dropdown 1. $f_d = 2(250)/0.03 = 16.67$ kHz. 2. $v_{\text{blind},1} = 15$ m/s, $v_{\text{blind},2} = 30$ m/s — yes, both are realistic. 3. They move to 150 m/s and 300 m/s — well outside the common ground-clutter band, much safer against low/slow threats. 4. Match the platform's own ground-clutter Doppler line, so the false return hides on the zero-Doppler clutter ridge and gets cancelled with it. (EP counters include geometry checks and inverse-cancellation logic.) ::: :::: ## Wrap-Up Doppler, $f_d = 2 v_r/\lambda$, is the second observable after range. Single delay-line MTI, $y[n] = x[n] - x[n-1]$, notches out zero-Doppler clutter via $|H(f)| = 2|\sin(\pi f T)|$, but creates blind speeds at $v_{\text{blind},n} = n\lambda\,\text{PRF}/2$ that the PRF choice must manage. Pulse-Doppler processing — an FFT across pulses — produces the range-Doppler map, with clutter as a vertical ridge and movers in their own cells. That closes the threat model. We now understand how a radar detects, ranges, resolves, and separates a moving target from clutter — the full Detect-and-Track problem the B-21 must defeat. Block 2 turns to the other side: the electronic-attack techniques that break these links.