Reading — Pulse Radars & Ambiguity

Reading — Pulse Radars & Ambiguity#

By the end of this lesson you should be able to:

Define pulse width (PW), pulse repetition interval (PRI), pulse repetition frequency (PRF), and duty cycle.
Compute range resolution from pulse width.
Compute the unambiguous range \(R_u\) and unambiguous velocity \(v_u\) from PRF, and explain the tradeoff between them.
Use the range–Doppler ambiguity invariant to explain why no single PRF measures both cleanly.

Why pulses#

A continuous-wave transmitter cannot easily tell when an echo left, so it struggles to measure range. A pulsed radar solves this by transmitting a short burst, then listening. The round-trip time of the echo gives the range directly:

\[ R = \frac{c\,\Delta t}{2}, \]

the factor of 2 again being the two-way path. The radar sends a regular train of pulses and processes the echoes between them.

Four parameters describe the train:

Parameter	Symbol	Meaning
Pulse width	PW	Duration the transmitter is on.
Pulse repetition interval	PRI	Time between pulse starts.
Pulse repetition frequency	PRF	\(1/\text{PRI}\) — pulses per second.
Duty cycle	—	\(\text{PW}/\text{PRI}\) — fraction of time transmitting.

Range resolution#

Two targets are resolvable in range only if their echoes do not overlap. That sets the range resolution by the pulse width:

\[ \Delta R = \frac{c\,\text{PW}}{2}. \]

A shorter pulse resolves finer detail — a 1 µs pulse gives \(\Delta R = 150\) m; a 0.1 µs pulse gives 15 m. But a shorter pulse carries less energy, hurting detection range. (Pulse compression resolves this tension; that is a later topic.)

The ambiguities#

Here is the central tension of pulsed radar. The PRF you choose to listen between pulses sets two unambiguous limits at once — and they pull in opposite directions.

Unambiguous range. The radar must receive an echo before it sends the next pulse, or it cannot tell which pulse the echo belongs to. The longest unambiguous round trip is one PRI, so

\[ R_u = \frac{c}{2\,\text{PRF}}. \]

A low PRF gives a long \(R_u\).

Unambiguous velocity. Doppler (Lesson 5) is a frequency shift, and the pulse train samples it at the PRF. By the sampling theorem, the radar can measure Doppler unambiguously only up to \(\pm\text{PRF}/2\), which in velocity is

\[ v_u = \frac{\lambda\,\text{PRF}}{4}. \]

A high PRF gives a large \(v_u\).

So low PRF is good for range but bad for velocity; high PRF is the reverse. You cannot have both.

Key Concept

PRF is a single knob that sets two limits that fight each other: \(R_u = c/(2\,\text{PRF})\) shrinks as PRF rises, while \(v_u = \lambda\,\text{PRF}/4\) grows. There is no PRF that makes both large.

The invariant#

Multiply the two limits and the PRF cancels:

\[ R_u \cdot v_u = \frac{c}{2\,\text{PRF}}\cdot\frac{\lambda\,\text{PRF}}{4} = \frac{c\,\lambda}{8}. \]

The product is a constant fixed only by wavelength — independent of PRF. You can trade range coverage for velocity coverage, but their product is conserved. This invariant is the cleanest statement of why no single waveform measures everything.

X-band numbers#

At X-band (\(\lambda = 3\) cm):

PRF	\(R_u\)	\(v_u\)
1 kHz	150 km	7.5 m/s
10 kHz	15 km	75 m/s
200 kHz	0.75 km	1500 m/s

Notice the product \(R_u \cdot v_u\) is the same in every row. Real radars solve this by using multiple PRFs and resolving the ambiguities across them — the foundation of low/medium/high-PRF modes.

Regime	PRF	Good for	Pays with
Low	low	unambiguous range	badly ambiguous velocity
Medium	medium	balance	ambiguous in both; resolved across PRFs
High	high	unambiguous velocity, clutter rejection	badly ambiguous range

Range folding: a B-21 example#

When a target is beyond \(R_u\), its echo arrives after the next pulse has gone out, and the radar folds it back to an apparent range:

\[ R_\text{apparent} = R_\text{true} \bmod R_u. \]

Suppose a B-21 is at a true range of 200 km, and a radar uses PRF = 800 Hz. Then

\[ R_u = \frac{3\times10^{8}}{2\times 800} = 187.5\ \text{km}. \]

Since 200 km exceeds \(R_u\), the echo folds:

\[ R_\text{apparent} = 200 - 187.5 = 12.5\ \text{km}. \]

The radar paints the bomber at 12.5 km — wildly wrong, and a known vulnerability the adversary must design around (and one EA can exploit).

Quick Exercise

An X-band radar (\(\lambda = 3\) cm) uses PRF = 2 kHz.

What are \(R_u\) and \(v_u\)?
A target is at a true range of 220 km. Where does it appear?
The radar switches to PRF = 4 kHz to double \(v_u\). What happens to \(R_u\), and what does that cost?

Solution

\(R_u = c/(2\cdot2000) = 75\) km. \(v_u = (0.03\cdot2000)/4 = 15\) m/s.
\(220 \bmod 75 = 220 - 2(75) = 70\) km apparent.
\(R_u\) halves to 37.5 km while \(v_u\) doubles to 30 m/s — the invariant \(R_u v_u = c\lambda/8\) holds. The cost is much worse range ambiguity.

Wrap-Up#

Pulsed radar measures range from echo delay, with resolution \(\Delta R = c\,\text{PW}/2\) set by pulse width. The PRF sets two competing limits — \(R_u = c/(2\,\text{PRF})\) and \(v_u = \lambda\,\text{PRF}/4\) — whose product \(c\lambda/8\) is invariant, so no single PRF is unambiguous in both. Targets beyond \(R_u\) fold to a false apparent range.

We have treated the echo as a single number, but each pulse also carries a phase, and across pulses that phase moves whenever the target does. The next lesson turns that motion into the Doppler shift and the moving-target processing that pulls the B-21 out of clutter.

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