# Reading — Pulse Radars & Ambiguity

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

1. Define **pulse width (PW)**, **pulse repetition interval (PRI)**, **pulse repetition frequency (PRF)**, and **duty cycle**.
2. Compute **range resolution** from pulse width.
3. Compute the **unambiguous range** $R_u$ and **unambiguous velocity** $v_u$ from PRF, and explain the tradeoff between them.
4. 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.

:::{admonition} Key Concept
:class: 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).

::::{admonition} Quick Exercise
:class: quick-exercise

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

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

:::{admonition} Solution
:class: dropdown

1. $R_u = c/(2\cdot2000) = 75$ km. $v_u = (0.03\cdot2000)/4 = 15$ m/s.
2. $220 \bmod 75 = 220 - 2(75) = 70$ km apparent.
3. $R_u$ 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.