ECE 495 — Electromagnetic Warfare and Survivability

By the end of this lesson, you will be able to:

• Define PW, PRI, PRF, and duty cycle for a pulse radar
• Compute range resolution and unambiguous range from PW and PRF
• Explain the range-Doppler ambiguity tradeoff
• Match a radar's PRF regime (low / medium / high) to its mission

A CW radar always transmits — it can't tell when an echo left. Pulse radars transmit a short burst, then listen.

• Range is read directly from echo delay: $R = c \cdot t_{\text{echo}} / 2$
• The Tx is off while the Rx listens — no self-interference
• Energy on target is shaped by duty cycle, not just peak power

• PW — pulse width, the time the radar transmits
• PRI — pulse repetition interval; PRF $= 1/$PRI
• Duty cycle $=$ PW $\cdot$ PRF $=$ PW $/$ PRI — average over peak

Two targets are resolvable only if their echoes do not overlap in time.

• PW $= 1~\mu$s $\Rightarrow \Delta R = 150$ m
• PW $= 0.1~\mu$s $\Rightarrow \Delta R = 15$ m
• Shorter PW gives finer range resolution — but less energy per pulse

Pulse compression buys finer $\Delta R$ without giving up energy — a separate lesson.

The radar must hear the echo before it transmits the next pulse:

• PRF $= 1$ kHz $\Rightarrow R_u = 150$ km
• PRF $= 10$ kHz $\Rightarrow R_u = 15$ km
• Higher PRF gives more pulses per second — but shrinks $R_u$

An echo from $R > R_u$ arrives after the next pulse has gone out, so it wraps into the following interval and appears at

PRF $= 800$ Hz, so the unambiguous window closes at 187.5 km. The bomber is outside it.

The echo lands in the following interval, so the radar reports it against the wrong pulse:

• Looks like a short-range threat — wrong range, wrong tactical picture
• Mitigations: PRF jitter, multiple PRFs, pulse identification

A closer at $v_r$ shifts the echo by $f_d = 2 v_r / \lambda$ — that is L5's headline, borrowed early. The pulse train samples this shift at the PRF, so Nyquist limits unambiguous Doppler to $\pm$ PRF $/ 2$:

• Doppler beyond $\pm v_u$ aliases into a different velocity bin
• Higher PRF buys more unambiguous Doppler — but it costs us range ambiguity

Multiply the two ambiguity limits and the PRF cancels:

• Product is independent of PRF — a hard physical limit
• Pick more $R_u$, you give up $v_u$ at the same rate (and vice versa)
• The whole game is choosing which side of this tradeoff matches the mission

• Modern radars dwell in multiple regimes within one scan
• EW implication: PRF tells you the radar's intent

Type this with me. X-band threat radar, $\lambda = 3$ cm.

c = 3e8;  lambda = 0.03;                     % X-band
PRF = [800 2e3 5e3 1e4 2e4];                 % Hz, low -> high
Ru  = c ./ (2*PRF) / 1e3;                    % unambiguous range, km
vu  = lambda * PRF / 4;                      % unambiguous velocity, m/s
fprintf(' PRF(Hz)   Ru(km)  vu(m/s)    Ru*vu\n');
fprintf('%8.0f %8.1f %8.1f %8.0f\n', [PRF; Ru; vu; Ru.*vu]);
need_R = 200;  need_v = 250;   % B-21: see to 200 km, closing 250 m/s
ok = (Ru >= need_R) & (vu >= need_v);
fprintf('\nPRFs meeting both: %d of %d\n', sum(ok), numel(ok));

Read the last two columns, then the last line. What did the sweep just prove?

So what would you change — the list of PRFs, or the question you asked of it?