3. The Radar Equation: Where Did All the Energy Go?

3.1 A back-of-envelope derivation

Let us carefully track every factor that affects the strength of the received echo. We will use only physics we already know.

Step 1: transmit power and antenna gain. The radar transmits power $P_t$. An isotropic antenna spreads $P_t$ uniformly over the surface of a sphere, so at distance $R$ the power per unit area is $P_t / (4\pi R^2)$. A directional antenna with gain $G$ does better in the favored direction, focusing the power onto a smaller patch:

$\text{Power density at target} = \frac{P_t G}{4\pi R^2} \quad\text{[W/m}^2\text{]}$

Step 2: target reflection. The target intercepts an effective "radar cross section" $\sigma$ (in m²), which is defined as the area that, if it scattered isotropically all the power that hit it, would produce the actually-observed reflected power. The total power scattered by the target is therefore

$P_{scatter} = \frac{P_t G}{4\pi R^2}\, \sigma \quad\text{[W]}$

Step 3: spreading on the way back. The target re-radiates this power, again roughly isotropically (this is the conservative assumption baked into the definition of $\sigma$). At distance $R$ from the target back to the radar:

$\text{Power density at radar} = \frac{P_{scatter}}{4\pi R^2} = \frac{P_t G \sigma}{(4\pi)^2 R^4} \quad\text{[W/m}^2\text{]}$

Step 4: receiving antenna effective area. The radar receives whatever power falls on its effective antenna area $A_e$. From Chapter 13, $A_e = G \lambda^2/(4\pi)$ for an antenna of gain $G$ at wavelength $\lambda$ (assuming the same antenna is used for transmit and receive, which is the usual case). So the received power is

$P_r = \frac{P_t G \sigma}{(4\pi)^2 R^4} \cdot \frac{G \lambda^2}{4\pi} = \frac{P_t G^2 \lambda^2 \sigma}{(4\pi)^3 R^4}$

This is the radar equation, in its most useful form:

$\boxed{P_r = \frac{P_t\, G^2\, \lambda^2\, \sigma}{(4\pi)^3\, R^4}}$

If we set $P_r$ equal to the minimum detectable signal $S_{min}$, we can solve for the maximum detection range:

$R_{max} = \left[\frac{P_t\, G^2\, \lambda^2\, \sigma}{(4\pi)^3\, S_{min}}\right]^{1/4}$

That fourth-root is the most consequential exponent in radar engineering.

3.2 Why R^4? The double-spreading penalty

The fourth-power dependence is what makes radar hard. It comes from spreading twice, once on the way out and once on the way back.

In a one-way communications link, signal falls off as $1/R^2$ (Friis equation). Doubling the range means receiving one quarter the power; you can usually recover by doubling the antenna gain or the transmit power.

In a two-way radar link, signal falls off as $1/R^4$. Doubling the range means receiving one sixteenth the power. To compensate by transmitter power alone, you need 16 times more power. To compensate by antenna gain alone, you need 4 times more gain (it appears squared, so $4^2 = 16$). Either way, the system gets bigger, hotter, and more expensive in a way that punishes range hard.

This is why long-range surveillance radars are enormous. A 4-fold range increase from 50 km to 200 km means $4^4 = 256$ times more transmit-receive product. You see the consequences in real radar: the largest air-defense radars push tens of megawatts of peak power through arrays of dozens of meters across, all to extract a useful signal from targets at hundreds of kilometers. By contrast, automotive radars at 30 to 200 meters use milliwatts.

Analogy. Imagine throwing handfuls of tennis balls at a small mirror in the dark and counting how many bounce back through your fingers. Throw ten thousand balls, the mirror sees a small fraction of them, reflects them back in random directions, and only a tinier fraction returns to your hand. Move further from the mirror and the fraction at each step gets worse. The fraction reaching the mirror falls as $1/R^2$ because of geometric spreading. The fraction returning falls as another $1/R^2$ for the same reason. The product is $1/R^4$.

3.3 The minimum detectable signal: noise

What sets $S_{min}$? Receiver noise. A receiver with bandwidth $B$, system noise figure $F$, at temperature $T_0 = 290$ K, has a noise floor of

$N = k T_0 B F$

where $k = 1.38 \times 10^{-23}$ J/K is the Boltzmann constant. To detect a target reliably, the received signal must exceed the noise by some signal-to-noise ratio $\text{SNR}_{min}$:

$S_{min} = k T_0 B F \cdot \text{SNR}_{min}$

Modern radars achieve noise figures of 1 to 3 dB in cooled or carefully designed front-ends. The Friis formula tells us the first amplifier dominates the noise, which is why radars put a low-noise amplifier (LNA) immediately after the antenna; some experimental radars cool the LNA cryogenically.

A typical $\text{SNR}_{min}$ for reliable detection (probability of detection 0.9, probability of false alarm $10^{-6}$) is around 13 dB. Higher false-alarm tolerance lets you drop SNR; a higher bar lifts it.

3.4 Pulse integration: getting more out of multiple pulses

A real radar fires many pulses at the same target during a dwell. Coherently combining the returns from $N$ pulses gives an SNR improvement of up to $N$ (a factor of $\sqrt{N}$ in voltage, $N$ in power). Non-coherent integration, where you only combine magnitudes, gives less, around $\sqrt{N}$ in power for large $N$.

Coherent integration is essentially a discrete Fourier transform across the slow-time pulse-to-pulse dimension. You sum the complex echoes weighted by phase rotations corresponding to your hypothesized Doppler. If the target's true Doppler matches your hypothesis, all $N$ contributions add in phase and amplitude scales as $N$. Random noise adds incoherently, scaling as $\sqrt{N}$. SNR improves as $N/\sqrt{N} = \sqrt{N}$ in voltage, $N$ in power.

The cost: integration time. To get $N$ samples at a 1-kHz PRF takes $N$ ms. A long dwell on one target means less time scanning the rest of the volume. Modern phased arrays manage this by adaptive scheduling, dwelling on confirmed targets while still scanning for new ones.

3.5 Worked example: solving the radar equation

A typical L-band air surveillance radar:

• $P_t = 1$ MW peak, 1-μs pulse, 1-kHz PRF.
• $G = 33$ dBi $= 2000$ (large parabolic, ~4 m).
• $\lambda = 0.23$ m (1.3 GHz). $\sigma = 1$ m² (medium fighter).
• Receiver: $F = 3$ dB, $B = 1$ MHz, SNR = 13 dB.

$S_{min} = kT_0 BF \cdot \text{SNR}_{min} \approx 1.6\times 10^{-13}$ W $= -98$ dBm. Numerator $P_t G^2 \lambda^2 \sigma \approx 2.1\times 10^{11}$. Denominator $(4\pi)^3 \cdot S_{min} \approx 3.2 \times 10^{-10}$. Then $R_{max}^4 \approx 6.6\times 10^{20}$ m⁴, so $R_{max} \approx 160$ km. With pulse integration over a dwell, $R_{max}$ extends further; practical radars top out around 400-500 km because of the Earth's horizon and atmosphere.

import numpy as np
 
def radar_max_range(P_t, G_dB, freq, RCS, F_dB, B, SNR_min_dB, N_pulse=1):
    """
    Solve the radar equation for max detection range.
    P_t : peak transmit power, W
    G_dB : antenna gain, dBi
    freq : carrier frequency, Hz
    RCS : target radar cross section, m^2
    F_dB : receiver noise figure, dB
    B : receiver bandwidth, Hz
    SNR_min_dB : minimum SNR for detection, dB
    N_pulse : number of pulses coherently integrated
    """
    c = 3e8
    k = 1.38e-23
    T0 = 290
    G = 10 ** (G_dB / 10)
    F = 10 ** (F_dB / 10)
    SNR_min = 10 ** (SNR_min_dB / 10)
    lam = c / freq
    # Coherent integration: SNR grows by N
    S_min = k * T0 * B * F * SNR_min / N_pulse
    R4 = P_t * G**2 * lam**2 * RCS / ((4 * np.pi)**3 * S_min)
    return R4 ** 0.25
 
# Example: ASR-9-like radar
print(radar_max_range(P_t=1e6, G_dB=33, freq=1.3e9,
                      RCS=1, F_dB=3, B=1e6, SNR_min_dB=13,
                      N_pulse=20) / 1000, "km")
# → ~340 km, which is plausible for the ASR-9 class.