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Radar Equation

Derivation and application of the radar range equation.

Overview

The radar equation relates transmitter power, antenna gain, target characteristics, and receiver sensitivity to determine detection capability.

Basic Radar Equation

Power Density at Target

Power density at range \(R\) from an isotropic radiator:

\[ S_i = \frac{P_t}{4\pi R^2} \]

With transmit antenna gain \(G_t\):

\[ S_t = \frac{P_t G_t}{4\pi R^2} \]

Power Reflected by Target

The target intercepts and re-radiates power proportional to its radar cross section (RCS) \(\sigma\):

\[ P_{reflected} = S_t \cdot \sigma = \frac{P_t G_t \sigma}{4\pi R^2} \]

Power Density at Receiver

The reflected power spreads again over \(4\pi R^2\):

\[ S_r = \frac{P_t G_t \sigma}{(4\pi)^2 R^4} \]

Received Power

The receiver antenna with effective area \(A_e\) captures:

\[ P_r = S_r \cdot A_e = \frac{P_t G_t \sigma A_e}{(4\pi)^2 R^4} \]

Using \(A_e = G_r \lambda^2 / 4\pi\):

\[ P_r = \frac{P_t G_t G_r \lambda^2 \sigma}{(4\pi)^3 R^4} \]

For monostatic radar (\(G_t = G_r = G\)):

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

SNR Form

Including noise and losses:

\[ SNR = \frac{P_t G^2 \lambda^2 \sigma}{(4\pi)^3 R^4 k T_s B_n L_s} \]

Where: - \(P_t\) = Peak transmit power (W) - \(G\) = Antenna gain (linear) - \(\lambda\) = Wavelength (m) - \(\sigma\) = Target RCS (m²) - \(R\) = Range (m) - \(k\) = Boltzmann constant = 1.38×10⁻²³ J/K - \(T_s\) = System noise temperature (K) - \(B_n\) = Noise bandwidth (Hz) - \(L_s\) = System losses (linear)

Range Form

Solving for range at minimum detectable SNR:

\[ R_{max} = \left[\frac{P_t G^2 \lambda^2 \sigma}{(4\pi)^3 k T_s B_n L_s \cdot SNR_{min}}\right]^{1/4} \]

Pulse Integration

Coherent Integration

For \(N\) coherently integrated pulses:

\[ SNR_N = N \cdot SNR_1 \]

The integration gain is \(N\) (linear) or \(10\log_{10}(N)\) dB.

Non-Coherent Integration

For non-coherent integration:

\[ SNR_N \approx \sqrt{N} \cdot SNR_1 \]

More precisely, the integration gain depends on the required \(P_d\) and \(P_{fa}\).

Detection Theory

Single-Pulse Detection

For a Gaussian noise background and non-fluctuating target (Swerling 0), the detection probability relates to SNR through:

\[ P_d = \frac{1}{2}\text{erfc}\left[\text{erfc}^{-1}(2P_{fa}) - \sqrt{SNR}\right] \]

Required SNR

For given \(P_d\) and \(P_{fa}\):

\[ SNR_{req} \approx \ln\left(\frac{1}{P_{fa}}\right) + \ln\left(\frac{1}{1-P_d}\right) \]

More accurate:

\(P_d\) \(P_{fa} = 10^{-6}\) \(P_{fa} = 10^{-9}\)
0.5 10.8 dB 12.6 dB
0.9 13.1 dB 14.9 dB
0.99 16.4 dB 18.2 dB

Swerling Target Models

Model PDF Decorrelation
0 Constant (non-fluctuating) -
1 Exponential (Rayleigh amplitude) Scan-to-scan
2 Exponential Pulse-to-pulse
3 Chi-squared, 4 DOF Scan-to-scan
4 Chi-squared, 4 DOF Pulse-to-pulse

SNR Penalty

Fluctuating targets require additional SNR:

Model Typical Penalty vs. SW0
SW1 +3 to +8 dB (depends on \(P_d\))
SW2 +2 to +5 dB
SW3 +1 to +3 dB
SW4 +1 to +2 dB

Power-Aperture Product

Radar performance fundamentally scales with:

\[ PA = P_t \cdot A = P_t \cdot \frac{G\lambda^2}{4\pi} \]

For a given target and detection requirement:

\[ R_{max}^4 \propto PA \]

Trade-off: Higher power OR larger aperture.

Example Calculation

Given: - Frequency: 10 GHz (\(\lambda\) = 0.03 m) - Array: 16×16 elements, G = 30 dBi - TX power: 10 W/element, 256 elements → 2560 W peak - Target RCS: 1 m² - Range: 100 km - Noise temp: 400 K - Losses: 4 dB (\(L_s\) = 2.51) - Pulse width: 10 μs → \(B_n\) ≈ 100 kHz

Calculate:

  1. Wavelength: \(\lambda\) = 0.03 m

  2. Gain (linear): \(G\) = 10^(30/10) = 1000

  3. Numerator: $$ P_t G^2 \lambda^2 \sigma = 2560 \times 1000^2 \times 0.03^2 \times 1 = 2.3 \times 10^6 $$

  4. Denominator: $$ (4\pi)^3 R^4 k T_s B_n L_s = 1984 \times 10^{20} \times 1.38 \times 10^{-23} \times 400 \times 10^5 \times 2.51 $$ $$ = 2.74 \times 10^6 $$

  5. SNR (single pulse): $$ SNR = \frac{2.3 \times 10^6}{2.74 \times 10^6} = 0.84 = -0.75 \text{ dB} $$

  6. With 10-pulse coherent integration: $$ SNR_{10} = -0.75 + 10 = 9.25 \text{ dB} $$

  7. Required SNR for \(P_d\) = 0.9, \(P_{fa}\) = 10⁻⁶, SW1 ≈ 17 dB

  8. SNR Margin: $$ M = 9.25 - 17 = -7.75 \text{ dB} \quad \text{(insufficient)} $$

Need more power, more elements, or shorter range.

Range Dependencies

Quantity Range Dependence
\(P_r\) \(R^{-4}\)
\(SNR\) \(R^{-4}\)
Double range -12 dB SNR
Half range +12 dB SNR

See Also