How do I calculate the power required for a self-protection jammer on an aircraft?

The power required for a self-protection jammer is calculated from the jam-to-signal (J/S) ratio needed to deny the victim radar accurate target information. For a self-protection jammer (SPJ), the jammer is co-located with the target (on the aircraft being protected): (1) J/S for self-protection: J/S = (P_j × G_j × 4π × R² × B_r) / (P_t × G_t × sigma × B_j). Where P_j = jammer transmit power (W), G_j = jammer antenna gain toward the radar, R = range from jammer to radar (= range from target to radar, since SPJ is co-located), P_t = radar transmit power, G_t = radar antenna gain, sigma = target RCS (m²), B_r = radar bandwidth, and B_j = jammer bandwidth. For a noise jammer: B_j ≥ B_r, so B_r/B_j ≤ 1. For a DRFM (deceptive) jammer: B_j = B_r, so B_r/B_j = 1 (maximum efficiency). (2) Required J/S: for effective jamming: J/S > 0 dB (jammer power exceeds radar echo at the radar receiver). For reliable masking: J/S > 6-10 dB. For RGPO/VGPO deception: J/S = 1-6 dB is sufficient (the deceptive signal only needs to be comparable to the skin return). (3) Simplification for SPJ: since the jammer is on the target (R_jammer = R_target = R): J/S = (P_j × G_j × 4π × R²) / (P_t × G_t × sigma). Solving for P_j: P_j = (J/S × P_t × G_t × sigma) / (G_j × 4π × R²). The required jammer power decreases with range R² (as the aircraft gets further from the radar, less jamming power is needed because the radar echo also decreases). (4) Example: radar: P_t = 1 MW (90 dBm), G_t = 35 dBi) target: sigma = 1 m² (0 dBsm). Jammer: G_j = 6 dBi, required J/S = 10 dB. Range R = 50 km. P_j = (10 × 10^9 × 3162 × 1) / (4 × 4π × (50000)²) = ... working in dB: P_j(dBm) = J/S + P_t + G_t + sigma(dBsm) - G_j - 11 - 20×log10(R). P_j = 10 + 90 + 35 + 0 - 6 - 11 - 94 = 24 dBm = 250 mW. At 50 km: only 250 mW of jammer ERP is needed. At 10 km: the required power increases dramatically because the radar echo is much stronger at shorter range.