How do I calculate the angle of arrival accuracy of a direction finding system from its baseline length?

The angle of arrival (AOA) accuracy of a phase interferometer DF system is directly related to the baseline length, wavelength, and signal-to-noise ratio: (1) Phase measurement error: the standard deviation of the phase measurement for a single pulse: σ_φ = 1 / √(2 × SNR) (radians), where SNR is the linear signal-to-noise ratio. For SNR = 20 dB (100): σ_φ = 0.071 rad = 4.05°. For SNR = 30 dB (1000): σ_φ = 0.022 rad = 1.28°. (2) AOA error from phase error: the AOA is θ = arcsin(λ × Δφ / (2π × d)). Taking the derivative: dθ/dΔφ = λ / (2π × d × cos(θ)). The AOA standard deviation: σ_θ = σ_φ × λ / (2π × d × cos(θ)). At broadside (θ = 0°, cos(θ) = 1): σ_θ = σ_φ × λ / (2π × d) = λ / (2π × d × √(2 × SNR)). (3) Key relationships: longer baseline (larger d): better accuracy (σ_θ decreases as 1/d). Higher SNR: better accuracy (σ_θ decreases as 1/√SNR). Higher frequency (shorter λ): better accuracy for a fixed physical baseline (σ_θ decreases with λ). Off-broadside (large θ): accuracy degrades (the 1/cos(θ) factor increases; at θ = 60°, accuracy is 2× worse than broadside). (4) Example: baseline d = 100 mm, frequency = 10 GHz (λ = 30 mm), SNR = 20 dB. σ_θ = 30 / (2π × 100 × √200) = 30 / (628.3 × 14.14) = 30 / 8886 = 0.0034 rad = 0.19°. This is excellent accuracy. With a shorter baseline (d = 15 mm = λ/2): σ_θ = 30 / (2π × 15 × 14.14) = 30 / 1333 = 0.0225 rad = 1.29°. The longer baseline provides 6.7× better accuracy. (5) Multi-pulse averaging: if N pulses are averaged: σ_θ_avg = σ_θ / √N. Averaging 100 pulses improves accuracy by 10×.