Principles of Modern Radar - Volume 2 1891121537

2.2 Stretch Processing 33The separation between the peak and first null isδ = ∣∣ peak − null∣ ∣ = 2π|d|(2.20)Equation (2.20) represents the Rayleigh resolution in radians per second, and the resolutionin hertz isδ f = 1(2.21)|d|Consider two point targets separated in time by δt d . The corresponding difference inbeat frequencies is δt d β/τ. Setting the frequency resolution achieved by the compressedresponse in equation (2.17) equal to the difference in beat frequenciesand solving for the time-delay resolutionor in terms of range resolution is1|d| = β τ δt d (2.22)δt d = 1 1(β1 − |t ) (2.23)d|τδR = c2β1(1 − |t ) (2.24)d|τwhere c is the speed of light. Now, consider a scatterer located at a time delay equal to thecenter of the range window (i.e., t d = t rcv or t d = 0). The resultant range resolution isδR td =0 = c(2.25)2βFor scatterers located at time delays less than or greater than t rcv , range resolution degradesby a factor of 1/(1 − |t d |/τ) as shown in equation (2.24). The loss in resolution occursbecause the receive oscillator extends only over a time interval equal to the original pulselength. A return at a range other than the center of the range window mixes with theoscillator, producing a tone whose duration is less than the original pulse length. Theshortened sinusoid achieves less frequency resolution and correspondingly less rangeresolution.2.2.4.6 SNR LossIn addition to the degradation in resolution, a loss in SNR occurs when the sinusoid (ormixing product) is shortened in time:SNR Loss = 1 − |t d|τ0 ≤ |t d |