考虑定义K_θ = (2π/c) F sinθ。

Recall that K_θ= (2π/c) F sinθ.

K_θ的微分为dK_θ = (2π/c) · [sinθ·dF + Fcosθ · dθ]。

The totaldifferential of K_θ is then dK_θ = (2π/c) · [sinθ· dF +Fcosθ · dθ].

对于固定的雷达频率，为了确定视线角上的解相关间隔，则dF = 0，dK_θ= (2π/c) ·Fcosθ· dθ，因此ΔK_θ ≈ (2π/c) · Fcosθ· Δθ。

To determine thedecorrelation interval in angle for a fixed radar frequency, set dF = 0 to givedK_θ = (2π/c) · Fcosθ · dθ, so that ΔK_θ ≈ (2π/c) ·Fcosθ · Δθ.

相似地，对于固定的视线角θ，为了确定频率上的解相关间隔，则dθ = 0，ΔK_θ ≈(2π/c) · sinθ · ΔF。

Similarly, the frequencystep required to decorrelate the target is obtained by fixing the aspect angle θ so that dθ = 0, leading to ΔK_θ ≈ (2π/c) · sinθ ·ΔF.

将以上各式结合式（2.61）得到回波幅度解相关的视线角或者频率为：

Combining theserelations with Eq. (2.61) then gives the desired result for the change in angleor frequency required to decorrelate the echo amplitude:

注意：Lcosθ是目标长度在雷达视线正交方向上的投影，而Lsinθ是目标长度在雷达视线方向上的投影。

Note that Lcosθ isthe projection of the target extent orthogonal to the radar boresight, while Lsinθis the projection along the radar boresight.

从上式可以看出，视线角上的解相关间隔与雷达观察的目标宽度有关，而频率上的解相关间隔与深度有关。

Thus, thedecorrelation interval in aspect angle is driven by the width of the target asviewed from the radar, while the interval in frequency is driven by the depth.

能够应用于更多任意散射目标的通用表达式为：

A more general pairof expressions that can be applied to more arbitrary many scatterer targets isthen

其中L_w和L_d分别为从雷达观察方向的目标宽度和深度。

where L_wand L_d are the target width and depth, respectively, as viewed fromthe radar.

例如，假设作为雷达目标的汽车长度大约为5米。

As an example,consider a target the size of an automobile, about 5 m long.

从目标正侧面进行观察，在L波段的1GHz，目标回波的视线角解相关角度为(3 × 10⁸)/(2× 5 × 10⁹) = 30 mrad，大约为1.7°，而W波段（95GHz）的解相关角度只有0.018°。

At L band (1 GHz),the target signature can be expected to decorrelate in (3 × 10⁸)/(2× 5 × 10⁹) = 30 mrad of aspect angle rotation relative to abroadside view (so the width is 5 m),about 1.7°, while at W band (95 GHz), this is reduced to only 0.018°.

从迎头进行观察，深度为5m的解相关频率为30MHz。

The frequency steprequired for decorrelation from a head-on aspect (depth of 5 m) is 30 MHz.

解相关频率与雷达工作的发射频率无关。

This result does notdepend on the nominal transmitted frequency.

——本文译自Mark A. Richards所著的《Fundamentals of Radar Signal Processing（Second edition）》

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