无论是雷达、目标还是两者都在运动，都没有区别，因此可以假定雷达是静止的，目标是运动的，并且所有的测量都是在雷达的参照系中进行的，这样不失一般性。
It makes nodifference whether the radar, the target, or both are moving such that therange between the two is R(t), so it can be assumed without loss of generalitythat the radar is stationary and the target is moving, and that allmeasurements are made in the frame of reference of the radar.
在以上条件下，接收信号可以表示为：
Under theseconditions the received signal can be shown to be (Cooper, 1980; Gray andAddison, 2003)
其中k表示所有相关的雷达距离方程幅度因子，h(t)满足以下函数
where k absorbs allradar range equation amplitude factors and h(t) is the function that satisfies
式（2.86）中h(t)上的点表示时间导数。
The dot over h(t) inEq. (2.86) denotes the time derivative.
在完全导电的表面上，边界条件将使信号的正负号产生反转（即180°相移）。
The minus sign (180°phase shift) is required by the boundary conditions at a perfectly conductingsurface.
函数h(t)的单位为秒，是指为了在时刻t、距离R(t)处截获运动目标，雷达应当发射脉冲的时刻。
The function h(t),which has units of seconds, is the time at which a wave must have been launchedin order to intercept the moving target at time t and range R(t).
例如，如果R(t)为常数R0，则h(t) = t – R0/c。
For example, if R(t)is a constant R0 , then h(t) = t – R0/c.
在该假设下可以认为，从雷达发射机到目标的波形传播过程中，任何一个特定位置的距离变化是可以忽略的。
This holds that therange change during the short flight of any particular point in the waveformfrom the transmitter to the target is negligible.
恒速目标的情况是特别有趣的。
The case of aconstant-velocity target is of special interest.
根据式（2.86）和（2.87），令R(t) = R0–vt，并定义βv≡ v/c。
Returning to theexact result of Eqs. (2.86) and (2.87), let R(t) = R0–vt and define βv≡ v/c.
那么很容易得到
It is easy to showthat
——本文译自Mark A. Richards所著的《Fundamentals of Radar Signal Processing（Second edition）》
更多精彩文章请关注微信号：