From my understanding, and correct me if I'm wrong, the bistatic radar equation assumes that the transmitter and the receiver are separated by a distance $L$, and that the transmitter and receiver are static, while the target is moving at a distance $R_r$ from the receiver and a distance $R_t$ from the transmitter. $$P_r = {{P_t G_t G_r \sigma \lambda^2}\over{{(4\pi)}^3 R_t^2R_r^2}}$$

Hence my question, does the bistatic radar equation is different when the transmitter, the receiver, and the target are all moving? Does the equation change only when one want to consider Doppler effect?


The bistatic radar equation is valid for a point target in vacuum with effectively infinite bandwidth transmit/receive hardware and as such it is time independent, and thus velocity independent as well. In reality when anything is moving (transmitter, target, receiver) things do change, for example, the multipath environment, scattering cross section, glint, etc., and the received power as calculated from this equation is less important than the time varying enviroment.

  • $\begingroup$ Thanks for the answer. I thought the time was going to show up in the equation when changing the $\lambda$ for considering Doppler effect. $\endgroup$ – m_power Jun 29 '17 at 15:22
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    $\begingroup$ Oh I see. In that you are right but think of the scale how much in practice $\lambda$ actually changes. If the target moves at 10km/sec, a very fast target, then it is only about $2\times 10/(0.3\times 10^6)$, or maybe ~6o ppm, and it is completely negligible when compared to any other effect. $\endgroup$ – hyportnex Jun 29 '17 at 16:03
  • $\begingroup$ and note too that if you are concerned about the the frequency dependence of the Friis's equation then you should not use antenna gain in it because that is also frequency dependent for a given aperture $A_{eff}$: $G \sim \frac{4\pi A_{eff}}{\lambda ^2}$ $\endgroup$ – hyportnex Jun 29 '17 at 16:54
  • $\begingroup$ Yet, the shift in frequency is still high enough to be measurable for calculating velocity or range of the target? $\endgroup$ – m_power Jun 29 '17 at 18:10
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    $\begingroup$ Yes, because frequency measurement resolution (and accuracy) depends on the length of time the target is illuminated and not on anything else. 1.5m/sec corresponds to 1kHz (0.1ppm) shift at 10GHz (X-band radar), and you can get very good estimate of 1kHz shift with 1msec coherent integration. Doppler shift cannot be used to measure range, at least not with a single receiver. $\endgroup$ – hyportnex Jun 29 '17 at 20:01

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