I have a doubt about the working principle of a simple Radar.

Consider a stationary object, which does not move with respect to our transmitter. We send an electromagnetic wave to it, and it will reflect it with the same frequency (since it is not moving). This wave will return to the transmitter with a certain phase shift from which we can evaluate the distance between the object and the transmitter.

But we know that in general the composition of an incident wave and its reflected wave generates a standing wave. Therefore, why at the transmitter we get the incident and reflected waves (from whose phase shift we evaluate the distance), instead of a unique standing wave?

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    \$\begingroup\$ Most radars are pulsed so that when the reflected signal reaches the transmitting antenna, the incident wave is gone. \$\endgroup\$ – Barry Jun 17 '19 at 18:46
  • \$\begingroup\$ A standing wave is just a superposition of counter-propagating travelling waves. You can't have one or the other. You have both. \$\endgroup\$ – The Photon Jun 17 '19 at 20:12

This wave will return to the transmitter with a certain phase shift from which we can evaluate the distance between the object and the transmitter.

No. The radar sends a pulse, and times the reflection. It's the round trip time delay that's used to evaluate the distance to the object.

If you have a left-travelling wave and a right-travelling wave, then measure the total amplitude at various points with a detector sensitive to both waves, say a dipole and detector diode, then you indeed see a standing wave.

Although there may be a standing wave pattern when both waves are viewed like this, the two component waves are still there, they do not interfere, and each can be received separately. The radar receiver is set up with a reflector to receive only the return waves.

  • \$\begingroup\$ I think the OP might be referring to modulated continuous wave radar or the radar version of this this: rp-photonics.com/… radartutorial.eu/02.basics/Continuous%20Wave%20Radar.en.html \$\endgroup\$ – DKNguyen Jun 17 '19 at 19:47
  • \$\begingroup\$ @Neil_UK: why are you saying that the two component waves do not interfere? If the object does not move, they have the same frequency, so they could interfere in principle. \$\endgroup\$ – Kinka-Byo Jun 18 '19 at 19:41
  • \$\begingroup\$ The do not affect the other wave, each wave is indivudually retrievable by a reciever designed to look in just that direction. Perhaps we are using the word interference in slightly different senses. Wave interference, as (say) ripples on a pond interfere to produce peaks and troughs, is the same thing. However the waves pass through each other and don't affect each other. If you receive both waves combined, then you see a standing wave pattern. If you recieve either wave by itself, then you see a travelling wave. \$\endgroup\$ – Neil_UK Jun 18 '19 at 20:19
  • \$\begingroup\$ @Kinka-Byo Yes, a phase shift reading won't tell you the unambiguous distance. Typically radar has always used time delay to indicate distance. Obviously this needs some form of wideband signal to discriminate time. Early radars used a pulse, later ones use FM, but modern digital ones can correlate any waveform. There's always a tradeoff between resolution and maximum unambiguous range \$\endgroup\$ – Neil_UK Jun 27 '19 at 18:22
  • \$\begingroup\$ Perfect, thank you very much! \$\endgroup\$ – Kinka-Byo Jun 27 '19 at 18:24

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