Radar Question

Depends upon the type of RADAR. If it is a tracking or navigation RADAR the frequency is the same. If it is Doppler such as weather or traffic control, then the frequency will be altered by the speed of the object. For example if you have RADAR over land all you will see are hills trees buildings etc. Aircraft, cars, boats etc will all be masked by the surrounding background. Doppler RADAR eliminates that by displaying only objects that change frequency (moving). I recall a report of an object that only occasionally appeared on the screen. It turned out to be a flagpole moving in the wind

The frequency shift from a reflecting target will be twice the shift from an emitting target, because the range to the latter is two way.

Range of reflecting target is two way.

OK, I think I understand, that if you are continuously tracking, one doesn't need frequency shift for target velocity. The velocity can be determined from successive plots.

And I was not aware that distance of target, in any way, effected shift????.

Let me ask the question in a different manner. My question is about the F shift.

Let me set up two scenarios.

No. 1. A reflective target is moving away from radar No. 1. There will be a red(decreasing F) shift, to the returning signal, in proportion the the receding velocity.

No. 2. Now attach a radar No. 2. to the target. Will the No. 2. radar signal, have the same frequency shift, as the radar No. 1. returning reflective signal?

Oh, you're saying that the shift interaction of the reflection is two way, and additive, because the reflection has to turn around?

Pardon me.

Let me try to explain a bit better...

Say the radar sends out a signal of 1 GHz, 1,000,000,000 cycle/sec, which has a wavelength of about 1 ft. If you are travelling toward this signal at a speed of 1 ft/sec, in 1 second you will see 1,000,000,001 cycles, the 1 billion transmitted plus the 1 you ran up on.

If you reflect these 1,000,000,001 cycles back, the radar receiver will receive 1,000,000,002 cycles, including the one extra cycle due to your motion.

In other words, the signal gets doppler shifted twice, once when it reaches the reflector and secondly, when it is received by the radar.

@Rixter wrote: "Say the radar sends out a signal of 1 GHz, 1,000,000,000 cycle/sec, which has a wavelength of about 1 ft. If you are travelling toward this signal at a speed of 1 ft/sec, in 1 second you will see 1,000,000,001 cycles, the 1 billion transmitted plus the 1 you ran up on.

If you reflect these 1,000,000,001 cycles back, the radar receiver will receive 1,000,000,002 cycles, including the one extra cycle due to your motion."

One should conceptually rather follow the relativistic thinking: in Doppler radar the only thing that matters is the relative speed (v) between the radar transceiver and the reflecting target. It does give the result that you calculated, but it comes about in a somewhat different way.

Fractional Doppler shift Δλ/λ_o= sqrt[(1+v²/c²)/(1-v²/c²)] ~ 2v/c when v is much smaller then c.

Oops, I have goofed the equation.

To avoid confusion, it should have been this one:

Δλ/λ_o= (1+v/c)/(1-v/c) - 1 ~ 2v/c when v is much smaller then c.

This is a very bad way to explain it, but if you can see your way round the wrong parts it might help you see what's really happening.

Think of each "part" of the waveform as being an individual particle

If the whole wave form is moving at say v towards an object moving at u, then the approach speed is (v+u). When the "particles" are reflected they are moving apart at (v+u), and because the object is still moving at u the "particles" are moving back towards the transmitter at (v+2u). The frequency shift is doubled.

Those are interesting examples. I prefer a moving line bouncing off a wall example. The movement of the surface, can change the length, of the reflected line. Due to the interaction time, or duration of the reflection.

It's interesting to note, that a reflection is an instant lossless change in the direction and momentum.(energy)

The frequency(energy) does not change with a stationary surface. And that a moving surface, can amplify energy.

It should not surprise anyone then, that emission could possibly be an instant event.

Let's replace the train whistle with a transceiver. And place another transceiver at the station.

With the whistle set up, we can put the whistle at the station, the the people on train, will experience to same shift, as the people at the station, when whistle is on train.

The audio Doppler shift is reciprocal. Because sound is continuous. The movement of the emitter, or the absorber, has the same effect.

Now, try it with radios. One radio is moving, one is stationary. Will the people on the train here the same shift as the people at the station? Will we get the same reciprocal effects?