> >
> > You do know that a signal reflected from a moving target undergoes two
> > Doppler shifts? The target serves both as a moving receiver and a
> > moving source. So the combined effect is the observed "Doppler" at the
> > radar is simply f' = f * (c+v)/(c-v) where f is the transmitted
> > frequency and f' is the received frequency, c is the speed of light and
> > v is the relative radial velocity.
> >
> > Clay
> >
>
> For added clarity, that should be f' = f*(c+v_r)/(c-v_t) where v_r is
> the radial velocity relative to the receiver and v_t is the radial
> velocity relative to the transmitter. For monostatic systems these are
> the same, but for multistatic systems with a large baseline length
> relative to range they can be very different.
>
> Cheers
>
> Marc

It depends on your point of view. Since I used relative velocity, the
formula I gave is correct. It is the formula resulting from Relativity.
If both your radar source and the target are moving, then use
Einstein's velocity addition formula to find the relative velocity. A
lot of people get hung up on using a 1-v/c factor. This only pops out
of the equation after using a handful of low velocity approximations.
Why use approximations, when the correct formula is not that much more
complicated?
Clay

Reply by Marc Brooker●October 25, 20062006-10-25

Clay wrote:

> birdforsale@gmail.com wrote:
> The problem is that the
>> Doppler shift represented in the second FFT doesn't seem to
>> correspond to the Doppler shift provided by the target. I don't
>> understand in this design how the Doppler shift is translated from the
>> target to the final FFT for display. I hope I have explained my self
>> well enough and I would appreciate any insight as to what I need to
>> change to correct the Doppler shift.
>
> You do know that a signal reflected from a moving target undergoes two
> Doppler shifts? The target serves both as a moving receiver and a
> moving source. So the combined effect is the observed "Doppler" at the
> radar is simply f' = f * (c+v)/(c-v) where f is the transmitted
> frequency and f' is the received frequency, c is the speed of light and
> v is the relative radial velocity.
>
> Clay
>

For added clarity, that should be f' = f*(c+v_r)/(c-v_t) where v_r is
the radial velocity relative to the receiver and v_t is the radial
velocity relative to the transmitter. For monostatic systems these are
the same, but for multistatic systems with a large baseline length
relative to range they can be very different.
Cheers
Marc

Reply by Clay●October 25, 20062006-10-25

birdforsale@gmail.com wrote:
The problem is that the

> Doppler shift represented in the second FFT doesn't seem to
> correspond to the Doppler shift provided by the target. I don't
> understand in this design how the Doppler shift is translated from the
> target to the final FFT for display. I hope I have explained my self
> well enough and I would appreciate any insight as to what I need to
> change to correct the Doppler shift.

You do know that a signal reflected from a moving target undergoes two
Doppler shifts? The target serves both as a moving receiver and a
moving source. So the combined effect is the observed "Doppler" at the
radar is simply f' = f * (c+v)/(c-v) where f is the transmitted
frequency and f' is the received frequency, c is the speed of light and
v is the relative radial velocity.
Clay

Reply by Rune Allnor●October 24, 20062006-10-24

birdfors...@gmail.com skrev:

> I am new to DSP and am currently implementing processing on Radar
> systems. I am presently taking a window around the returned/simulated
> target return and digitizing the return at 80MHz, and then push the
> data through a 256 point FFT with a Hanning window. The return has
> been mixed to an IF of 20MHz and I always take my data point from that
> bin, because the target will never provide enough Doppler shift to move
> out of that frequency bin. After I have 1024 of those peaks I then do
> another FFT, that is then used as the display. The problem is that the
> Doppler shift represented in the second FFT doesn't seem to
> correspond to the Doppler shift provided by the target. I don't
> understand in this design how the Doppler shift is translated from the
> target to the final FFT for display. I hope I have explained my self
> well enough and I would appreciate any insight as to what I need to
> change to correct the Doppler shift.

Maybe I misundertand something, but you might want to mix the
return down to baseband before digitizing. If sampled at 80 MHz,
one bin in a 256 pt DFT amounts to 20 kHz, or a "Warp number"
of 0.004 (i.e. that the target need to travel at a speed of 0.4%
of the speed of light -- 12000 km/s or so -- to shift one bin).
It seems to me as if you might want to have a look through the
overall design of your system.
Rune

Reply by ●October 24, 20062006-10-24

I am new to DSP and am currently implementing processing on Radar
systems. I am presently taking a window around the returned/simulated
target return and digitizing the return at 80MHz, and then push the
data through a 256 point FFT with a Hanning window. The return has
been mixed to an IF of 20MHz and I always take my data point from that
bin, because the target will never provide enough Doppler shift to move
out of that frequency bin. After I have 1024 of those peaks I then do
another FFT, that is then used as the display. The problem is that the
Doppler shift represented in the second FFT doesn't seem to
correspond to the Doppler shift provided by the target. I don't
understand in this design how the Doppler shift is translated from the
target to the final FFT for display. I hope I have explained my self
well enough and I would appreciate any insight as to what I need to
change to correct the Doppler shift.