range-doppler map to range-velocity map conversion

Question

UPDATE 2.0

I think I've managed to produce the correct range-velocity map. I'm struggling with analytically calculating the range which I expect to see on the range-velocity map. Could someone please give me a direction for how to calculate this? I have one target which moves away from the radar in a constant velocity.

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I'm trying to implement the pulsed-doppler radar algorithm. I've went through most of the steps and generated range-doppler map where the y axis is the pulse number and the x axis is the range. I'm trying to convert the map into range-velocity map by using the relation $$ v = (c \cdot f_d)/(2 \cdot f_c) $$ where $c$ is the speed of light, $f_d$ is Doppler frequency and $f_c$ is the carrier wave frequency.

What I did was multiply the frequency axis in the range [-fs/2,fs/2] by c/(2*fc) where fs is the Nyquist frequency. The number of points which I divided the frequency axis into are fs/(number-of-samples). But when I use imagesc to plot the range-velocity map, it looks like the velocity is zero. I believe my problem is when generating the velocity axis, since when plotting all the other stages the results look good. what could be the problem?

Thank you very much once again!

Answer 1

In a range-Doppler map (RDM), you actually have two sample rates that define the two dimensions of the matrix:

1. Fast-time dimension. This is usually established by the ADC to sample the pulse return. This tends to be the fastest of the two sample rates, hence the name. This is what is typically given by the symbol $f_s$ and establishes the range dimension of the RDM.
2. Slow-time dimension. The second dimension is made up of the number of pulses collected. The fastest you can hope to gather these is at the $PRI$, and thus the sample rate across this dimension is the $PRF$.

So now the RDM captures two Nyquist bands:

1. $[-f_s/2, fs/2)$ for the fast-time dimension
2. $[-PRF/2, PRF/2)$ for the slow-time dimension

You can see directly from (2) that $\pm PRF/2$ is the maximum Doppler frequency you can measure unambiguously. You can apply the frequency-Doppler conversion formula to get the velocity limits that you can measure.

$$v_{max} = \pm\frac{f_{d_{max}}\lambda}{2} = \pm\frac{PRF\lambda}{4}$$

Example

We're going to start with some already-simulated target returns from a moving target. The pertinent parameters are

• Target range of 800 m and velocity of 250 m/s
• Wavelength $\lambda$ of 0.03 m
• PRF of 50 kHz
• Collected 500 fast-time samples and 256 slow-time samples (which is the number of pulses)

We then form the 500x256 RDM:

The mapping still has to be done in order to determine the target's range and velocity. Let's assume we've already done the range mapping, but now we need to do velocity.

Let's check that this system can unambiguously measure this velocity at the given PRF:

$$v_{max} = \pm\frac{PRF \lambda}{4} = \pm\frac{(50 \space kHz)0.03}{4}$$ $$v_{max} = \pm 375 \space m/s$$

It can, so now lets map the Doppler axis to velocity. Assuming that we did a N-point DFT in the slow-time dimension, the frequency bin size is now:

$$\Delta f = \frac{PRF}{N}$$

And using the Doppler-velocity mapping we get the velocity bin size:

$$\Delta v = \Delta f\frac{\lambda}{4}$$

We now have the bin size itself, and with a little work you can use it to transform the slow-time dimension from $[-PRF/2, PRF/2)$ to $[-v_{max}/2, v_{max}/2)$:

Update

You should be careful in defining the range axis with a negative frequency. This type of range mapping applies only to a very specific type of system that uses stretch-LFM processing, which I doubt you're doing.

For more traditional radars, it makes sense that you can't have a negative range so introducing $-f_s$ is not appropriate. Now, this was my fault since I did state that the fast-time dimension captures the band $[-f_s/2, fs/2)$. This is technically true in the mathematical sense, but is not used directly as shown to define the range axis. The velocity axis defined by $[-PRF/2, PRF/2)$ is still valid.

The calculations that you have won't work. Instead try something like this:

rangeAxis = (1:numRangeBins)*(c/(2*fs)) % Range bin size is c/(2*fs)
dopplerAxis = (-numDopplerBins/2:numDopplerBins/2 - 1).*PRF/numDopplerBins;
velocityAxis = dopplerAxis*lambda/2;

With the right values this should yield the correct range and velocity axes you're looking for.

Answer 2

First some background.

In radar the relationship between range rate, $v$ and Doppler frequency offset, $\Delta f$ is given by: $$v = -\frac{c}{2} \frac{\Delta f}{f_c}$$ where $c$ is the speed of light and $f_c$ is the carrier frequency. There may be a sign difference depending on conventions for positive range rate.

This expression is equivalent, to within a factor of 2 (see below) to that in Hunter Akins' answer.

Starting with: $$f_d = \frac{c+v_r}{c+v_s}f_c$$ where $f_d$ is the Doppler frequency, $v_r$ is the receiver velocity, and $v_s$ is the source velocity.

Assume that $v_s = 0$ and all of the motion is on the part of the receiver: $$f_d = \frac{c+v_r}{c}f_c $$

Solving for $v_r$: $$v_r = c \frac{f_d - f_c}{f_c}$$

The Doppler frequency offset is equal to the amount the Doppler frequency has been shifted from the carrier $\Delta f = f_c - f_d $: $$v_r = -c \frac{\Delta f}{f_c}$$

And, since in radar the signal is reflected from the target, the Doppler shift is twice the case for a one way path so the effective velocity for the receiver is only half that it would be for a one way signal: $$v = -\frac{c}{2} \frac{\Delta f}{f_c}$$

If you original frequency axis is Doppler space [-fs/2,fs/2] (fs being the sample rate, typically in these sorts of plots equal to the pulse repetition interval) then you can convert the axis labeling to the equivalent velocity by multiplying by -(c/(2 * carrier_freq)).

This corresponds to what you have stated you are doing. I suspect there is some other issue with you program. If you can post more details we can provide a better answer.

Answer 3

It seems like your formula is wrong. For example, if $f_{d} = f_{c}$, then $v=c/2$. The formula I'm familiar with is $$ f_{d} = \frac{c + v_{r}}{c + v_{s}} f_{c} , $$ where $v_{r}$ is the velocity of the receiver and $v_{s}$ is the velocity of the source (one dimensional). You need to estimate the frequency of max power for each received pulse. This frequency is $f_{d}$. Then probably assume $v_{s} = 0$ and use the fact that you now $f_{c}$ (since presumably you sent out the pulse). Then you can solve for $v_{r}$ for each received pulse.