Doppler Frequency Resolution for Radar

Question

Doppler frequency is calculated by the formula $f_{d} = 2*v_{r}/\lambda$, where $v_{r}$ is the radial velocity and $\lambda$ is the radar wavelength.

Normally FFT is used to detect the doppler frequency(A bank of FFT filters is used for Doppler frequency estimation).

Assume that sampling frequency is fixed as $4*f_{max}$, where $f_{max}$ is the maximum $f_{d}$ corresponding to the maximum radial velocity.

Now I have to determine the number of FFT points to be employed in the FFT calculation. How can I determine this.

In other words, which target parameter determines the Doppler frequency resolution.

Answer 1

Doppler resolution usually depends on the dwell time $T_{dwell}$, i.e., the time for which the radar is going to stare (look) at the target. If the pulse/sweep repetition interval is given by $T_{PRI}$ then $N = T_{dwell}/T_{PRI}$ is the number of pulses/sweeps in one dwell, which is also the limit of your FFT size. Your maximum Doppler corresponds to $N/2$th bin, therefore the Doppler resolution is $f_d/(N/2)$. Note that more pulses you are able to send to the target, better is the Doppler resolution.

Answer 2

Depends on the resolution you want for doppler frequency. Doppler frequency resolution determines how well you can observe targets of different radial velocities. If you want to resolve targets that are closely spaced in radial velocities you will need to have higher doppler resolution measurement. That means you will have to have more number of points in your FFT calculation. That means you have to observe the signal for a longer time. Fsampling/N will give you the frequency bin resolution where N is the number of points in FFT and Fsampling is the sampling frequency.

Answer 3

Similar to spectral estimation problems, if you need better resolution, (aka narrower mainlobe in your estimate), then you need to have a larger set of data to FFT. A larger set of data means you have a longer record, and thus, when FFT'd, your mainlobe becomes narrower.

The ability to discriminate between peaks of overlapping main lobes is what is commonly used as a metric for resolution.

If however you want to keep the frame size the same, but increase your frequency granularity, then you can zero-pad your data pre-FFT, (but post-windowed).

In pulse-doppler radar applications, that means you need more pulse-records whose constant-range samples are to be FFT'd across records. This in turn means you need to transmit more pulses. This also makes intuitive sense: If you want to estimate the velocity of a target, you need to probe it longer - aka - you need to send out more pulses - aka - more data, and thus, a larger record length.

Answer 4

The Doppler frequency resolution is not determined by any characteristic of the moving target. It is determined by the Radar system parameters. If you are using a FFT to to determine the Doppler shift, this assumes that you're transmitting a sinusoidal pulse. The length of the pulse then determines the Doppler resolution for that single pulse. This assumes the radial velocity of the target is relatively constant for the pulse - usually this is not too bad an assumption

You can increase the resolution by coherently integrating multiple pulses. There are design constraints on the Pulse Repetition Frequency (PRF). The PRF needs to be high enough to sample the Doppler frequency bandwidth of interest. PRF needs to be low enough to allow the pulse to propagate to the target range and back, or else you will suffer from pulse ambiguities. Note that at a fixed PRF you will also have blind velocities.

The number of pulses you integrate is referred to as the Coherent Processing Interval (CPI). As mentioned elsewhere, if your target does not maintain a constant radial velocity throughout the CPI, this will result is some spectral spreading across your FFT bins.

Answer 5

Doppler frequency for a target is calculated by taking the FFT over the slow time for the range bin the target is in. The key to understanding velocity resolution for pulsed radar is that coherent processing interval (CPI - the total amount of time spanned by slow time samples) acts as a rectangular windowing function on your slow time samples.

A target moving at a constant velocity sampled for an infinite number of slow time samples would be represented by a impulse (delta) at a frequency defined by the equation below. Since we only are able to collect coherent samples over the CPI we've effectively multiplied (in time) our signal by a rectangular windowing function. Using the multiplication property of the Fourier transform (multiplication in time is equivalent to convolution in frequency) we realize that windowing has the effect of "smearing" the spectrum of our target. Specifically we will have "copies" of our windowing function spectra centered at the frequency of our target.

With this understanding we can calculate a velocity resolution by calculating the minimum distance in frequency our windowing function can be spaced before we can no longer discriminate between the two copies.

To derive an equation for velocity resolution we start with the fact that the scaled (by $\alpha$) rectangular window function forms a Fourier pair with the sinc function. This can be derived from basic Fourier transform properties.

1. $$ rect(\alpha t) \Leftrightarrow \frac{1}{\alpha}*sinc\left(\frac{\pi*f}{\alpha}\right) $$

The scale factor $\alpha$ is related to the CPI by the following.

2. $$ \alpha = \frac{2}{CPI} $$

In our case f is equal to twice the ratio of the target velocity to carrier wavelength.

3. $$ f_{target} = \frac{2V_{target}}{\lambda}$$

Now we realize if we have two targets they will be represented in frequency by two $sinc$ functions separated by the difference in their velocities.

We choose the distance of the first null in the sinc function as the minimum distance we can separate two $sinc$ functions and still resolve them. This intuitively makes sense to me, but I can't claim to understand exactly why this is the minimum distance we can resolve two overlapping sinc functions. Note there is a analogous problem in optics that leads to the Rayleigh Criterion where the first null of the sinc function is used is the minimum separation distance to resolve two sources so I'm inclined to accept this choice.

Given that nulls of the sinc function occur at integer multiples of $\pi$ it is easy to derive that the first null of the target spectra sinc occurs at the following.

4. $$ f_{null} = \alpha \Rightarrow f_{DopplerRes} = \alpha$$

Substituting equations 2 & 3 into equation 4 we get the following.

5. $$ \frac{2v_{DopplerRes}}{\lambda}=\frac{2}{CPI} $$

Hence.

6. $$ v_{DopplerRes}=\frac{\lambda}{CPI} $$