The Physics

The so-called deep-water equation for the speed of water waves is $$v = \sqrt \frac{g \lambda}{2 \pi}$$ where $g$ is gravitational accelecation and $\lambda$ is the wavelength of the water waves.

This equation is good when the water is deeper that about 10 wavelengths.

The water waves will have a wavelength half that of the radar waves, $ \lambda_{water} = \frac{1}{2} \lambda_{radar} $. $$v = \sqrt { \frac{g}{2 \pi} \frac {\lambda_{radar}}{2}}$$ so in terms of the radar frequency the equation becomes $$ v = \sqrt \frac{g c}{2 \pi \times 2 f}$$ with $f$ the radar fequency and $c$ the speed of light. $$ v = \sqrt \frac{g c}{4 \pi f}$$ In knots this reduces to [f in MHz]. $$ v = \frac{30}{\sqrt{f}}(knots)$$
The radar waves will reflect from waves advancing toward the radar and also those receding from the radar. This will produce a positively Doppler shifted set of echoes and a negatively Doppler shifted set.

These will form a pair of lines extending in range, called Bragg lines in honour of the father/son scientists who worked in Adelaide on X-ray crystallography, William Henry Bragg [the father] and William Lawrence Bragg [the son].

The above equation can be used to calculate the expected Doppler shift of the Bragg lines, which comes out as ... $ D = \pm .1 \sqrt f (Hz)$ with $f$ in $MHz$.