# FMCW radar scattering direction from a special case of a rough dielectric surface

I have some misunderstandings about radar backscatter from rough dielectric surfaces and was hoping somebody could enlighten me.

I have one specific question first concerning a thought experiment. A radar wave is incident (far field, monostatic) on a surface with a twin-delta height distribution. A 'base' surface, and a 'raised' surface, with the two heights differing by half the wavelength of the radar (I appreciate the radar in fact has a bandwidth, but let's just say the central wavelength). Can the radar 'see' this surface? It seems like half the radar energy would come back from the lower surface in anti-phase with the other half, the amplitude would cancel out at the sensor and the surface would be missed? If this is the case, is the effect of the radar bandwidth enough to cancel it out?

I ask this question because it seems like many rough surfaces that are more complex could be broken down into many pairs of surface segments that differ in range by $$\lambda$$/2? Would all these paired surfaces be invisible?

Assuming that the angle of incidence is perpendicular to the surface if the depth is $$\lambda/2$$ then the radar echo from the reflecting surface (be it metal or dielectric) will be in phase because of the round-trip, but they will be out of phase if the depth is $$\lambda/4$$. This has nothing to do with the modulation (here FMCW) but you need that the bandwidth be narrow relative to the carrier frequency and also ignored any reflection from the edges of the grooves and other (waveguide) propagation effects in the grooves themselves that may make the guide wavelength different from the free space wavelength.

• Ah yes of course, I forgot about the round trip! Were the heights different by $\lambda$/4 then I think from your answer that (barring the effects you mention), the surface would be 'invisible'? Just wondering if you have any thoughts about my conjecture at the end of the post regarding a surface where you can pair up segments that differ in range by $\lambda$/4 to make an 'invisible surface'? Thanks so much for your help! – Robbie Mallett Oct 14 '20 at 17:21
• that is a lot more complicated question, because if at a certain setup it were "invisible" due to the cancellation then it could become visible if you moved the radar somewhat to the side where the return waves will be somewhat or mostly in phase. Also the waves scatter off the groove edges and can be visible. – hyportnex Oct 14 '20 at 17:31
• Yes, I was thinking of minimising the number of groove-edges to one, perhaps making the surface viewed by the radar just a step function (h = 0 for x < 0, h= $\lambda$/4 for x> 0). – Robbie Mallett Oct 14 '20 at 17:34