Problems in calculating signal to volume clutter ratio

A problem with volume clutter, e.g. rain, is that the volume illuminated may not be completely filled, in which case the fraction filled must be known, and the scatterers may not be uniformly distributed. Consider a beam 10° in elevation. At a range of 10 km the beam could cover from ground level to a height of 1750 metres. There could be rain at ground level but the top of the beam could be above cloud level. In the part of the beam containing rain the rainfall rate will not be constant. One would need to know how the rain was distributed to make any accurate assessment of the clutter and the signal to clutter ratio. All that can be expected from the equation is an estimate to the nearest 5 or 10 dB.

Beam filling

Figure 2. Illustration of high- and low-angle surface-clutter illumination

For a target close to the Earth's surface such that the earth and target are in the same range resolution cell one of two conditions are possible. The most common case is when the beam intersects the surface at such an angle that the area illuminated at any one time is only a fraction of the surface intersected by the beam as illustrated in Figure 2.

Pulse length limited case

For the pulse length limited case the area illuminated depends upon the azimuth width of the beam and the length of the pulse, measured along the surface. The illuminated patch has a width in azimuth of

${\displaystyle \ 2R\tan \theta /2}$.

The length measured along the surface is

${\displaystyle \ (c\tau /2)\sec \psi }$.

The area illuminated by the radar is then given by

${\displaystyle \ A=2R(c\tau /2)(\tan \theta /2)\sec \psi }$

For 'small' beamwidths this approximates to

${\displaystyle \ A=R(c\tau /2)\theta \sec \psi }$

The clutter return is then

${\displaystyle \ C={\frac {P_{t}G^{2}\lambda ^{2}}{(4\pi )^{3}R^{4}}}A\sigma ^{o}}$ Watts

Substituting for the illuminated area ${\displaystyle A}$

${\displaystyle \ C={\frac {c}{2^{7}\pi ^{3}}}{\frac {P_{t}G^{2}\lambda ^{2}}{R^{3}}}\tau \theta \sec \psi \sigma ^{o}}$ Watts

where ${\displaystyle \sigma ^{o}}$ is the back scatter coefficient of the clutter. Converting ${\displaystyle \theta }$ to degrees and putting in the numerical values gives

${\displaystyle \ C=1300{\frac {P_{t}G^{2}\lambda ^{2}}{R^{3}}}\tau \theta ^{o}\sec \psi \sigma ^{o}}$ Watts

The expression for the target return remains unchanged thus the signal to clutter ratio is

${\displaystyle \ {\frac {S}{C}}={\frac {1}{1300}}{\frac {R^{3}}{P_{t}G^{2}\lambda ^{2}}}{\frac {1}{\tau \theta \sec \psi \sigma ^{o}}}{\frac {P_{t}G^{2}\lambda ^{2}}{(4\pi )^{3}R^{4}}}\sigma }$ Watts

This simplifies to

${\displaystyle \ {\frac {S}{C}}=4\times 10^{-7}{\frac {\cos \psi }{R\tau \theta }}{\frac {\sigma }{\sigma ^{o}}}}$

In the case of surface clutter the signal to clutter now varies inversely with R. Halving the distance only causes a doubling of the ratio (a factor of two improvement).

Problems in calculating clutter for the pulse length limited case

There are a number of problems in calculating the signal to clutter ratio. The clutter in the main beam is extended over a range of grazing angles and the backscatter coefficient depends upon grazing angle. Clutter will appear in the antenna sidelobes, which again will involve a range of grazing angles and may even involve clutter of a different nature.

Beam width limited case

The calculation is similar to the previous examples, in this case the illuminated area is

${\displaystyle \ A=\pi R^{2}\tan ^{2}\theta /2}$

which for small beamwidths simplifies to

${\displaystyle \ A\approx \pi R^{2}\theta ^{2}/4}$

The clutter return is as before

${\displaystyle \ C={\frac {P_{t}G^{2}\lambda ^{2}}{(4\pi )^{3}R^{4}}}A\sigma ^{o}}$ Watts

Substituting for the illuminated area ${\displaystyle A}$

${\displaystyle \ C={\frac {P_{t}G^{2}\lambda ^{2}}{(4\pi )^{3}R^{4}}}\pi R^{2}(\theta /2)^{2}\sigma ^{o}}$ Watts

This can be simplified to:

${\displaystyle \ C={\frac {P_{t}G^{2}\lambda ^{2}}{4^{4}\pi ^{2}R^{2}}}\theta ^{2}\sigma ^{o}}$ Watts

Converting ${\displaystyle \theta }$ to degrees

${\displaystyle \ C={\frac {P_{t}G^{2}\lambda ^{2}}{4^{4}R^{2}}}(\theta ^{o}/180)^{2}\sigma ^{o}}$ Watts

The target return remains unchanged thus

${\displaystyle \ {\frac {S}{C}}={\frac {4^{4}R^{2}}{P_{t}G^{2}\lambda ^{2}}}(180/\theta ^{o})^{2}{\frac {1}{\sigma ^{o}}}{\frac {P_{t}G^{2}\lambda ^{2}}{(4\pi )^{3}R^{4}}}\sigma }$

Which simplifies to

${\displaystyle \ {\frac {S}{C}}=5.25\times 10^{4}{\frac {1}{\theta ^{o2}R^{2}}}{\frac {\sigma }{\sigma ^{o}}}}$

As in the case of Volume Clutter the Signal to clutter ratio follows an inverse square law.

General problems in calculating surface clutter

The general significant problem is that the backscatter coefficient cannot in general be calculated and must be measured. The problem is the validity of measurements taken in one location under one set of conditions being used for a different location under different conditions. Various empirical formulae and graphs exist which enable an estimate to be made but the results need to be used with caution.