Download Debunking SAR Moving Target Mythology

Spotlight Synthetic Aperture Radar and Moving Targets
Synthetic aperture radar processing works optimally when there is no motion in the scene
being collected against. It is well known that target motion parallel to the SAR velocity
vector (along track motion) causes the target image to be blurred. However, just from a
general scan of the vast literature on the subject, there appears to be some ambiguity
concerning whether it’s constant velocity “towards the SAR” that causes a simple
focused transverse shift of the target from its actual path or motion perpendicular to the
SAR velocity vector (cross track motion) that causes that simple shift. (It seems to be
generally agreed that it’s motion towards the SAR that causes the shift but it’s not so
clear what “towards the SAR” means.) This is a simple analysis showing that it’s
constant velocity perpendicular to the SAR velocity vector (i.e., cross track motion) that
causes the focused shift1 and it does not matter whether or not the SAR is at the “point of
perpendicularity” at any time during the data collection.
Figure 1 diagrams a spotlight-mode collection geometry in coordinates relevant to an
actual SAR image. x is the nominal cross range direction and y is the image range
direction (not to be confused with “radar range”). Figure 1 shows a squinted collection.
A broadside collection would consist of the SAR flight path being parallel to the x-axis.
Although Figure 1 implies ground plane geometry, it can easily be thought of as slant
plane geometry. Just take the SAR ground track to be the actual elevated SAR path and
all other parameters to apply in the slant plane instead of the ground plane.
This is sometimes referred to as “defocusing” because the image is displaced from where it’s “supposed”
to be. However, that terminology is confusing. Defocusing implies blurring, and blurring corresponds to
different parts of the image shifting in different directions, not the entire image shifting in one direction.
Constant velocity cross track motion does not cause blurring. There is no defocus. It’s perfectly focused,
just at the wrong spot.
1
(Speed = Vs)

2
s
t
Vt
(x0,y0)
Figure 1: SAR Collection Geometry (in image coordinates)
The radar range to the SAR from scene center along the y-axis is R in the slant plane and
its projection in the ground plane is Rg (not shown in Figure 1 for clarity). There are
different ways of defining the squint angle, s. The definition used here relates to the
angle between the direction to scene center from the SAR at its central aperture position
and a vector perpendicular to its velocity vector. It is merely represented in a somewhat
different manner from that in the figure. Nevertheless, that is still the angle that s
signifies. (Think of “opposite interior angles,” and it may be important in subsequent
development to remember that Figure 1 depicts a negative squint collection.) As the SAR
moves in its path, it radiates pulses at equally spaced intervals of time that correspond to
not quite equally spaced azimuth angles, . Hence, a left-looking SAR would start
collecting at  =  and stop collecting at  = -, while a right-looking SAR would begin
at  = -. Each azimuth angle is associated with its own SAR elevation angle, or
“instantaneous grazing angle.”
Consider a point target at (x0, y0), as shown in Figure 1, at the start of the SAR
collection.2 With Vx and Vy the components of the target velocity, Vt, the scattered field
from the target at azimuth angle  and elevation angle  at time t can be represented as
E(k,) = Aexp[ik(x0 + Vxt)coscos + ik(y0 + Vyt)cossin],
(1)
where k is the standard 4/ and the standard term corresponding to range from scene
center to the SAR has been suppressed. (If Figure 1 is taken to apply to the slant plane,
take  to be zero in equation 1.) The important quantity in equation 1 is the phase factor
in the exponent. With kx = kcossin and ky = kcoscos, and Vx = Vtsint and Vy =
Vtcost, that phase factor is
(kx,ky) = kx(x0 +Vttsint) + ky(y0 + Vttcost).
(2)
The remaining task is to relate time to wavevector components kx and ky. First, note that
if Vt = 0 the fourier transform of E(kx,ky), i.e., the image of the stationary target, is
I(x,y) = A(x – x0)(y – y0)
under the ludicrous assumption of infinite bandwidth. A more realistic finite bandwidth
would lead to a product of sync functions centered at (x0, y0).
But Vt = 0 isn’t what this is about. There are three points in Figure 1 that form a right
triangle. Two of those points are scene center and the SAR position. The third point is
where the line segment of length D intersects the SAR’s ground track (or the SAR
velocity vector in case of slant plane geometry) perpendicularly. With a right-looking
SAR (positive) speed of Vs, simple kinematics yields
2
Reference to a point source is not fundamentally important. It just makes fourier theory more explicitly
useful.
Vst = D[tan( - s) + tan( + s)],
(3)
(For a left-looking SAR, replace Vs and  with the negatives of those parameters.) An
involved reader can easily reduce equation 3 to
D(tan + tan)sec2s
t = _________________________.
Vs(1 + tantans)(1 - tantans)
But tan = kx/ky, and the above equation simplifies to
D(kx + kytan)
t = _______________________________.
Vs(kxsins + kycoss)(coss - tansins)
(4)
The wavevector components in the denominator of equation 4 look troublesome. The
nonlinear dependence of time on wavevector components, leading to such behavior in the
phase function of equation 2, does not make things look good for avoiding smearing in
the image. The reader continuing to be involved may likely see what the target velocity
angle t needs to be in order to eliminate that wavevector term in the denominator, but
let’s just make things explicit.
Equations 2 and 4 give
DVt(kx + kytan)(kxsint + kycost)
(kx,ky) = kxx0 + kyy0 + _______________________________.
Vs(kxsins + kycoss)(coss - tansins)
(5)
Equation 5 makes it very clear what the target velocity angle needs to be to keep the radar
phase return linear with wavevector components and avoid smearing in the image. That
angle needs to be s, the collection squint angle, or s + 180. That puts the target on a
path perpendicular to the SAR velocity vector (and its ground track). No other velocity
direction will eliminate the nonlinearity. (Nothing in the development indicates that the
SAR is required to have been at the point of perpendicularity at any time during the data
collection.) In particular, simply moving towards the SAR’s central collection position
will not produce a mere focused shift in image position unless the direction of motion just
happens to also be perpendicular to the SAR velocity vector (i.e., a broadside collection).
In the case of t = s, simple fourier theory indicates that the target displacements in cross
range (x) and range (y) can be expressed with the vector
DVt(x + ytan)
S = _________________,
Vs(coss - tansins)
(6)
where x and y are the standard axis unit vectors. (Negate that vector if t = s + 180.) Of
more interest might be the displacements perpendicular to the target motion (i.e., the
along track shift) and along the target motion (i.e., the cross track shift). A unit vector
perpendicular to the target velocity and parallel to the ground (or slant3) plane is given by
P = xcoss - ysins
(7a)
and a unit vector in the direction of the target motion is
L = xsins + ycoss
(7b)
Dot products between equation 6 and the above two unit vectors give for the transverse
(along track) and along target motion (cross track) displacements, respectively,
Sat = DVt/Vs
3
Just remember to use parameters defined in the slant plane.
(8a)
and
Sct = (DVt/Vs)tan(s + ).
(8b)
It is not surprising that the target is displaced along its line of motion in the image from
where it was at the start of the collection. After all, it was moving along that line in the
first place. However, the result of equation 8b is actually more curious than the
transverse displacement of equation 8a. Sct/Vt is the time it would take the SAR to travel
from the collection starting point to the point corresponding to a broadside collection,
whether or not it actually travels along any part of that path during the specific data
collection. For negative squint and s larger in magnitude than  (the usual squinted
case), i.e., the situation depicted in Figure 1, this time is negative. The SAR would have
to reverse course from where it was at t = 0 (-) to make that trip, and the along target
displacement is behind where the target was at when the collection started.
It was stated above that no target velocity direction other than s or s + 180 will keep
the returned radar phase from being nonlinear in wavevector components. That statement
was true in context. However, in the context of 3D space, it is not true. There are two
other directions that will cause a simple focused shift in target position. Those directions
are vertically upwards and downwards. Consider vertical motion with target speed Vt at
fixed transverse position (x0, y0). Taking z = 0 at t = 0, the radar phase is
DVt(kx + kytan)kz
(kx,ky) = kxx0 + kyy0 + _______________________________.
Vs(kxsins + kycoss)(coss - tansins)
(9)
The wavevector component kz is related to the other two wavevector components via the
equation for the slant plane. From Jakowatz, et. al.’s book on Spotlight Mode Radar,4
4
s is Jakowatz’s g. This analysis assumes that Figure 1 refers to ground plane geometry.
kz = tan(kxtans + ky),
(10)
where  is the slant plane grazing angle. Equation 10 expands to
kz = tan(kxsins + kycoss),
where  is the slant plane slope angle (tan = tan/coss; see Jakowatz, et. al.’s book),
again eliminating the wavevector nonlinearity in the radar phase:
DVttan(kx + kytan)
(kx,ky) = kxx0 + kyy0 + __________________.
Vs(coss - tansins)
(11)
The cross range and range displacements for vertical motion can be expressed as the
vector
DVt(xtan + ytan)
S = _________________,
Vs(coss - tansins)
(12)
Equations 7a and 7b still apply as the along track and cross track unit vectors,
respectively, so the along track and cross track displacements are
Sat = DVttan/Vs
(13a)
and
Sct = (DVttan/Vs)tan(s + ).
(13b)
Equations 13 look like a simple scaling of equations 8 by the tangent of the slope angle.
The net resultant displacement for vertical motion is (DVttan/Vs)sec(s + ), and that
resultant for motion in the ground plane is the same expression without the tan factor.
So, the gist of the above is that constant velocity target motion towards (or away from)
the SAR is what causes an unblurred shift in target position as long as “towards (or away
from) the SAR” refers to the SAR’s velocity vector, not the SAR itself. Further, it
doesn’t matter in which dimension the motion is, just as long as the motion is
perpendicular to the SAR velocity vector.
References
C.V. Jakowatz, Jr., D.E. Wahl, P.H. Eichel, D.C. Ghiglia, P.A. Thompson, SpotlightMode Synthetic Aperture Radar: A Signal Processing Approach, Kluwer Academic
Publishers, Boston, 1996.