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       The term synthetic aperture radar (SAR) derives from the fact that the
motion of an aircraft (airplane, satellite, UAV, etc.) is used to artificially create,
or synthesize a very long linear array. The reason for creating a long array is to
provide the ability to resolve targets that are closely spaced in angle, or cross
range (usually azimuth). This, in turn, is driven by one of the main uses of
SARs: to image the ground or targets. In both cases we need to be able to
resolve very closely spaced scatterers. Specifically, we desire resolutions in the
order of a few feet. To realize such resolutions in the range coordinate we
would use wide bandwidth waveforms. To realize such resolutions in cross
range we need very long antennas.
      To get an idea of what we mean by “long” antenna, let’s consider an
example. Suppose we are trying to image a ground patch at a range of 20 Km.
We decide that to do so we want a cross range resolution of 1 m. We can
approximately relate cross range distance,  y , to antenna beamwidth,  B , and
range, R , by
        y  R B

as shown in Figure 1. For  y  1 m and R  20 Km we get  B  5  10 5 rad or
about 0.003°!

            Figure 1 – Relation of Cross-Range distance to Beamwidth

      The beamwidth of a linear array with uniform illumination can be
approximately related to antenna length by

       B   L .                                                           (0)

If we assume that the radar of the above example operates at X-band and that
  0.03 m we get
       L    B  600 m .

Clearly it would not be practical to use a real antenna that is as long as six
football fields. Instead, a SAR synthesizes such antenna by using aircraft
motion and signal processing.

       Before we discuss SAR processing we want to discuss some properties of SAR.
We will start with a review linear arrays since a SAR synthesizes a linear array. Suppose
we have a M  2 N  1 1 element linear array as shown in Figure 2. We have a target
located at some xn , yn that emits an E-field Eo e j 2 fct that eventually reaches each antenna
element. We can write the E-field at the kth element as
                         Eo j 2 f c t rk            Eo  j 2 rk  j 2 fct
           Ek  t         e                 c
                                                         e          e         .     (1)
                         rk                            rk

                                     Figure 2 – 2N+1 Element Linear Array

The resulting voltage at the output of the kth element is
                        Vo  j 2 rk  j 2 fct
           vk  t        e          e         .                                    (2)

We can write rk as

    We choose this form of M to simplify some of the notation to follow.

       rk  xn   yn  kd   xn  yn  k 2d 2  2kdyn
             2                  2    2   2

           ro2  k 2d 2  2kdro sin 

where we use yn  ro sin  .

       Since ro             kd we can approximate rk as

       rk  ro2  2kdro sin   ro 1                      sin   ro  kd sin  .   (3)

We next substitute Equation 3 into the exponent of vk  t  and assume that
rk  ro in the Vo rk term. This gives,

                    Vo  j2 ro  j 2 fct j 2kd sin
       vk  t        e         e        e             .

       To form the total output of the array we sum the vk  t  to get
                                        Vo  j 2 ro  j 2 fct N j 2kd sin
       v t       
                  k  N
                           vk  t  
                                           e          e         e
                                                               k  N
                                                                               .     (4)

We next use Equation 4 to form the antenna radiation pattern as

                             1  sin  M d sin   
       R    o2 v  t  
                                                   
               Vo            M  sin  d sin   
                                                   
where we included the r02 Vo2 term to normalize it out of R   .

       When we formulated the antenna radiation or gain pattern as above we
were interested in how R   varied with target angle,  . We found that the
peak of R   occurred at   0 as shown in Figure 3.

              Figure 3 – Normalized Radiation Pattern vs. Target Angle

       As an extension to the above we can devise a means of steering the beam
by including a linear phase shift across the array elements as shown in Figure
4. This phase shift is included in the equation for v  t  in the form
                                       Vo  j 2 ro  j 2 fct N  j 2kd sinS j 2kd sin
       v t      ak vk t  
                  k  N               ro
                                          e          e         e
                                                              k  N
                                                                                e             .

This leads to a more general R   of

                            sin  M d  sin   sin  S   
       R               
                            sin  d  sin   sin  S   
                                                                .                                 (5)
                M                                             
                                                             
Again, for standard array theory, we were interested in how R   varies with 
for a fixed  S . In this case the peak of R   would occur at  S , as shown in
the example of Figure 5.

                Figure 4 – Linear Array with Phase Shifters

Figure 5 – Normalized Radiation Pattern vs. Target Angle – Beam Steered to -
                                 0.01 deg

       In SAR theory we need or reorient ourselves by thinking of the target
angle,  , as being fixed and examining how R   varies with  S . In other
words, we consider a fixed,  , and plot R S  . An example plot of R S  for
  0.01 (i.e. the target location is fixed at 0.01 deg) is shown in Figure 6. In
this plot, R S  peaks when the beam is steered to an angle of 0.01°.

   Figure 6 – Normalized Radiation Pattern vs. Beam Steering Angle – Target
                             located at 0.01 deg

       Figure 7 contains a plot of R S  for the case where there is a target at -
0.02° and a second target at 0.01°. Further, the second target has half the RCS
(radar cross-section) of the first target. Here we note that the plot of R S  tells
us the location of the two targets and their relative amplitudes. This is one type
of information we want when we form SAR images.

    Figure 7 – Normalized Radiation Pattern vs. Beam Steering Angle – Two
                    Targets located at -0.02 and 0.01 deg

       It will be noted that the plot of R S  is analogous to a plot of the power
out of a stretch processor, P  r  . P  r  is the other information we use to we
form SAR images. Specifically, we will compute R S  and P  r  for various
values of  S and r and then plot R S  P  r  as intensities on a rectangular
grid. The discrete values of  S and r will be separated by the angle resolution
of the SAR array and the range resolution of the stretch processor.
      With the above background we now want to start addressing the issues
associated with forming R S  in practical situations. We begin by modifying
the above array theory so that it more directly applies to the SAR problem.
       In standard array theory we generate a one-way antenna pattern
because we consider an antenna radiating toward a target (the transmit
antenna case) or a target radiating toward an array (the receive antenna case).
In SAR theory we need to consider a two-way problem since we transmit and
receive from each element of the synthetic array. If we refer to Figure 2, we can
think of each element as the position of the SAR aircraft as it transmits

successive pulses. When the aircraft is located at kd the normalized transmit
“voltage” is
         v  t   e j 2 fct .2

The resultant received signal (voltage) from a target at xn , yn is

                               v  t  2rn ,k c   2 Sn e c  n ,k   2 Sn e
                          PSn                        P j 2 f t  2 r c  P  j 4 rn ,k  j 2 fct
         vn ,k  t       2
                                                                                         e         (6)
                         rn ,k                      rn ,k               rn ,k

where PSn is the return signal power and is determined from the radar range
equation. rn , k is the range to the nth target when the aircraft is at kd .

We note that the difference between Equations 2 and 6 is that the latter
includes an r 2 term in the denominator and has twice the phase shift. If we
modify Equation 3 as

         rn ,k  rn2,0  2kdrn ,0 sin   rn ,0 1          sin   rn ,0  kd sin 
                                                      rn ,0
and repeat the math shown on pages 3 and 4, we get the following equation for
the radiation pattern of a SAR antenna

                    PSo 1  sin  2 M  d  sin   sin  S   

         R  S   2
                                                                .                               (7)
                    rn,0 M  sin    sin   sin  S   
                                   2 d
        Figure 8 contains plots of R S  for the standard linear array (Equation
5) and the SAR array (Equation 7). In both cases the peak is normalized to
unity. The notable difference between the two plots is that the width of the
main beam of the SAR array is half that of the standard linear array. This leads
to one of the standard statements in SAR books that a SAR has twice the
resolution of a standard linear array. In fact, this is not quite true. If we were
to consider the two-way antenna pattern of a standard linear array we would
find that its beamwidth lies between the one-way beamwidth of a standard
linear array and the beamwidth of a SAR array. The reason that the two-way
beamwidth of a standard linear array is not equal to the beamwidth of a SAR
array has to do with the interaction between “elements” in the two arrays. In a
standard linear array each receive element receives returns from all of the
elements of the transmit array: In the SAR array, each receive “element” only
receives returns from itself.

  In our initial discussions we will be concerned only with cross-range imaging and can thus use a CW
signal. We will consider a pulsed (LFM) signal when we add the second dimension.

   Figure 8 – Normalized Radiation Patterns for a Standard Linear Array (top
                      plot) and a SAR Array (bottom plot)

      If we adapt the equations on page 1 we have, for the SAR array, that
      B   2L .                                                      (8)

If we combine this with the equation for cross-range distance we get, again for
the SAR array,
       y  R 2 L                                                     (9)
which is also termed the cross-range resolution of the SAR. This equation
indicates that the cross-range resolution of a SAR can be made arbitrarily small
(fine) by increasing the length of the SAR array. In theory, this is true for a
spotlight SAR. In the case of strip map SAR, the size of the actual antenna on
the SAR aircraft (the “element” of the SAR array) is the theoretical limiting
factor on resolution. In either case, there are several other factors related to
phase coherency that place further limits on the cross-range resolution.

       At this point we want to discuss the basic differences between spotlight
and strip map SAR and show how the actual antenna limits resolution for the
strip map SAR.
       Figures 9 and 10 contain illustrations of the geometry associated with
strip map and spotlight SAR, respectively. With strip map SAR, the actual
antenna remains pointed at the same angle while the aircraft flies past the area
being imaged. This angle is shown as 90° in Figure 9 but can be any angle. For
spotlight SAR, the actual antenna is steered to constantly point towards the
area being imaged. The term strip map derives from the fact that this type of
SAR can continually map strips of the ground as the aircraft flies by. The term
spotlight derives from the fact that the actual antenna constantly illuminates,
or spotlights, the region being imaged. A spotlight SAR must map a strip of
ground in segments.

                      Figure 9 – Strip Map SAR Geometry

                      Figure 10 – Spotlight SAR Geometry

       As might be deduced from Figure 9 a limitation of the strip map SAR
geometry is that the length of the synthetic array is limited by the fact that the
region imaged must remain in the actual antenna beam as the aircraft flies by
it. For the case of spotlight SAR the antenna is always pointed at the region
being imaged so that the length of the synthetic array can, in theory, be as large
as desired. In practice, the length of the synthetic array for the spotlight SAR is
limited by other factors such as range cell size. Since the cross range
resolution of a SAR is related to the length of the synthetic array, spotlight
SARs can usually attain finer cross range resolution than strip map SARs
      The theoretical limit on cross range resolution for a strip map SAR can
be deduced with the help of Figure 11. As illustrated in this figure, and
discussed above, the point to be imaged must be in the actual antenna beam
from the beginning to the end of the aircraft motion. The cross range span of
the main beam of the actual antenna is
       L  rn ANT                                                      (10)

where rn is the perpendicular range from the aircraft flight path to the point
being imaged. From the geometry of Figure 11 it clear that the point being
imaged will remain in the main beam of the actual antenna as the aircraft
traverses a distance of L . Thus the length of the synthetic array is L .

                     Figure 11 – Resolution Limit for Strip Map SAR

       Using the second equation on page 1, we can write the beamwidth of the
actual antenna as
        ANT   LANT                                                   (11)

where LANT is the horizontal width of the actual antenna. If we substitute
Equation 11 into Equation 10 we get
       L  rn  LANT                                                    (12)

which we can combine with Equation 9 to yield

                     rn 
        y                  LANT 2 .                                  (13)
                 2rn  LANT
Thus, the finest cross range resolution one can expect from a strip map SAR is
half of the horizontal width of the actual antenna. This cross range resolution
only applies to the case where a point is being imaged. The resolution for a
finite sized area will be slightly worse.
      Figure 12 illustrates a case where the width of the region to be imaged is
w . It can be observed from this figure that L '  L  w . From this we conclude
that the modified cross range resolution is
                rn     y
        y'                 y.                                      (14)
                2L ' 1  w L

In practice the term w L will be small so that  y '   y . As an example of this,
lets us consider the earlier example where we had a synthetic antenna length of
 L  600 m . From Equation 9, the resulting resolution for a point target is, in
theory  y  0.5 m . Suppose we wanted to image an area with a width of 50 m.
For this case we would need to shorten the synthetic array to
 L '  L  w  550 m . As a result, from Equation 14, the resolution would be
0.546 m instead of 0.5 m.

      Figure 12 – Effect of Finite Area Width on Strip Map SAR resolution


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