2.5 SAR SIGNAL CHARACTERIZATION
In order to formulate a SAR processor we first need to characterize the
signal that the SAR processor will operate upon. Although our previous
discussions treated SAR cross-range imaging as an antenna problem, for the
rest of the development we will cast the problem in the Doppler domain. Past
experience in this course indicates that the Doppler formulation is easier to
understand than the antenna formulation.
2.5.1 Derivation of the SAR Signal
As we have done so far, we will initially consider only the cross-range
problem. We will later extend the discussions to the down-range and crossrange problem. Since we are considering the cross-range problem we start by
considering a CW transmit signal of the form
vT  t   e j 2 fct .
(24)
If we consider the geometry of Figure 13 the signal returned from the
“nth” scatterer located at xn , yn is
vnRF  t  
PSn
P j 2 t 2 r t  c
vT  t  2rr  t  c   2 Sn e  r 
r t 
rn  t 
2
n
(25)
where
rn  t   xn2   yn  d  t  
2
3
(26)
and d  t  is the y position of the aircraft at some time t . If we assume that the
aircraft is flying at a constant velocity of V and t  0 occurs at y  0 of Figure
13 we get
d  t   Vt .
(27)
We assume that the total time for the aircraft to travel a distance of L is
TL and that the aircraft starts at  L 2 when t  TL 2 . With this we get
L  TLV .
(28)
We note that the area to be imaged has a cross-range width of w and a down
range length of l . The region is centered in cross-range at y  0 and in downrange at x  ro .
In Equation (25), PSn is the normalized signal power associated with the
nth scatterer. It is determined from the radar range equation, without the R 4
term (since it is incorporated into Equation (25)).
For now we are assuming that the aircraft (platform) is at an altitude of zero. The extension to
non-zero altitude is straightforward.
3
14
If we assume that w , l and L are small relative to ro we can replace
rn2  t  in the denominator of Equation (25) by ro2 and write
vnRF  t  
PSn j 2 fct  j 4 rn t  
e
e
ro2
(29)
where   c f c is the wavelength of the transmit signal.
Figure 13 – Geometry Used to Develop Signal Representation – Cross-range
Imaging
Since the information needed to form the image is in the second
exponential term we will eliminate the first exponential term by heterodyning
(which is done in the actual radar). We also eliminate the ro2 term through
normalization to yield the baseband signal
vn  t   ro2e j 2 fc r vnRF  t   PSn e  j 4 rn t   .
(30)
If we have N s scatterers in the image region the resulting baseband signal
would be
Ns
Ns
n 1
n 1
v  t    vn  t    PSn e j 4 rn t  
(31)
15
2.5.2 Examination of the Phase of the SAR Signal
Since the information we seek is in the phase of vn  t  we want to
examine it. To proceed we need to examine rn  t  . We can write
rn  t   xn2  yn2  2 ynVt  V 2t 2  rn2  2 ynVt  V 2t 2  rn 1  2 ynVt rn2  Vt rn  . (32)
2
We note that rn  ro , yn
roVt  L 2, L 2 . This means that the
ro and Vt
second and third terms of the last square root are small relative to 1. This, in
turn, allows us to write


rn  t   rn 1  ynVt rn2  12 Vt rn   rn  ynVt rn  12 V 2t 2 rn .
2
(33)
If we substitute this into Equation (30 we get
vn  t   PSn e  j 4 rn  e


2
2
j 4  ynV  rn t  j 2 V  rn t
e
.
(34)
2.5.2.1 Linear Phase, or Constant Frequency, Term
The first exponential is a phase caused by range delay to the scatterer
that we must live with. The second term is a linear phase term or a term that
we associate with frequency. In fact, the frequency represented by this term is
f yn  2 ynV  rn .
(35)
This tells us that vn  t  has a constant frequency term that depends upon the
scatterer cross range position, yn .
f yn also depends upon the aircraft velocity,
V , and the radar wavelength,  . However both of these are known (and fixed).
Finally, f yn also depends upon rn . If we assume that all of the scatterers are at
the same xn  ro (which we can do since we are only concerned with the crossrange problem) and note that yn
ro we get the previous assertion that
rn  xn2  yn2  ro .
(36)
This discussion tells us that, if we can determine f yn , we can determine yn from
yn 
 f yn ro
2V
.
(37)
2.5.2.2 Quadratic Phase, or LFM, Term
The third exponential of Equation (34) is a quadratic phase, or linear
frequency modulation, term and causes problems. We can write the quadratic
phase as
16
 V2  2
t .
  rn 
Q  t   2 
(38)
With the previous assumption that rn  ro , Q  t  is approximately the same for
all scatterers. This means that we can remove it by a mixing or heterodyning
process.4 If we do this we will be left with only the magnitude, constant phase
term and the yn -dependent frequency term. This is what we want.
2.5.3 Extracting the Cross-Range Information
Once we remove the quadrtic phase we have
vIn  t   e


j 2 V 2  ro t 2
vn  t   PSn e  j 4 rn  e
j 2 f yn t
(39)
for a single scatterer. For the more general case of N s scatterers we have
vI  t   e

 v t 
   PSn e j 4 r  e j 2 f t .
Ns
j 2 V 2  ro t 2
n
(40)
yn
n 1
The forms of Equations (39) and (40) tell us that we can extract the information
we want by taking the Fourier transform of vIn  t  or, more generally, vI  t  .
From our experience with Fourier transforms, this will give us a response that
peaks at the frequencies f yn . The heights of the peaks will be proportional to
PSn . If we use Equation (37) to plot this as intensity ( PSn ) vs. yn we have a
one dimensional image.5
To verify the above, and begin to understand SAR resolution properties,
we will find the Fourier transform of vIn  t  . Indeed
VIn  f  

 v t  e
 j 2 ft
In
dt .
(41)

To perform the integration we need to realize that vn  t  and thus vIn  t  is
measured only over t   TL 2, TL 2, . Thus, we assume that vIn  t  is zero
outside of these limits and write
VIn  f  
TL 2

vIn  t  e
 j 2 ft
TL 2
 PSn e  j 4 rn  sinc
dt 
TL 2

PSn e  j 4 rn  e
TL 2
 f  f T 
yn
j 2 f yn t
dt
(42)
L
where
Note that this is similar to stretch processing wherein we remove the quadratic phase in the
mixer.
5 Again, note the similarity to stretch processing
4
17
sinc  x   sin  x   x  .
(43)
Figure 14 contains a plot VIn  f

vs.
 f  f T
yn
L
. It will be noted that the
response has a peak at f  f yn  0 , or at f  f yn and that the peak has a height
of VIn  f

PSn . The width of the peak is 1 TL which means that the SAR
image will have a resolution of
 f  1 TL .
(44)
Figure 14 – Plot of VIn  f

PSn vs.
 f  f T
yn
L
If we change the horizontal axis to y using the relation (see Equation
(37))
y
 fro
(45)
2V
we get the plot of VIn  y  vs.  y  yn  of Figure 15. This plot has a peak at
y  yn with a height of VIn  y   PSn . From Equation (44), if the resolution of
Figure 14 is  f  1 TL the resolution of Figure 15 is
y
 ro
2V
f 
 ro 1
2 TLV

 ro
2L
(46)
since L  TLV . This is the same as the resolution we obtained from the linear
array approach (see Equation (18)).
18
PSn vs.  y  yn   y
Figure 15 – Plot of VIn  y 
2.6 PRACTICAL IMPLEMENTATION
In the previous section we found that the processing methodology we
must use to form an image is to form a Fourier transform of vI  t  (or vIn  t  ).
However, this approach makes the tacit assumption that vI  t  is a continuous
function of time. Thinking ahead to when we will consider both cross-range
and down-range imaging we realize that the SAR will transmit a pulsed signal
rather than a CW signal. Because of this, we recognize that vI  t  will not be a
continuous-time signal but a discrete-time signal with samples spaced by the
radar PRI, T . If we also recognize that we will perform the processing with
digital signal processors (DSPs) we reinforce the realization that vI  t  is, in fact,
a discrete-time signal, vI  kT  or vI  k  .
2.6.1 A Discrete-Time Model
For a single scatterer (i.e. vIn  t  ) we get (from Equation (34))
vIn  k   PSn e  j 4 rn  e


2
j 4  ynV  rn  kT   j 2 V  rn  kT 
e
2
.
(47)
After we (digitally) remove the quadratic phase term, the signal we process to
form the image is
vIn  k   PSn e j 4 rn  e
j 4  ynV  rn Tk
.
(48)
19
Since vIn  k  (and vI  k  for multiple scatterers) is a discrete-time signal we use
the discrete-time Fourier transform (DFT) to find its Fourier transform.
Specifically we find
VIn  f  

 v k  e
k 
 j 2 fkT
.
In
(49)
As with the continuous-time Fourier transform, we limit the sum by considering
that we gather data only from  L 2 to L 2 or for t  TL 2 . If we use t  kT the
limits on k become
k  TL 2T  K L
(50)
where it is understood that we round, or truncate, TL 2T to the nearest integer.
If we use this in Equation (49), with Equation (48) and Equation (35), we
get
VIn  f   PSn e  j 4 rn
 PSn e  j 4 rn

KL

e


j 2 f yn  f Tk
k  K L


sin  K L  f  f yn  T

sin   f  f yn  T

.
(51)
We note that this is similar to Equation (42).
Figure 16 contains a plot of VIn  f



vs. f  f yn TL . As can be seen, it has
a peak at f  f yn  0 as did Figure 14. However, it also has peaks at
f  f yn  1 T . In fact, if we recall the theory associated with discrete-time
signals and the DFT we recognize that VIn  f

will have peaks at f  f yn   n T
where n is an integer. All peaks except the one corresponding to n  0 are
ambiguities and are undesirable. In terms of SAR, they result in what are
termed ghost images. We want to be sure we choose T , and the characteristics
of the SAR antenna, to avoid these ghosts since they can result in misleading
SAR images. The SAR antenna was mentioned because it acts as a spatial antialiasing filter.
20
Figure 16 – Plot of VIn  f

PSn vs.
 f  f T
yn
L
Using a Discrete-time signal
with T  0.1TL
2.6.2 Other Considerations
As we did before, we want to change the horizontal axis of Figure 16 to
cross-range distance rather than frequency. To do so we use Equation (45).
This results in the plot of VIn  y  shown in Figure 17. The ambiguities are
shown in this figure and are located at
yambig  yn  
 ro 1
2V T
.
(52)
The above tells us that we want to choose the PRI such that all scatterers lie
within ±1/2 ambiguity. That is we want to choose the PRI such that all yn
satisfy
yn 
1  ro 1
.
2 2V T
(53)
Since all scatterers of interest lie within the imaged area this says that we want
to choose the PRI such that
w
 ro 1
2V T
.
(54)
In fact we usually choose the PRI such that
w
 ro 1
2V T
(55)
21
so as to be sure that the SAR antenna beam adequately attenuates targets
outside of the imaged region. After we form the larger image, because we will be
using a larger effective w , we will truncate it to include only the region of
interest.
Figure 17 – Plot of VIn  y 
PSn vs.  y  yn   y Using a Discrete-time signal
with T  0.1TL
We can turn Equation (54) around and use it to find an upper bound on
PRI. Specifically we solve Equation (54) for T to yield
T
 ro 1
(56)
2V w
or from Equation (55)
T
 ro 1
2V w
.
(57)
If we consider an earlier example where ro  20 Km ,   0.03 m and w  50 m ,
and consider an aircraft velocity of V  50 m/s we get
T
120 ms ,
(58)
which is an easy constraint to satisfy. When we consider down-range imaging
we will impose a lower limit on T so as to satisfy unambiguous range operation.
However, that lower limit is generally well below the upper limit of Equation
(58).
We now want to summarize the above as an algorithm that we can
implement to form a cross-range image.
22
2.7 AN ALGORITHM FOR CREATING A CROSS-RANGE IMAGE

We assume we have a base-band, CW signal (see Equation (30) and
Equation (31))

We sample this signal at intervals of T and generate 2 K L  1 samples
where
 ro 1
(Equation (57))
o
T
o
K L  TL 2T (Equation (50))
o
TL  L V (Equation (28))
o
L  ro  2 y (Equation (46))
2V w
In these equations  , ro , V , w , and  y are known. The samples are
taken for kT between TL 2 and TL 2 or for k between  K L and K L .

Remove the quadratic phase by multiplying the sampled signal by
vh  k   e
V2 
2
j 2 
 kT 
  ro 
(59)
This gives vI  k  .

Compute the DFT of vI  k  . This is most easily done using an FFT. The
minimum FFT length is 2 K L  1 although we usually choose the FFT
length to be a power of 2 greater than 2 K L  1 . In “real” applications we
often adjust various SAR parameters so that 2 K L  1 is close to a power
of 2. For purposes of this class we choose an FFT length much greater (4
to 16 times) than 2 K L  1 so that the resulting frequency plot is smooth.

If LFFT is the length of the FFT, the frequency spacing between output
FFT taps is
f 
1
TLFFT
(60)
After the front and rear halves of the FFT outputs are swapped (think
fftshift) the frequencies of the taps are
f  mf   LFFT 2, LFFT 2  1 f

(61)
Transform the frequency scale to cross-range using
y
 ro
2V
f (Equation (45))
23
and plot the magnitude of the FFT output vs. y . This doesn’t produce
an image but a linear plot as shown in Figure 16.

To generate a pseudo image, create an array of zeros where the number
of columns is equal to the number of samples needed to cover the width,
w , of the image area and a comparable number of rows. If there are N row
rows, replace row N row 2 with the appropriate FFT outputs. Use the
resulting array to create the pseudo image.
2.8 EXAMPLE 1
To illustrate the above, we consider a specific example. The parameters
of this example are given in Table 1.
Table 1 – Parameters Used in SAR Example 1
Parameter
Value
Width of image area, w
50 m
Depth of image area, l
50 m
Range to image area center, ro
20 Km
SAR wavelength, 
0.03 m
Aircraft velocity, V
50 m/s
Synthetic array length, L
600 m
Number of scatterers, N s
3
Scatterer locations,  xn , yn  (m)
( ro ,0), ( ro ,20), ( ro ,-15)
Scatterer powers, PSn (w)
1, 0.25, 0.09
Given these we can compute some of the SAR parameters indicated in the
algorithm description. Specifically:
TL  L V  600 50  12 s and
T
 ro
2Vw

 0.03 20, 000   120 ms
.
 2  50  50 
(62)
(63)
We will choose a PRI of 50 ms, that is we choose
T  50 ms .
(64)
This gives
24
KL 
TL
12

 120 .
2T  2  0.05
(65)
With this the SAR starts sampling at t  6 s and samples until t  6 s . The
samples are taken every T  50 ms and a total of 2 K L  1  241 samples are
used. This means that we need, as a minimum, a 256 point FFT. In order to
produce a smooth plot we will use a 2048 point FFT.
We note that, since we chose T  50 ms the actual width of the area
included in the image is
wactual 
 ro
2VT

 0.03 20, 000   120 m .
 2  50  0.05
(66)
Once we form the image we will need to discard the FFT outputs outside of the
range of ±25 m (after the conversion from frequency to y position)
The resolution of the SAR image is
y
ro   20, 000  0.03

 0.5 m .
2L
 2  600 
(67)
This means that we ought to be able to distinguish scatterers separated by
about 1 m or greater, and maybe down to 0.5 m separation if their relative
powers and phases allow this.
Before processing the SAR signal using the previously discussed
algorithm we need to generate the SAR signal. To do so we use Equation (31)
with N s  3 . We generate 241 samples of v  t  starting at t  6 s and ending at
t  6 s in steps of T  0.05 s .
In my code I generated rn  t  , n  1, 2,3 , using Equation (26) and Equation
(27). I then used these rn  t  , along with the PSn values in Equation (30) to
compute the three vn  t  . Finally, I summed the three vn  t  to form v  t  .
Figure 18 is a linear plot of the VIn  y  for 25 m  y  25 m and Figure 19
is a pseudo image. To form the pseudo image I first created an array of zeros
that had 101 rows (which is l  y  1 ) and a number of columns equal to the
number of y values in the linear plot. I next loaded the VIn  y  values from the
linear plot in the 51st row of the array. Next I used the commands
imagesc(x, y, max(max( VIn  y  ))- VIn  y  );colormap gray
to generate the image. The use of max(max( VIn  y  ))- VIn  y  creates an image
with a white background, which is really a negative image. I did this to save on
printer ink.
25
Figure 18 – Linear plot of VIn  y  - Three Scatterers at -15, 0, 20 m
In examining Figure 18 we note that VIn  y  has three peaks at the y
positions of the scatterers. Further, the heights of the peaks are
PSn . The
image (Figure 19) shows three dots at the given scatterer positions. It will be
noted that the dots are different shades of gray, indicating different amplitudes.
To check the aforementioned resolution statement I moved the y
positions of the scatterers to -1, 0 and 1 m and reformed the linear plot and
image. These are shown in Figure 20 and Figure 21. The linear plot clearly
shows three peaks but the relative amplitudes are somewhat different than
those of Figure 18. This is due to the sidelobes of the Fourier transform
response function. The presence of the three scatterers can also be seen in the
image of Figure 21.
26
Figure 19 – Image of VIn  y  - Three Scatterers at -15, 0, 20 m
Figure 20 – Linear plot of VIn  y  - Three Scatterers at -1, 0, 1 m
27
Figure 21 – Image of VIn  y  - Three Scatterers at -1, 0, 1 m
As another interesting experiment I eliminated the quadratic phase
removal step of the SAR processing algorithm. The results are shown in Figure
22 and Figure 23 (I reverted to the original locations of the scatterers). As can
be seen, the peaks are spread and the image is blurred in the y direction. In
SAR lingo, we say that the image is not focused. In fact, the process of
removing the quadratic phase is sometimes termed focusing of the SAR image.
28
Figure 22 – Linear plot of VIn  y  - Three Scatterers at -15, 0, 20 m – Without
Quadratic Phase Removal
Figure 23 – Image of VIn  y  - Three Scatterers at -15, 0, 20 m – Without
Quadratic Phase Removal
29