1.0 STRETCH PROCESSING
1.1 INTRODUCTION AND BACKGROUND
Stretch processing is a way of processing large bandwidth waveforms
using narrow band techniques. For our present purposes we want to look at
stretch processing as applied to LFM waveforms. It turns out that the concepts
of stretch processing appear in other applications such as FMCW radar and, as
we will see later, SAR processing.
We consider a transmit waveform of the form
2
 t 
v  t   e j t rect  
 T 
(1)
where
1
rect  x   
0
x 1 2
x 1 2
,
(2)
 is the LFM slope and  T is the pulse width. The instantaneous phase of v  t 
is
  t    t 2
(3)
and the instantaneous frequency is
f  t   21 d  t  dt  t .
(4)
Over the duration of the pulse, f  t  varies from  T 2 to  T 2 . Thus, the
bandwidth of the LFM signal, v  t  , is
B   T .
(5)
We can also determine the bandwidth of v  t  by finding and plotting its
Fourier transform. Specifically,
Vf 

 v t  e
 j 2 ft
dt


1
2
e
 j f 2 
F


2

 f   T 2    F 
2

 f   T 2  
(6)
where
F  x   C  x   jS  x 
(7)
and C  x  and S  x  are the cosine and sine Fresnel integrals, respectively,
defined by
1
x
C  x    cos  2 t 2  dt
(8)
0
and
x
S  x    sin  2 t 2  dt .
(9)
0
A normalized plot of V  f

for an LFM bandwidth of B  500 MHz and a pulse
width of  T  100 s is shown in Figure 1. It will be noted that, indeed, the
bandwidth is 500 MHz.
Figure 1 – Spectrum of an LFM pulse with B=500 MHz and PW=100 µs
If we were to process the LFM pulse using a matched filter, the impulse
response of the matched filter would be
2
 t 
h  t   v  t   e j t rect  
 T 
(10)
where we have made use of the fact that rect  x  is an even function.
The form of h  t  means that the signal processor (i.e. matched filter)
would need to have a bandwidth of B   T . Herein lies the problem: large
bandwidth signal processors are still difficult and very costly to build. Two
methods of building LFM matched filters (LFM pulse compressors, LFM signal
processors) are SAW (surface acoustic wave) devices and digital signal
2
processors. According to the Skolnik Radar Handbook1 (page 10.11) one can
build SAW compressors for bandwidths up to 1 GHz. However, I have not
heard of hardware implementations with such devices. I would expect that the
upper limit on bandwidth for practical SAW LFM compressors is in the 10s, or
possibly low 100s, of MHz.
The bandwidth of digital signal processors is usually limited by the
sample rate of the analog-to-digital converters (ADCs) needed to convert the
analog signal to a digital signal. I think that the current limit on ADC rates is
300 MHz or so. If an upper limit on ADC sample rate is 300 MHz, then the
maximum bandwidth of a LFM signal processor would also be 300 MHz
(assuming complex signals and processors).
Stretch processing relieves the signal processor bandwidth problem by
giving up all-range processing to obtain a narrow-band signal processor. If we
were to use a matched filter we could look for targets over the entire waveform
pulse repetition interval (PRI). With stretch processing we are limited to a range
extent that is usually smaller than an uncompressed pulse width. Thus, we
couldn’t use stretch processing for search because search requires looking for
targets over a large range extent, usually many pulse widths long. We could
use stretch processing for track because we already know range fairly well but
want a more accurate measurement of it. We must point out that, in general,
wide bandwidth waveforms, and thus the need for stretch processing, is
“overkill” for tracking. Generally speaking, bandwidths of 1s to 10s of MHz are
sufficient for tracking
One of the most common uses of wide bandwidth waveforms, and stretch
processing, is in discrimination, where we need to distinguish individual
scatterers on a target. Another use we will look at is in SAR (synthetic aperture
radar). Here we only try to map a small range extent of the ground but want
very good range resolution to distinguish the individual scatterers that
constitute the scene.
In the above discussion, we have focused on the signal processor and
have argued, without proof at this point, that we can use stretch processing to
ease the bandwidth requirements on a signal processor used to compress wide
bandwidth waveforms. Stretch processing does not relieve the bandwidth
requirements on the rest of the radar. Specifically, the transmitter must be
capable of generating and amplifying the wide bandwidth signal, the antenna
must be capable of radiating the transmit signal and capturing the return
signal, and the receiver must be capable of heterodyning and amplifying the
wide bandwidth signal. This places stringent requirements on the transmitter,
antenna and receiver, but current technology has advanced to be point of being
able to cope with the requirements.
1.2 STRETCH PROCESSOR CONFIGURATION
Figure 2 contains a functional block diagram of a stretch signal
processor. It consists of a mixer, a LFM generator, timing circuitry and a
M. I. Skolnik, Radar Handbook, Second Edition, McGraw-Hill, New York, 1990, ISBN 0-07057913-X
1
3
spectrum analyzer. If the transmit signal is as given in Equation (1) the
normalized signal returned from a point scatterer at a range delay of  R is
2
 t  R 
r  t   PS v  t   R   PS e j t  R  rect 

 T 
where
(11)
PS is a scaling factor that we will use when we address signal-to-noise ratio
(SNR). PS is the peak signal power at the matched filter output and comes from the radar
range equation.
Figure 2 – Stretch Processor
The normalized heterodyne signal generated by the LFM generator is
2
 t  M
hs  t   e j t  M  rect 
 h

.

(12)
In the above  M is the range delay to which the stretch processor is “matched”
and is usually close to  R . Actually, usually is not a correct word. A more
precise statement is that  M must be close to the  R of the scatterers that we
wish to resolve.  h is the duration of the heterodyne signal and, as we will
show, satisfies  h   T .
Notional sketches of r  t  and hs  t  are shown in Figure 3. The
horizontal axis is time and the vertical axis is frequency. The frequency of each
signal is shown only over the time that the signal itself is not zero. Since r  t 
and hs  t  are LFM signals, we note that their frequencies increase linearly over
their respective durations. Furthermore, by design, both frequency vs. time
plots have the same slope of  . The top plot corresponds to the case where the
target range delay,  R , is greater than  M and the lower plot corresponds to the
case where the range delay is less than  M . It will be noted that when  R   M
4
the frequency of hs  t  is greater that the frequency of r  t  . When  R   M the
frequency of hs  t  is less that the frequency of r  t  . Further, the size of the
frequency difference between r  t  and hs  t  depends upon the difference
between  R and  M .
Figure 3 – Sketches of r  t  and hs  t 
Figure 3 also tells us how to set the value of  h , the duration of the
heterodyne signal. Specifically, we want to set  h so that r  t  is completely
contained within hs  t  for all expected values of  R relative to  M . From the
bottom plot of Figure 3 we conclude that we want
 RMIN   T 2   M   h 2 .
(13)
5
From the top plot we want
 RMAX   T 2   M   h 2 .
(14)
This leads to the requirement on  h that it satisfy
 h   R   T
(15)
where
 R   RMAX   RMIN
(16)
is the range delay extent over which we want to use stretch processing. If  h
satisfies the above constraint and
 M   RMIN   R   RMAX   M
(17)
then hs  t  will completely overlap r  t  and the stretch processor will offer
almost the same SNR performance as a matched filter. If the various timing
parameters are such that hs  t  does not completely overlap r  t  , the stretch
processor will experience a SNR loss proportional to the extent of r  t  that
doesn’t lie within the extent of hs  t  .
1.3 STRETCH PROCESSOR OPERATION
Given that r  t  and hs  t  satisfy the above requirements, we can write
the output of the mixer as
2
 t   M   j t  R 2
 t  R 
vo  t   hs  t  r   t   PS e j t  M  rect 
rect 
e

 T 
 h 
(18)
or
vo  t   PS e

2
j  M
 R2
e j 2 
R  M
t
 t  R 
rect 
.

 T 
(19)
The first exponential term of vo  t  is simply a phase term. However, the second
exponential term tells us that the output of the mixer is a constant frequency
signal with a frequency that depends upon the difference between the target
range delay,  R , and the range delay to which the stretch processor is tuned,
 M . Thus, if we can determine the frequency of the signal out of the mixer we
can determine the target range. Specifically, since
fm    R  M 
(20)
we get that
 R  fm    M .
(21)
6
The spectrum analyzer of Figure 2 is used to determine f m . Ideally, the
spectrum analyzer computes the Fourier transform of vo  t  . Thus, we can
write
Vo  f  

 j 2 ft
j
 vo t  e dt  PS e


e

j 2 f mt
 t   R   j 2 ft
rect 
dt
e
 T 
(22)
or
Vo  f    T PS e j e
j 2  f m  f  R
sinc   f  f m  T 
(23)
where
    M2   R2  .
The information of concern is contained in Vo  f
(24)
 , a normalized plot of
which is contained in Figure 4. As we would expect, the sinc function is
centered at f m and has a nominal width of 1  T . Thus we can measure f m but
not with perfect accuracy. This is consistent with the result we would get with
a matched filter. That is, the range measurement accuracy is related to the
width of the main lobe of the output of a matched filter. For an LFM signal with
a bandwidth of B the nominal width of the main lobe is 1 B .
Figure 4 – Plot of Vo  f  f m  for B=500 MHz and PW=100 µs
We would now like to examine the range resolution of the stretch
processor. Since the nominal width of the sinc function is 1  T we normally say
that the frequency resolution of at the output of the spectrum analyzer is also
1  T . Suppose we have a target that is at a range of  R1 and a second target at
7
a range of  R 2   R1 . The mixer output frequency associated with the two targets
will be
fm1    R1  M 
(25)
and
f m2    R 2  M  .
(26)
Suppose further that  R1 and  R 2 are such that
f m  f m 2  f m1  1  T .
(27)
That is, the frequencies are separated by a resolution cell of the stretch
processor. With this we can write
fm  fm2  fm1  1  T    R 2  M     R1  M 
(28)
or
 R res   R 2   R1  1  T  1 B
(29)
or that the stretch processor has the same range resolution as a matched filter.
Recall that with LFM we can use an amplitude taper, implemented by a
filter at the input or output of the matched filter, to reduce the range sidelobes
at the matched filter output. We can apply a similar taper to a stretch
processor by applying an amplitude taper to vo  t  before sending it to the
spectrum analyzer. We will look at this further when we discuss a specific
implementation of the spectrum analyzer.
1.4 STRETCH PROCESSOR SNR
At this point we want to compare the SNR at the output of a matched
filter to the SNR at the output of a stretch processor. Since neither processor
includes nonlinearities we can invoke superposition and treat the signal and
noise separately.
1.4.1 Matched Filter
For the matched filter case we can write signal voltage at the output of
the matched filter as
vsm  mm , f mm   PS e
jMF

 v t  v t    e

mm
j 2 f mmt
dt
(30)

where  mm and f mm are the range delay and Doppler frequency mismatch,
respectively, between the target return and matched filter. v  t  is given by
Equation (1). We are interested in the power out of the matched filter at
matched range and Doppler. That is, we want
8

Psm  vsm  0,0 vsm
 0,0 .
(31)
Substituting Equation (30) into Equation (31) yields
Psm 
PS e
jMF
2

 v t  v t  dt

 Ps T2 .
(32)

The noise voltage at the output of the matched filter is given by

vnm  t  
 h  x  n  t  x  dx
(33)

where n  t  is zero-mean, wide-sense stationary white noise with
E n  t  n    N o  t    .
(34)
No  kT0 FN is the noise power spectral density and   x  is the Dirac delta.
Since n  t  is a random process so is vnm  t  . Thus, the average noise power out
of the matched filter is given by

Pnm  E vnm  t 
2

2
 

 E   h  x  n  t  x  dx   N o T
 

(35)
where we have made use of Equations (33) and (10). With this, we get the SNR
at the matched filter output as
SNRm 
Psm Ps T

Pnm
No
(36)
which we recognize from radar range equation theory.
1.4.2 Stretch Processor
For the stretch processor we are interested in the signal power at the
target range delay,  R . Thus, we are interested in the output of the spectrum
analyzer at f  f m (This assumes that the stretch processor is matched to  R ,
i.e.,  M   R ). With this we get, using Equation (23),
Pss  Vo  f m   Ps T2 .
2
(37)
If the noise into the mixer part of the stretch processor is n  t  the noise
out of the mixer is
vnM  t   n  t  hs  t  .
(38)
9
Recall that the signal power was computed at the spectrum analyzer output
where f  f m . The noise signal at this spectrum analyzer output is
Vns  f m  

 n t  h t  e

 j 2 f m t
s
dt
(39)

and the average power at the output is

Pns  E Vns  f m 
2
 N 
0 h
(40)
where we have made use of Equations (34) and (12).
The SNR at the output of the stretch processor is
SNRs 
Pss Ps T2
.

Pns N 0 h
(41)
If we combine Equations (36) and (41) we get
SNRs  T
 .
SNRm  h
(42)
Thus, the stretch processor encounters a SNR loss of  h  T relative to the
matched filter. This means that we should be careful about using stretch
processing for range extents that are significantly longer of the transmit pulse
width. At first inspection it appears as if stretch processing could offer better
SNR than a matched filter, which would contradict the fact that the matched
filter maximizes SNR. This apparent contradiction is resolved by the stretch
processor constraint imposed by Eqution (15). Specifically,  h   R   T . The
constraint if Equation (42) also demonstrates another reason why stretch
processing should not be used in a search function: it would be too lossy.
1.5 STRETCH PROCESSOR IMPLEMENTATION
We next want to turn our attention to practical implementation issues.
The mixer, timing and heterodyne generation are reasonably straight forward.
We want to address how to implement the spectrum analyzer. The most
obvious method of implementing the spectrum analyzer is to use an FFT. To do
so, we need to determine the required ADC (analog-to-digital converter) sample
rate and the number of points to use in the FFT. To determine the ADC rate we
need to know the expected frequency limits of the signal out of the mixer2.
We will assume base-band processing in these discussions. In practice the mixer output will be
at some intermediate frequency (IF). The signal could be brought to base-band using a
synchronous detector or, as in some modern radars, by using IF sampling (i.e. a digital receiver).
In either case, the effective ADC rate (the sample rate of the complex, digital base-band signal)
will be as derived here.
2
10
If  RMIN   M and  RMAX   M are the minimum and maximum range delays,
relative to  M , over which stretch processing is performed then the
corresponding minimum and maximum frequencies out of the mixer are
fmMIN    RMIN  M 
(43)
and
f mMAX    RMAX  M  .
(44)
Thus, the expected range of frequencies out of the mixer is
fmMAX  fmMAX  f mMIN    RMAX   RMIN    R .
(45)
Thus the ADC sample rate should be at least f mMAX .
The FFT will need to operate on data samples taken between  RMIN   T 2
and  RMAX   T 2 or over a time window of at least
 RMAX   RMIN   T   R   T   h .
(46)
The total number of data samples processed by the FFT will be
N samp  f mMAX  h .
(47)
This means that the FFT length will need to be some power of 2 that is greater
than N samp .
As an example of the above calculations we consider the following
parameters.
-
 T  100  s
-
B  500 MHz
-
Stretch processing performed over 1500 m.
With this we get that
 RMAX   RMIN  10 s
(48)
and
 h   RMAX   RMIN   T  110 s .
(49)
To compute f mMAX we first need to compute  as
  B  T  5 MHz/s .
(50)
With this we get
fmMAX    RMAX  RMIN   50 MHz .
(51)
Thus, the minimum required ADC sample rate is 50 MHz. The number of
samples to be processed by the FFT is
11
N samp  f mMAX  h  5500 .
(52)
This means that we would want to use a 8192 point FFT. The most logical
method of getting to 8192 samples would be to increase the size of the range
window. This would cause both f mMAX and  h to increase.
If we continue the calculations we find that the time extent of the
heterodyne window is
 h   R   T  110 s .
(53)
The SNR loss associated with the use of stretch processing over a matched filter
is  h  T  110 100 or about 0.4 dB.
1.6 DOPPLER EFFECTS
We now want to examine the effects of Doppler frequency on the output
of the stretch processor. Since we have established the equivalency between
the stretch processor output and the output of a matched filter, we will
approach the discussion from the perspective of matched filter theory. We start
by developing extending the definition of v  t  from Equation (1) to include a
carrier term. We then specifically look at how range and rate affects the
returned signal, r  t  . After this we examine the matched filter response to r  t 
from the specific perspectives of range resolution degradation due to Doppler
frequency and range error due to Doppler frequency.
1.6.1 Expanded Transmit and Receive Signal Models
We extend the previous definition of the transmitted LFM pulse to
include the carrier term. Thus we write
2
 t 
v  t   e j 2 fct e j t rect  
 T 
(54)
where the first exponential is the carrier term and f c is the carrier frequency.
The signal returned from the target is
r  t   PS v  t   R  t    PS e
j 2 f c  t  R  t  
e
j  t  R  t  
2
 t  R t  
rect 
.
 T

(55)
It will be noted that the range delay,  R  t  , is now shown as a function of time
to account for the fact that range changes with time because the range-rate is
not zero. We will assume that the target range-rate is a constant. With this, we
can write
 R t  
2R t 
c

2 R0 2 R

t
c
c
(56)
12
Where R0 is the range at t  0 (the center of the transmit pulse in this case) and
R is the range-rate.
Substituting Equation (56) into Equation (55) results in
r  t   PS e

j 2 f c t  2 Ro c  2 R c t

e

j t  2 Ro c  2 R c t

2
 t  2 Ro c   2 R c  t 

rect 


T


 t  2 Ro c   2 R c  t 

 PS e j1 t  e j2 t  rect 


T


(57)
where 1  t  is a phase term that we associate with Doppler effects due to the
interaction of the target range-rate with the carrier and 2  t  is a phase term
that we associate with the interaction of range-rate with the LFM modulation. If
we expand 1  t  we get
1  t   2 f ct  4 R0 c  2  2 R   t  2 f ct  4 R0 c  2 f d t
(58)
where the first term on the right is the carrier term, the second is a phase shift
associated with the initial target position and the third term is the Doppler
frequency term. This term ( 1  t  ) is the same as we developed in EE619 when
we discussed Doppler frequency.
The second phase term can be written as


2  t    t  2R0 c   2R c  t .
2
(59)
In this case we want to examine the frequency or
f FM  t  
 2 R    2 R  2 R0 
1 d 2  t 
  1 

  1 
t 
2 dt
c 
c 
c 

2
 2R 
 2 R   2 R0 
  1 
 t   1 

   r t  f0
c
c
c






.
(60)
It will be noted that the LFM slope of the received signal,  r , is slightly different
from the LFM slope,  , of the transmit signal. As we will see, this slight
difference can degrade the range resolution of the LFM waveforms in some
cases.
It turns out that the increase in slope of the received LFM signal is
caused by a slight shortening of the pulse as it is “reflected” by the target. To
see this we consider a specific example. Suppose we have a 1 ms pulse and a
target moving at 7500 m/s. Let t 0 be the time that the leading edge of the
pulse reaches the target. During the time that the pulse is interacting with the
target, the target moves about (7500 m/s)×(0.001 s) or 7.5 m. This translates
to an effective round-trip time delay of 2×7.5/c or 50 ns. This means that the
13
length of the pulse returned to the radar is shorter than the transmit pulse by
50 ns. Since the frequency still varies the same amount over the duration of
the pulse, the LFM slope must increase.
1.6.2 Effect of Doppler Frequency on Range Resolution
To quantify the effect of the change in received waveform LFM slope on
range resolution we will consider specific examples. We will assume that the
radar uses a matched filter to perform pulse compression. We further assume
that the matched filter is matched to the transmit waveform plus some
frequency offset, f M , to account for the target Doppler frequency. As we did
earlier, we will shift our time reference so that the center of the received pulse is
at t  0 . Thus, we can write the normalized received signal as
2
 t 
r  t   e j 2 fd t e jr t rect 
.
 PWr 
(61)
The first exponential term is the Doppler term discussed above (see the
discussion related to Equation (58) – we have omitted the carrier frequency
term since we assume that it has been removed by a heterodyning process in
the receiver). We have temporarily ignored the term f 0 in Equation (60). We
will address this later.
We can write the matched filter impulse response as
2
 t
h  t   e j 2 fM t e j t rect 
 T

.

(62)
The response of the matched filter to the received signal is the convolution of
r  t  and h  t  . That is,

vo  t   r  t   h  t  
 r   h t    d  .
(63)

Substituting for r  t  and h  t  results in
vo  t  

   j 2 fM t    j t  2
 t  
j 2 f d  j r  2
e
e
rect
e
rect 
 d .

e

 T 
 PWr 
(64)
After considerable manipulation it can be shown that we can write vo  t  as
  2
K F

vo  t   


U  f

  F  2 
0
 L  f


U L
(65)
UL
14
In Equation (65):

F is the Fresnel integral

   r  

f  f d  f M

U  min  t   T 2, PWr 2

L  max  t   T 2,  PWr 2

K is a complex constant that we will normalize away .
Equation (65) applies only to the case where the received LFM slope and rect()
function widths are different (the mismatched case). If the LFM slope and rect()
function widths are the same, vo  t  can be derived from the ambiguity
function of v  t  and is
 t 
vo  t     t , f   K1  T  t sinc  f   t   T  t  rect 
.
 2 T 


(66)
In Equation (66), K1 is a complex constant that we will normalize away.
To see the effect of range-rate on the matched filter response we consider
two examples. In both cases we consider the LFM waveform of previous
examples. Specifically, we consider a waveform with a bandwidth of 500 MHz
and a pulse width of 100 µs. We assume further that the matched filter is
matched to the target Doppler. That is, f d  f M . For the first case we consider
a typical aircraft range-rate of -150 m/s and for the second case we consider a
ballistic missile with a (extreme) range-rate of -7500 m/s. Plots of the matched
filter outputs for the two cases are shown in Figure 5 and Figure 6. Each plot
contains a curve where we ignore the effects of range-rate on the LFM slope (the
red curve) and another curve where the range-rate effects are included (the blue
curve).
For the aircraft velocity case, it will be noted that the difference in LFM
slope caused by the range-rate does not have a significant effect on the output
of the matched filter. However, for the ballistic missile case, the difference in
LFM slopes causes a significant degradation of range resolution. This, in turn,
could cause problems in isolating closely spaced scatterers. To eliminate this
effect, the LFM slope of the matched filter should be changed to match the
expected LFM slope of the received waveform. Since changing the LFM slope of
the matched filter could be difficult, the alternative would be to change the LFM
slope of the transmit signal so that the LFM of the received signal matches that
of the matched filter.
15
Figure 5 – Matched Filter Response – Target Range-Rate = -150 m/s
Figure 6 – Matched Filter Response – Target Range-Rate = -7500 m/s
Since we expect that the stretch processor behavior will be similar to the
matched filter behavior, LFM slope mismatch should have the same effect on
the output of the stretch processor. That is, for large range-rates, and long,
large bandwidth waveforms, LFM slope mismatch will cause a degradation of
range resolution.
16
1.6.3 Effect of Doppler Frequency Mismatch on Range Error
We next want to examine the effect of Doppler mismatch between the
received signal and the matched filter. We consider the aircraft target example
from above. Figure 7 contains a plot of the matched filter output for the case
where the matched filter is matched to the target Doppler (the red curve) and
the case where the matched filter is matched to zero Doppler (the blue curve).
It will be noted that the mismatched Doppler case has a range error of 0.3 m,
which is the range resolution of the waveform. To understand the cause of this
range error it will be helpful to look at the matched filter response in terms of
the ambiguity function of the transmit signal. This ambiguity function is given
by
 t 

 2 T 
 LFM  t , f    T  t sinc   f   t   T  t   rect 
(71)
Figure 7 – Effects of Doppler Mismatch on Matched Filter Response
In Equation (71) we note that if f  f d , the peak of  LFM  t , f
td  fd   f d T 1 B  .

occurs at
(72)
From Equation (72) we note that a Doppler shift of f d  1  T relative to matched
Doppler will cause a range error equal to the range resolution of the LFM
waveform. Said another way, a Doppler mismatch equal to the reciprocal of the
uncompressed pulse width will cause a range error of one range resolution cell.
In the specific example above, if we assume the radar is operating at X-band
with   0.03 then a range-rate of -150 m/s causes a Doppler frequency of 10
KHz. It turns out that the uncompressed pulse width of our example was 100
µs so that 1  T  10 KHz . Thus, by the above, we expect that the peak of the
17
ambiguity function (matched filter response) will be at one range-resolution cell
instead of zero. Given that our waveform bandwidth was 500 MHz, the range
resolution of the waveform is 2 ns, or 0.3 m, which is where the peak is located!
As with the LFM slope, we expect that the response of the stretch
processor to Doppler mismatch will be the same as the matched filter. Suppose
the return signal, r  t  , into the stretch processor has a Doppler frequency of
f d . This will mean that the return signal into the mixer (see Figure 1) will be
rd  t   e j 2 fd t r  t  .
(73)
If we repeat the math of Section 1.3 using rd  t  in place of r  t  , the signal out
of the mixer will be
vod  t   e j 2 fd t vo  t  .
(74)
In other words, the frequency out of the mixer will be shifted by the Doppler
frequency (albeit with a negative sign). This, in turn, will cause the peak of the
spectrum analyzer output shift by  f d . That is, the output of the spectrum
analyzer will be Vod  f   Vo  f  f d  . If f d  1  T then the peak will shift by
 1  T . From the discussions of Section 1.3, a frequency deviation in the
spectrum analyzer output of 1  T corresponds to a range shift of 1 B , or one
range resolution cell. From this we observe that the response of the stretch
processor to Doppler is the same as for a matched filter. In other words, a
Doppler shift of 1  T cause the range to be in error by one range resolution cell.
18