Radar Systems and Applications - RRY080
Home Assignment 2
Nadeem Khan
September 24, 2022
Introduction
The SAR processing exercise aims to resolve scene reflectivity by compressing raw data. The data
to be used is given Table below:
ρmin
∆p
∆Xn
θ
D
λc
Vx
fR
VX
R
First Range Sample
Pixel size in the range
Pixel size in azimuth
Azimuth Beam-width
Antenna size in Azimuth
Center wavelength
Along track velocity
First guess
Slant Range
10446 m
4m
0.43 m
3o
1.3 m
0.0566 m
131 ms−1
2.32 m−1
To be Calculated
The given data SAR data matrix has 350 x 4096 in with rows and columns representing the
range and azimuth dimension respectively.
Synthetic Aperture
a)
Length of synthetic aperture Lf ar is given as:
θ
Lf ar = 2 ∗ ρmax × tan( )
2
To calculate ρmax , following equation has been used:
ρmax = ρmin + 350 × ∆ρ = 11846m
(1)
(2)
Excluding the first term one gets:
ρmax = 11842
(3)
Lf ar = 620.66m
(4)
Finally Lf ar came out to be:
1
b)
Number of indices for the matched filter are obtained by considering the basic distance-speed
relation:
Lf ar = vX × t
(5)
Using Lf ar from previous section and velocity from given data:
Lf ar
620.66
=
= 4.74s
Vx
131
(6)
fR = 2.32 × Vx = 303.92Hz
(7)
t=
From data given in table above:
Using fR and t, number of indices M are given as:
M = fR × t = 1440
(8)
c)
Length of aperture in range bin (row) 37 is given as:
ρ37 = ρmin + 37 × ∆ρ
(9)
Using ρmin = 10446 m and ∆ρ = 4 m:
ρ37 = 10594m
(10)
ρ37 × λc
= 555.25m
2× D
2
(11)
From this L37 is given as:
L37 =
Azimuth Bandwidth
a)
The 3-dB bandwidth (Baz ) of the range compressed data in azimuth can be calculated using a
spectrum plot in MATLAB and shown in Fig 1 below. The 3-dB azimuth bandwidth is found to
be 208 Hz.
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Figure 1: PSD of Azimuth Spectrum
b)
Using equation (12) and assuming the integration angle ϕ being equal to azimuth beam-width θ,
azimuth bandwidth is given as:
Baz =
4 × Vx
θ
× sin( ) = 242Hz
λc
2
(12)
c)
The 3-dB azimuth bandwidth is found to be 242 Hz. The difference in 34 Hz in bandwidth is due
to the noise in the real data and errors due to approximation. Moreover, in the above equation
assumption is made that integration angle ϕ is equal to azimuth beam-width θ, which doesn’t seem
to hold in this case and considering ϕ ≤ θ in case of strip-map mode, actual azimuth angle might be
different from θ. Therefore there is a difference in the result because in the strip mode case theta
is greater than phi.
Azimuth Resolution
a)
The resolution in azimuth (δaz ) calculated by using different methods are given as:
i
δaz =
ρmax × λc
= 0.54m
2Lf ar
3
(13)
ii
δaz−meas =
Vx
= 0.6298m
Baz
(14)
δaz−theor =
Vx
= 0.5413m
Baz
(15)
iii
δaz =
λc
= 0.54m
2 × θbeam
(16)
D
= 0.65m
2
(17)
iv
δaz =
b)
Synthetic aperture, azimuth bandwidth and theoretical azimuth band with give the same value for
resolution but resolution estimated through measured azimuth bandwidth and antenna size differs
from the other values. This is because of rough approximation of both antenna size and estimation
of azimuth bandwidth from measured data (as the data is quite noisy).
So calculating resolution from aperture, azimuth and theoretical bandwidth will give more accurate
results than using antenna size or estimated azimuth bandwidth.
c)
i)
No resolution doesn’t depend on constant velocity along track vx . Even from second and third
equation above there appears to be some dependency but considering equation (12) above there
will be cancellation of vx if expression of Baz is substituted into equation (14) and (15) and hence
Vx is not really required.
ii)
Yes, the antenna size is a critical part of equation D and somehow in equation A. In equation D its
azimuth resolution is directly proportional to antenna size so the smaller size of the antenna results
in finer resolution. But reducing the antenna size will effect the SNR and to fulfil the condition
that D is very large as compared to λ antenna size cannot be reduced to quite small value.
Matched Filtering
a)
In this task the squared magnitude of the online vector (row 37 in signal) has been plotted and is
shown in Fig below. From the graph, the signal around column 2180 is not visible. Moreover, it
has been observed from the graph that there is a lot of noise.
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Figure 2: Squared magnitude vs azimuth columns (Row 37)
b)
In this task one need to apply the filter as per the procedure mentioned in the manual. Accordingly,
the graph is plotted and is shown below in Fig 3. From Fig 3 it can easily be seen a sharp peak
around 2179.
Figure 3: Squared magnitude vs azimuth columns)
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c)
The matched filtering through time-domain convolution took 0.76 msec, whereas the matched filtering in frequency domain multiplication took 0.35 msec. From this, it can be deduced that
frequency-domain multiplication is much quicker. When using a large data vector, the frequencydomain multiplication remains the same, but the time-domain convolution does not remain the
same.
Figure 4: Squared magnitude vs azimuth columns)
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Figure 5: Squared magnitude vs azimuth columns)
Transformation in frequency domain is much faster as compared to time domain because convolution is much more complex operation as compared to multiplication especially when the order of
filter increases time to perform convolution will increase drastically but multiplication in frequency
domain will have lower effect.
Optimum Resolution
a)
The filtered has been applied and tested for numerous fR /Vx and thereafter shown in Fig below.
The table below is presented for bandwidth with various
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Figure 6: Resolution at different fr/Vx)
Figure 7: Bandwidth at different fr/Vx)
It has been noticed from figure that fR / Vx = 2.33 m−1 , the compressed signal has the narrowest
bandwidth, which is 1.45 columns, corresponding to 1.45 xn = 0.62 m.
b)
Using results from task 3a) of theoretical resolution, δ = 0.54. Here using
best match as comapred to other ratios.
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fR
vx
= 2.33 m−1 gives the
Image Construction
a)
In this task it is required to display the range of compressed data (signal) using the function
imageplot. The range of compressed data is displayed in Fig below.
Figure 8: Range compressed data plot)
b)
In this part a filter matrix with 350 rows and M columns has been set up. By using the optimal
fR/Vx ratio found in the previous part. SAR filter is applied on the image after completion of
script createHfilter provided along with the assignment. Subsequently, the focus image is displayed
and is shown in Fig 9 below. It is clear from Fig 9 that information is available in the image.
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Figure 9: Filtered Image Plot)
c)
In this part processed area is marked and is very much clear as compared to the first part of this
section and correspondingly plotted in Fig 10 below. It is obvious from the Fig 10, that it is oriented
differently as compared to the original image.
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Figure 10: SAR processed part of the original image)
Filter Analysis
a)
In this task 3-dB bandwidth of the image in azimuth is required to be determined using the function
spectrumplot. 11 obtained which represents the power spectrum over the number of rows as in part
2a. The 3-dB bandwidth is calculated as below.
100.2 − (−101.0) = 200.2Hz
11
(18)
Figure 11: Frequency (Hz)
b)
In this part the properties of the matched filter using a function filterplot has been plotted as shown
in Fig 12.
Figure 12: Plot of the matched filter
From the graph, it can deduce that it is a band-pass filter that allows to pass the frequency
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approximately 77%. From the graph it can be seen the angle is maximum at the centre frequency
(Phase).
c)
To elaborate on how filter helps in resolving azimuth dimension without varying the bandwidth one
can analyze the matched filter and compressed signal in η(slowtime) considering equation below.
saz = exp(
−j4πfc r(0)
−j2πfc (vx η)2
) × exp(
)
c
c × r(0)
(19)
From above equation, first term doesn’t effect image focus as it is constant in phase and the second
term results in linear frequency modulation as it has negative quadratic phase. The matched filter
resulting from this as given below:
haz (η) = exp(
j2π(vx η)2
) = exp(jπαη 2 )
λc ρ
(20)
has a positive quadrature equation.
From this when both of the above equations are convolved together the opposite polarity of quadrature will cancel out each other and filter will be more focused on azimuth in the image and resulting
bandwidth will be Baz = tψ × α where α is given as:
α=
2vx 2
λc ρ
(21)
Therefore, from equation (12) it can be concluded that the azimuth bandwidth is not changed when
signal is compressed in azimuth.
Speckle Reduction
a)
Speckle is basically due to the noise effect. The speckle noise can be minimized using various
techniques that are averaging or multi- looking. The image in this part is shown in 13 which is
much better than the previous image (fine enough, bright, and good pixel).
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Figure 13: Processed image with reduced speckle noise
b)
Theoretical resolution in azimuth σaz−9l is given by:
σaz−9l = N × ∆xn = 3.87m
14
(22)
c)
Figure 14: Range profile overview of averaged image
From figure 14 3-dB width is found as 2185 - 2176 = 9 which gives an absolute resolution of 9 ×
0.43 = 3.87 that is the same as the one obtained in previous section.
Matlab Code
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%% Given p a r a m e t e r s
clc ,
close all ,
clear all
load sarlab
rhomin = 1 0 4 4 6 ;
delp
= 4;
delxn = 0 . 4 3 ;
theta = 3;
D
= 1.3;
lamc
= 0.0566;
vx
= 131;
frvx
= 2.32;
%% Task 1 −−−−− a
rhomax = rhomin + 350∗ d e l p ;
Lfar
= 2∗ tan ( deg2rad ( t h e t a / 2 ) ) ∗rhomax ;
%% Task 1 −−−−− b
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t = L f a r / vx ;
M = round ( f r v x ∗vx∗ t ) + 1 ;
%% Task 1 −−−− c
rho37 = rhomin + 37∗ d e l p ;
L37 = 2∗ tan ( deg2rad ( t h e t a / 2 ) ) ∗ rho37 ;
%% Task 2 −−−− a
spectrumplot ( signal , s i z e ( s i g n a l ( : , 1 ) ) )
imageplot ( s i g n a l )
s a v e a s ( g c f , ’ 2 a . png ’ )
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%% Task 2 −−−− b
p s i = deg2rad ( t h e t a ) ;
Baz = ( ( 4 ∗ vx ) / lamc ) ∗ s i n ( p s i / 2 ) ;
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%% Task 3 −−−− a
Baz meas = 2 0 8 ;
sigmaaz L = ( rhomax∗ lamc ) / ( 2 ∗ L f a r ) ;
sigmaaz B meas = vx/ Baz meas ;
s i g m a a z B t h e o = vx/Baz ;
s i g m a a z t h e t a = lamc / ( 2 ∗ s i n ( t h e t a ) ) ;
sigmaaz D = D/ 2 ;
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%% Task 4 −−−− a
figure (1) ;
p l o t ( abs ( o n e l i n e ) . ˆ 2 ) , g r i d on
y l a b e l ( ’ Magnitude [ dB ] ’ )
x l a b e l ( ’ Azimuth Column ’ )
t i t l e ( ’ Range p r o f i l e ’ ) ;
s a v e a s ( g c f , ’ 4 a . png ’ ) ;
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%% Task 4 −−−− b
figure (2)
j = l i n s p a c e (−M/ 2 , M/ 2 , M) ;
h = exp ( ( i . ∗ 2 . ∗ p i . ∗ ( j . / f r v x ) . ˆ 2 ) . / ( lamc . ∗ rho37 ) ) ;
h37 MF = conv ( o n e l i n e , h , ’ same ’ ) ;
p l o t ( abs ( h37 MF ) . ˆ 2 ) ;
y l a b e l ( ’ Magnitude [ dB ] ’ )
x l a b e l ( ’ Azimuth Column ’ )
t i t l e ( ’ Range p r o f i l e ’ ) ;
g r i d on ;
s a v e a s ( g c f , ’ 4b . png ’ )
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%% Task 4 −−−− c
% Time
t = 0;
for j = 1:10
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tic
h37 MF = conv ( o n e l i n e , h , ’ same ’ ) ;
toc
t = t + toc ;
end
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%a v g t i m e f f t = t . / 1 0 ;
figure (3) ;
p l o t ( abs ( h37 MF ) . ˆ 2 ) ;
y l a b e l ( ’ Magnitude [ dB ] ’ )
x l a b e l ( ’ Azimuth Column ’ )
t i t l e ( ’ Range p r o f i l e ’ ) ;
g r i d on ;
s a v e a s ( g c f , ’ 4 c1 . png ’ )
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% Frequency
t
= 0;
f o r j =1:10
tic
h37 MF = c i r c s h i f t ( i f f t ( f f t ( o n e l i n e , l e n g t h ( o n e l i n e ( 1 , : ) ) ) . ∗ f f t ( h ,
l e n g t h ( o n e l i n e ( 1 , : ) ) ) ) , − f l o o r (M. / 2 ) ) ;
toc
t = t + toc ;
end
%a v g t i m e f f t = t . / i ;
figure (4) ;
p l o t ( abs ( h37 MF ) . ˆ 2 ) ;
y l a b e l ( ’ Magnitude [ dB ] ’ )
x l a b e l ( ’ Azimuth Column ’ )
t i t l e ( ’ Range p r o f i l e ’ ) ;
g r i d on ;
s a v e a s ( g c f , ’ 4 c2 . png ’ )
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% %% Task 5 −−−− a
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figure (5) ;
frvx arr = [2.31 , 2.32 , 2.33 , 2 . 3 4 ] ;
C = linspace (1 , length ( oneline ( 1 , : ) ) , length ( oneline ( 1 , : ) ) ) ;
m = l i n s p a c e (−M/ 2 , M/ 2 , M) ;
for j = 1: length ( frvx arr )
h i
= exp ( ( 1 i . ∗ 2 . ∗ p i . ∗ (m. / f r v x a r r ( j ) ) . ˆ 2 ) . / ( lamc . ∗ rho37
));
o n l n f i l t c o n v = conv ( o n e l i n e , h i , ’ same ’ ) ;
[ d b f i l t c o n v i n t , t ] = i n t e r p o l a t e s i g n a l ( o n l n f i l t c o n v , C, 4 9 ) ;
subplot (2 ,2 , j ) ;
p l o t ( t , abs ( d b f i l t c o n v i n t ) . ˆ 2 )
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y l a b e l ( ’ Magnitude [ dB ] ’ )
x l a b e l ( ’ Azimuth Column ’ )
t i t l e ( ”Range p r o f i l e ” + ” f { r }/ v {x} = ” + s t r i n g ( f r v x a r r ( j ) ) ) ;
g r i d on ;
end
s a v e a s ( g c f , ’ 5 a . png ’ )
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%% Task 6 −−−− a
figure (6) ;
imageplot ( s i g n a l )
s a v e a s ( g c f , ’ 6 a . png ’ )
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%% Task 6 −−−− b
figure (7)
createHfilter
f i l t s i g n a l = s a r f i l t e r ( Hfilter , signal ) ;
imageplot ( f i l t s i g n a l )
s a v e a s ( g c f , ’ 6b . png ’ )
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%% Task 7 −−−− a
figure (8)
spectrumplot ( f i l t s i g n a l , length ( s i g n a l ( : , 1 ) ) )
s a v e a s ( g c f , ’ 7 a . png ’ )
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%% Task 7 −−−− b
figure (9)
filterplot ( Hfilter )
s a v e a s ( g c f , ’ 7b . png ’ )
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%% Task 8 −−−−− a
figure (10) ;
N = 9;
w = o n e s ( 1 ,N) . /N;
d b f i l t s i g n a l = s q r t ( ( conv2 ( abs ( f i l t s i g n a l ) . ˆ 2 , w, ’ same ’ ) ) ) ;
imageplot ( d b f i l t s i g n a l )
s a v e a s ( g c f , ’ 8 a . png ’ ) ;
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%% Task 8 −−−− c
m = linspace (1 , length ( oneline ( 1 , : ) ) , length ( oneline ( 1 , : ) ) ) ;
d b f i l t s i n g a l 3 7 = abs ( d b f i l t s i g n a l ( 3 7 , : ) ) . ˆ 2 ;
[ d b f i l t c o n v i n t , t ] = i n t e r p o l a t e s i g n a l ( d b f i l t s i n g a l 3 7 , m, 4 9 ) ;
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figure (11)
spectrumplot ( d b f i l t s i n g a l 3 7 , s i z e ( d b f i l t s i n g a l 3 7 ( : , 1 ) ) )
s a v e a s ( g c f , ’ 8 c1 . png ’ ) ;
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figure (12)
p l o t ( t , abs ( d b f i l t c o n v i n t ) . ˆ 2 )
y l a b e l ( ’ Magnitude [ dB ] ’ )
x l a b e l ( ’ Azimuth Column ’ )
t i t l e ( ’ Range p r o f i l e ’ ) ;
s a v e a s ( g c f , ’ 8 c2 . png ’ ) ;
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createHfilter.m,
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% createHfilter
%
% This i s an i n c o m p l e t e s c r i p t t h a t c r e a t e s t h e H f i l t e r .
%
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Ronear =10446;
d e l t a R =4;
NOfRows=350;
%D i s t a n c e t o n e a r r a n g e [m]
%P i x e l s i z e i n r a n g e d i r e c t i o n [m]
%Number o f rows
lambda = 0 . 0 5 6 6 ;
%Wavelength [m]
n=−M/ 2 :M/ 2 ;
%Update n a f t e r you d e t e r m i n e t h e f i l t e r l e n g t h !
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PRF v = 2 . 3 3 ;
%PRF d i v i d e d by a i r c r a f t s p e e d [mˆ −1]
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H f i l t e r=z e r o s ( NOfRows , l e n g t h ( n ) ) ;
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f o r row =1:NOfRows
R = Ronear + ( row −1)∗ d e l t a R ;
H f i l t e r ( row , : ) = exp ( ( i . ∗ 2 . ∗ p i . ∗ ( n . / PRF v ) . ˆ 2 ) . / ( lamc . ∗R) ) ;
end
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