Radar Erim Denklemi Konuları Pekiştirelim

Transkript

5/14/2013
Ankara
Radar Systems and Remote Sensing
Research Group
TOBB ETÜ – Turgut Özal - Bilkent
TOBB UNIVERSITY OF
ECONOMICS AND
TECHNOLOGY
DERS 3
Radar Erim Denklemi
Yrd. Doç. Dr. Sevgi Zübeyde Gürbüz
ELE 465: Radar Sinyal İşleme Temelleri
ELE 565: Radar ve Sonar Sistemleri
Konuları Pekiştirelim
2
Menzil, menzil çözünürlüğü
 Açısal çözünürlüğü
 Doppler frekansı
 Minimum detectable range
 Maximum unambiguous range
 Radar Erim Denklemi
 Radar Kesiti

Ankara Radar Systems and Remote Sensing Research Group © SZG 2012
1
5/14/2013
Soru 1
3

Consider a radar transmitting pulses of duration t seconds at a
“pulse repetition frequency” of PRF pulses per second (pps). In
terms of PRF and/or t, what is the maximum range at which a target
can be located so as to guarantee that the leading edge of the echo
from that target on one pulse is received before transmission begins
for the next pulse? (This range is called the maximum unambiguous
range or the first range ambiguity.) What is the unambiguous range
if PRF = 3000 pps (often written, somewhat carelessly, as 3000 Hz)
and t = 10 ms? What is the range if we require the trailing edge of
the first pulse be received before the leading edge of the next pulse
is transmitted (which is more realistic), and we also require a 10 ms
switching time between transmit and receive (still more realistic)?
Çözüm
4




Diyelim ki hedefimizin uzaklığı R
t = 2R/c zaman sonra gönderdiğimiz darbe geri gelecektir
Bir sonra ki darbeyi 1/PRF = PRI sonra gönderiyoruz,
dolayısıyla 2R/c < PRI olmasını istiyoruz
→ Rmax = 0.5c/PRF = 50 km
Daha gerçekçi bir hesap yapabiliriz: Darbe uzunluğunu (t) ve
de switching süresini hesaba katalım
1
2R

 t  t switch
PRF
c
c 1

Rmax  
 t  t switch  → Rmax = 47 km
2  PRF

2
5/14/2013
Soru 2
5

Suppose the parameters of a radar are such that
the power of the echo from a particular target is
just detectable at a range of 50 miles. If the
target RCS is reduced by 10 dB, what will be the
new detection range? By how many dB must
the RCS be reduced to reduce the detection
range to 5 miles?
Çözüm
6

Radar Range Denklemini sadece RCS ve menzili açığa
vuracak şekilde yazalım:
Pr  k


R4
RCSin 10 dB küçülmesi demek alınan gücün 1010/10  0.1
kat düşmesi demek. Hedefin hala sezilebiliyor olması
için alınan gücün aynı kalmasını istiyoruz. Yanı oranın
değişmemesi için R^4 da 0.1 kat azalması gerek.
4
 Rnew 

  0.1  Rnew  500.14  28 miles
 Rold 
3
5/14/2013
Çözüm’e Devam
7

Eğer menzil 0.1 kat azalırsa, alınan gücün
aynı kalması için RCS (0.1)^4 kat
azalmalı:

RCS 40 dB azalmalı
Soru 3
8

According to the Smithsonian Air and Space
Museum, the radar cross section (RCS, s) of a
B-52 bomber is about 1,000 m2, while that of a
B-2 stealth bomber is 10-6 m2. If a given radar
system could detect the B-52 at a range of 100
km, at what range could the same radar system
detect the B-2 stealth bomber? Assume that
atmospheric losses are negligible. If the B-2
flies at 550 mph, how much warning time would
the radar give?
4
5/14/2013
Çözüm
9

İki uçaktan yansıyan güçler öncelikle aynı olmalı:
  
  
Pr  k 4 
 k 4 
 R  stealth  R  B 52

Atmosferik kayıpları ihmal edersek, her iki durum için
hesaplanan ‘k’ faktörü aynı olacaktır:
106
103
 
 



4
4
 R4 
 4
Rstealth
  stealth  R  B 52
105
 
Rstealth  562 m
Çözüm’e Devam
10

Uyarı süresi:

550 mph = 246 m/s
 t  562 / 246  2.24 sec
5
5/14/2013
Radar Erim Denklemi
11
Pr 




2 2
PG

t
 4 
3
R Ls La  R 
4
W
Vericinin gücü, dalga boyu önceden bilinir
Kayıplar için modeller, ölçümler bulunmakadır
Anten özelliklerinden G hesaplanabilir
RCS’in modellenmesi gerekir
12
ANTENLER
6
5/14/2013
Koordinat Sistemi
13
P   R, , 
z
Radar works
in spherical
coordinates

boresight direction = range
y
antenna
 (elevation angle)
x
 (azimuth angle)
Aperture Antennas
14


Parabolic dishes, flat plates, etc.
Far-field antenna radiation pattern is
the Fourier transform of the aperture
current distribution:
Dy 2
E   




A y
 2 y

 j  sin  
e
dy
1.8 m diameter parabolic dish reflector
antenna (image courtesy of Darius Ltd.)
 Dy 2

Important parameters are

peak gain
 peak sidelobe
 mainbeam (3 dB) width
7
5/14/2013
Uniform Current Distribution Antenna
Simplest case

y
E()
 Dy 

x
 Dy 


sin  Dy  sin  


E   
 Dy  sin 


normalized magnitude of radiation pattern
15
1
0.9
0.8
3 dB width
0.7
Peak
Sidelobe
Level
0.6
0.5
0.4
0.3
0.2
0.1
0
-100
-80
-60
-40
-20
0
20
40
60
80
100
angle off boresight,  (degrees)
For a Typical Antenna:
16

Gain: G 
=

26,000
33
7.9
33
(3 ,3 in degrees)
(3 ,3 in radians)
(Slightly different
scale factors are
common in other
sources)
 1.4 

  Dy 
3 dB Beamwidth: 3  2sin 1 
Bazı kişiler 0.89 faktoru düşürüp
sadece lambda / D kullanmaktadır –
bu 4dB beamwidth’e tekabul eder.
 0.89

Dy
radians
8
5/14/2013
Array Antennas
17


Uniform array of
individual elements
Conducive to digital
beamforming

d
a0
•••••
incoming signal
a1
•••••
a2
a3
a N 1
E

Antenna pattern is
product of array factor
and element pattern:
E    AF   Eel  
N 1
j 2   nd sin 
AF    E0  an e 
n 0
Eel    cos 

Linear array case:
AF    E0
sin  N  d   sin  
sin  d   sin  
Example Array Antennas
18
APG-68 in F-16
F-22 Solid State
Active Array
9
5/14/2013
Soru 4
19

In terms of Dy and l, what is the peak-tofirst null beamwidth (called the Rayleigh
beamwidth) in radians of the antenna
pattern for an aperture antenna with
constant illumination?
Çözüm
20

Hatırlayın, aperture antenna için
E   




1st null occurs when E()=0, i.e. Numerator = 0
 Dy
 



sin  Dy  sin  


 Dy  sin 
  

1

 sin   0    sin 


D

y


Since peak occurs at  = 0, peak-to-first-null-beamwidth is
  


 Dy 
  sin 1 
10
5/14/2013
Soru 5
21

Compute an exact expression for the
cross-range resolution, approximated in
the previous class as
Çözüm
22

Sum of the legs of two triangles:

 
2  R sin 
2
 This
is approximately R
11
5/14/2013
Soru 6
23

What is the maximum 3 dB beamwidth 3
in degrees such that the approximation for
the cross-range resolution, R3, has an
error of no more than 1%?
Çözüm
24

Matematiksel olarak şartımız:
 
 
R3  2 R sin 3   0.01 2 R sin 3 
2
2

Numerik çözümle:
 3 
  3  0.488 rad  28
2
3  2.02 sin

Bu approximation’un büyük beamwidth’ler
için geçerli olduğunu gösteriyor.
12
5/14/2013
Soru 7
25

Find the received power Pr expected from a
radar and target having the following
parameters: RF frequency = 95 GHz (W band),
transmitted power = 100 W, antenna beamwidth
= 2º in azimuth and 5º in elevation, system
losses = 5 dB, target range = 3 km, target RCS
= 20 m2. Use Fig. 1-3 and assume operation
near sea level to estimate atmospheric losses.
What is the ratio in decibels of the received
power to the transmitted power?
Çözüm
26
26000
Gain: G 

 = c/f = 3.16 mm

Ls = 5 dB = 10^(5/10) = 3.16

Atmospheric Loss at 95 GHz...
33

26000
 2600
2(5)

13
5/14/2013
Çözüm’e Devam
27
*
Çözüm’e Devam
28



Atmospheric Losses = 0.4 dB/km at sea level
Range is 3 km for a round trip distance of 6 km
So La(R) = 6 x 0.4 = 2.4 dB = 10^(2.4/10) = 1.74
Pr 
Pr 
2 2
PG

t
 4 
3
R 4 Ls La  R 
W
100(2600) 2 (3.16 10 3 ) 2 (20)
 1.53 10 13W
(4 )3 (3000) 4 (3.16)(1.74)
Pr 1.53 10 13

 1.53 10 15  10 log(1.53 10 13 )  148dB
Pt
100
14
5/14/2013
Soru 8
29

By how many dB will the received power
be reduced (not the absolute power
received, but the reduction in power) in
problem #7 if the weather changes from
clear to a heavy rain of 25 mm/hr?
Çözüm
30


@heavy rain: 13 dB/km → La = 78 dB
@clear: La = 2.4 dB ► Reduction = 75.6 dB
15
5/14/2013
31
RANGE EQUATION
FOR
VOLUME AND AREA
TARGETS
For Distributed Targets
32

Remember:
Pr 
Pt  2
 4 3 Ls





V  R0 ,0 ,0 
P 2  , 
R 4 La  R 
d   R , ,  
16
5/14/2013
Volume Scattering
33



Observed RCS is presumed to be due to scatterers
uniformly distributed throughout the resolution cell
RCS  expressed in terms of RCS per cubic meter,
called volume reflectivity, denoted h :
 Units are m2/m3 = m-1
 d = h dV = h R2dR dW
 dW = differential solid angle
Range equation becomes
Pr 
Pt  2h
 4 
3
Ls





V  R0 , 0 ,0 
P 2  , 
dR d W
R 2 La  R 
Integrate over Range
34

Assume the variation in atmospheric loss over
the range extent of a resolution cell is negligible
 La(R0)
 pull this term outside of integral
 La(R)

Remaining integral over range terms is
R 
R
 0 2


R  R
0
Pr 
2
R
R
 dR 

 2 2
2
R02
 R  R0   R 2 
Pt  2h R
 4 
3


R Ls La  R0  W
2
0
P 2  ,  d W
17
5/14/2013
Integrate over Angle
35

If we approximate the main lobe of the
antenna pattern P(,) as a Gaussian, the
integral becomes
 P 2


 , d d 
33
8ln 2
G 2  0.5733G 2
Frequently simply assume a constant gain
over main lobe, zero elsewhere, giving
 P 2

 2nd
 , d d  G 2 33
approximation is 2.5 dB higher than first
Range Eq. For Volume Scattering
36
Pr 

2 2
PG
 hR 33
t
 4 
3
R02 Ls La  R0 
Note power decreases as R2, not R4 as
in point scatterer case
 because
spreading of antenna beam
causes resolution cell to grow as R2
18
5/14/2013
Area Scattering
37


Observed RCS presumed due to scatterers
uniformly distributed on a 2-D surface
RCS  expressed in terms of RCS per
square meter, called area reflectivity or
“sigma-nought”, denoted 0 :
 units are m2/m2 = dimensionless!
 But often expressed in dB as 10log10(0), called dBsm


d   D  R  R0   0 dA

dA = differential surface area
Range equation becomes
Pt  2
Pr 
P 2  ,  0 dA
3

4
 4  R0 Ls La  R0  A R , , 
Ankara Radar Systems and Remote Sensing Research
Group © SZG 2012
0
Two Cases of Area Scattering - 1
38



Assume planar reflecting surface (flat earth)
A scatterer does not contribute significantly to the receiver output unless
it is both illuminated by the pulse and within the antenna mainbeam
Range extent of contributing scatterers depends on relative size of
illuminated area and pulse length
z
x
z=0
Beam-limited
range extent
R
0
z = -h
R03

ground plane
R03
sin 
19
5/14/2013
Two Cases of Area Scattering - 2
39
x
z=0
Pulse-limited
range extent
R0

z = -h
ground plane
R
cos 
Pulse- vs. Beam-Limited
40

If the beam footprint on the ground is
greater than the pulse footprint, the
resolution cell extent in range is pulse
limited
 and
vice-versa
beam-limited:
R
tan   3
R0
pulse-limited:
R
tan   3
R0
20
5/14/2013
Beam-Limited Area Range Equation
41

Using constant-gain mainlobe approximation:
R0
R02
dA  R0 d 
d 
d d 
sin 
sin 
Pr 

2 2 0
PG
   33
t
 4 
3
R02 Ls La  R0  sin 
Note RCS varies as R2
 cell
gets wider in both range and cross-range as R
increases
Pulse-Limited Area Range Equation
42

Using constant-gain mainlobe antenna
approximation:
dA  R0 d 
Pr 

R R
R
d  0 d d
cos 
cos 
2 2 0
PG
  R 3
t
 4 
3
R03 Ls La  R0  cos 
Note RCS varies as R3
 cell
gets wider in cross-range only as R increases
21
5/14/2013
Simplified Approach
43

These results are equivalent to assuming a particular
form for  in the point target range equation:
clutter: 0 times the scattering area
 volume clutter: h times the scattering volume
 area
Case
Volume Scatterer
Beam-Limited Area
Scatterer
Pulse-Limited Area
Scatterer
RCS 
h R  R3  R3  h R 2 R33
0 2
 R   R 33
 0 R3   3  
sin 
 sin  
 0 RR3
 R 
 0 

R


3

cos 
 cos  
Teşekkür
44

Bu slaylarda yoğun olarak Prof. Mark
Richards’in sunumundan yararlanılmıştır,
kendisine teşekkür ederiz.
22

Radar Erim Denklemi Konuları Pekiştirelim

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