target strength
DESCRIPTION
reflected wave. a. incident wave. Target Strength. At r = 1 yd. Factors Determining Target Strength. the shape of the target the size of the target the construction of the walls of the target the wavelength of the incident sound the angle of incidence of the sound. R 2. R 1. - PowerPoint PPT PresentationTRANSCRIPT
Target Strength
reflected wave
incident wave
a
2i rI 4 r I
r
i
ITS 10log
I
scattering cross section
2TS 10log 10log
4 r 4
At r = 1 yd.
Factors Determining Target Strength
• the shape of the target
• the size of the target
• the construction of the walls of the target
• the wavelength of the incident sound
• the angle of incidence of the sound
Target Strength of a Convex Surface
i 1 2dP I ds ds
R1
R2
1d1ds
2d
2ds
i 1 1 2 2dP I R d R d
Incident Power
Large objects compared to the wavelength
Reflected Intensity
12 11R
11
r
1 1ds r2d
1 2 1 2dA ds ds r2d r2d
i 1 1 2 2 i 1 2r 2
1 2
dP I R d R d I R RI
dA r2d r2d 4r
R1
R2
1d1ds
2d
2ds
r
i
ITS 10log
I
1 2R RTS 10log
4
(At r = 1 m)
Special Case – Large Sphere
1 2R R a
2a aTS 10log 20log
4 2
a
Note: 2a
4 4
2a
Large means circumference >> wavelength
ka 1
TS positive only if a > 2 yds
Example
• An old Iraqi mine with a radius of 1.5 m is floating partially submerged in the Red Sea. Your minehunting sonar is a piston array and has a frequency of 15 kHz and a diameter of 5 m. 20 kW of electrical power are supplied to the transducer which has an efficiency of 40%. If the mine is 1000 yds in front of you, what is the signal level of the echo. Assume spherical spreading.
Scattering from Small Spheres (Rayleigh Scattering)
22 2r
4 2i
I V 3cos 1
I r 2
4 225TS 10log ka a
36
ka 1
Scattering from Cylinders
L
2a
22 2
2
aL sin cosTS 10log
2 1yd
2 Lsin
2
2
aL 1TS 10log
2 1yd
o0
Dimensions (L,a) large compared to wavelength
Gas Bubbles
• Damping effect is due to the combined effects of radiation, shear viscosity and thermal conductivity. A good approximation is
• where fk is the frequency in kHz.
3
22
20
0
1
resonant frequency
damping term
bs
a
f
f
f
0
3
5
31 3.251 0.1
2
1000 kg/m
hydrostatic pressure in Pa 10 1 0.1
depth in meters
adiabatic constant for air ( 1.4)
w
w
w
w
Pf z
a a
P z
z
0.30.03 for 1 kHz< 100 kHzk kf f
Fish
• Main contribution for fish target strength comes from the swim bladder.
• This gas-filled bladder shows a very strong impedance contrast with the water and fish tissues. It behaves either as a resonator (frequencies of 500 Hz-2 kHz depending on fish size and depth) or as a geometric reflector (> 2 kHz). This swim bladder behaves very similar to gas bubbles. The difference in target strength between fish with and without swim bladder can be 10-15 dB.
• A semi-empirical model most often used is:
• Love (1978)• This formula is valid for dorsal echoes at wavelengths smaller than
fish length L.
19.1log 0.9 log 24.9fish kTS L f
421aa
2r
4
4
2
7.61V
2
ar
ra
2
449
2
2aL
22 2sinaL cos
2
2a
4
2
22
cossin
ab
2
2
a
bc
2
2
12
cos2
Ja
Formt
TS=10log(t)Symbols Direction of incidence Conditions
Any convex surface
a1a2 = principal radii of
curvaturer = rangek = 2/wavelength
Normal to surfaceka1, ka2 >>1
r>a
Large Sphere a = radius of sphere Anyka>>1r>a
Small SphereV = vol. of sphere = wavelength
Anyka<<1kr>>1
Infinitely long thick cylinder
a = radius of cylinderNormal to axis of cylinder
ka>>1r > a
Infinitely long thin cylinder
a = radius of cylinderNormal to axis of cylinder
ka<<1
Finite cylinder
L = length of cylindera = radius of cylinder
Normal to axis of cylinder ka>>1
r > L2/a = radius of cylinder = kLsin At angle with normal
Infinite Plane surface Normal to plane
Rectangular Platea,b = sides of ractangle = ka sin
At angle to normal in plane containing side a
r > a2/kb >> 1a > b
Ellipsoida, b, c = semimajor axis of ellipsoid
parallel to axis of aka, kb, kc >>1r >> a, b, c
Circular Platea = radius of plate = 2kasin At angle to normal
r > a2/ka>>1
Example
• What is the target strength of a cylindrical submarine 10 m in diameter and 100 m in length when pinged on by a 1500 Hz sonar?
2 4 6 8 10
-40
-20
20
40TS
10o5o