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
Large Spheres (continued)
2 2 2 2r12
i
I 1a a cot J ka sin
I 4 r 2
a
0o180o
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
Example
• What is the target strength of a single fish 1m in length if the fish finder sonar has a frequency of 5000 Hz?