<< Chapter < Page Chapter >> Page >

h = ( x , y ) H τ size 12{h= \( x,y \) in H rSup { size 8{τ} } } {} if R min < ( x x r ) 2 + ( y y r ) 2 < R max size 12{R rSub { size 8{"min"} }<sqrt { \( x - x rSub { size 8{r} } \) rSup { size 8{2} } + \( y - y rSub { size 8{r} } \) rSup { size 8{2} } }<R rSub { size 8{"max"} } } {}

Where we are assuming that the depth of the target is small when compared to its ( x , y ) size 12{ \( x,y \) } {} coordinates, the receiver is located at ( x r , y r ) size 12{ \( x rSub { size 8{r} } ,y rSub { size 8{r} } \) } {} . R min size 12{R rSub { size 8{"min"} } } {} is the range at which the echo is noise, not reverberation limited, and R max size 12{R rSub { size 8{"max"} } } {} is the farthest range of interest. For this problem, h size 12{h} {} is an index into the target range from the sonar.

Sonar receiver model

The sonar transmits the waveform m ( t ) size 12{m \( t \) } {} for each ping. In most sonar transmitters, the transmitted waveform is narrow-band, that is, the waveform bandwidth is much smaller than its center frequency, f size 12{f} {} . This is true because efficient sonar transmitters use resonant mechanical and electrical components to provide maximum electrical to sound power transfer. An approximation therefore is to model the transmitted waveform as an amplitude modulated carrier:

m ( t ) = sin ( ft ) w ( t ) size 12{m \( t \) ="sin" \( 2πital "ft" \) w \( t \) } {} , t = ( 0, T ) size 12{t= \( 0,T \) } {}

We will assume that the target is motionless, so that Doppler effects can be ignored. We will assume that the sonar receiver is a single sensor, with no directionality characteristics. For each target location hypothesis h = ( x , y ) size 12{h= \( x,y \) } {} we know approximately the received echo time series:

g ( t h ) = Bm ( t 2R / c ) size 12{g \( t \lline h \) = ital "Bm" \( t - 2R/c \) } {}

The amplitude B size 12{B} {} is related to the propagation loss out to the target hypothesis location, and the reflection characteristics of the target. The time delay 2R / c size 12{2R/c} {} corresponds to the time it takes for the transmission waveform to reach the target and return to the sonar. R size 12{R} {} is the range to the target and c is the effective speed of sound, when including refraction and boundary reflections.

The received echo is band-limited to approximately the same frequency band as the transmission. The receiver bandwidth may be greater than the transmitted bandwidth due to Doppler frequency shifts, but for the present, we are assuming that the target is not moving. Sonar receivers use heterodyne techniques to reduce the data storage of the ping history. The sonar receiver multiplies the ping history by a carrier signal e j2π ft size 12{e rSup { size 8{ - j2πital "ft"} } } {} to shift the positive frequency part of the received echo closer to DC. The resulting signal is then low pass filtered to eliminate the shifted negative frequency part of the ping history. Since the original ping history was real, the negative frequency part of the signal spectra carries no additional information. The result is a complex signal with a lower bandwidth, but retains all of the echo related information of the original ping history. This heterodyne process can be done in the analog or digital domain.

A target echo passing through the heterodyne part of the sonar receiver becomes:

r ( t h ) = Ae w ( t 2R / c ) size 12{r \( t \lline h \) = ital "Ae" rSup { size 8{jθ} } w \( t - 2R/c \) } {}

The phase shift θ size 12{θ} {} corresponds to the phase shift due to heterodyne operation; the uncertainty in propagation conditions; and the summation of multi-path arrivals with almost the same time delay, etc.

We will assume that the target echo amplitude, Ae size 12{ ital "Ae" rSup { size 8{jθ} } } {} ,is a complex Gaussian random variable with zero mean and with standard deviation σ 2 ( h ) . size 12{σrSup { size 8{2} } \( h \) "." } {} We are modeling the echo as having the same waveform as the transmission, but with an uncertain phase and amplitude. This is assuming that the target echo amplitude obeys Swerling target type I statistics with unknown phase.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Signal and information processing for sonar. OpenStax CNX. Dec 04, 2007 Download for free at http://cnx.org/content/col10422/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Signal and information processing for sonar' conversation and receive update notifications?

Ask