# 4.2 Active sonar detection in ambient noise  (Page 2/4)

 Page 2 / 4

$h=\left(x,y\right)\in {H}^{\tau }$ if ${R}_{\text{min}}<\sqrt{\left(x-{x}_{r}{\right)}^{2}+\left(y-{y}_{r}{\right)}^{2}}<{R}_{\text{max}}$

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

The sonar transmits the waveform $m\left(t\right)$ 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, $f$ . 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:

$m\left(t\right)=\text{sin}\left(2\pi \text{ft}\right)w\left(t\right)$ , $t=\left(0,T\right)$

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 $h=\left(x,y\right)$ we know approximately the received echo time series:

$g\left(t\mid h\right)=\text{Bm}\left(t-2R/c\right)$

The amplitude $B$ is related to the propagation loss out to the target hypothesis location, and the reflection characteristics of the target. The time delay $2R/c$ corresponds to the time it takes for the transmission waveform to reach the target and return to the sonar. $R$ 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 ${e}^{-\mathrm{j2\pi }\text{ft}}$ 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:

$r\left(t\mid h\right)={\text{Ae}}^{\mathrm{j\theta }}w\left(t-2R/c\right)$

The phase shift $\theta$ 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, ${\text{Ae}}^{\mathrm{j\theta }}$ ,is a complex Gaussian random variable with zero mean and with standard deviation ${\sigma }^{2}\left(h\right)\text{.}$ 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.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
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
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
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.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
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.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
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
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.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!