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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
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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I'm not sure why it wrote it the other way
I got X =-6
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oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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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
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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.
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Source:  OpenStax, Signal and information processing for sonar. OpenStax CNX. Dec 04, 2007 Download for free at http://cnx.org/content/col10422/1.5
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