Download - Ultrasonic Sensor for Level Measurement
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Assessment on the performance of sensor
Ultrasonic sensor for level measurement
Submitted by
Ashish Rawat
Amruta bahulikar
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Contents
i. Review of basics of ultrasonic sensorii. Block Schematic
iii. Front enda.Configurable parametersb.designc.Installation related issuesd.Environment condition
iv. Backenda.Governing equationsb.Extended equationsc.Compensation
v. Sensor response as Transmittervi. Sensor response as Receiver
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Review of Basics of ultrasonic sensors
In order to understand the capabilities and limitations of ultrasonic instruments, onemust
understand the conditions that determine the characteristics of sound: temperature,
reflection, propagation and absorption. Temperature compensation is required, because
thevelocity of sound is proportional to the square root of temperature and in case of air, it
changes by about 2ft/s (0.6m/s) for each degree of centigrade change in temperature. The
speed of travel rises with temperature, and it amounts to about 0.18% per degree
centigrade.
In order to measure time travel of theecho of an ultrasonic pulse, it is essential that some
of sonic energy be reflected. It is also important that reflecting surface must be flat,
because angle of reflection equals the angle of incidence. Intensity of sound decreases with
square of distance. Therefore echo becomes exponentially weaker as the range of
instrument is increased. The decrease in sound energy is caused not only the dispersion
(travelling distance) but also the absorption in the substance through which it travels. It can
be seen that an ultrasonic transmitter is subject to many interferences which will affect the
strength ofecho. The continuous level detector designs can be categorized as under liquidsensor and above liquid sensors. Most design use20 KHz or higher oscillator circuit. Some
designs incorporate filters or discriminatory circuitry and electronics to prevent false
reading that might be caused by noise.
Generation of ultrasound
There are many methods which can be employed to create oscillating surface that
generates ultrasound. The piezoelectric sensor is the best option because of its very highstrain sensitivity of 5.0 V/
Piezoelectric Materials and their PropertiesCertain single crystal materials exhibit the following phenomenon: when the crystal is
mechanically strained, or when the crystal is deformed by the application of an external
stress, electric charges appear on certain of the crystal surfaces; and when the direction of
the strain reverses, the polarity of theelectric charge is reversed. This is called the direct
piezoelectric effect, and the crystals that exhibit it are classed as piezoelectric crystals.
Piezoelectric CeramicsMany of today's applications of piezoelectricity use polycrystalline ceramics instead
of natural piezoelectric crystals. Piezoelectric ceramics are more versatile in that their
physical, chemical, and piezoelectric characteristics can be tailored to specific applications.
Piezoceramic materials can be manufactured in almost any shape or size, and the
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mechanical and electrical axes of thematerial can be oriented in relation to the shape of
the material. These axes are set during poling (the process that induces piezoelectric
properties in the material). The orientation of the DC poling field determines the
orientation of themechanical and electrical axes.
Piezoelectric constants
1.Piezoelectric charge constants(d constant)The piezoelectric d constant is a measure of the charge density per unit stress or the strainper unit field
Piezoelectric coefficients with double subscripts link electrical and mechanical
quantities. The first subscript gives the direction of theelectrical field associated with the
voltage applied or the charge or the voltage produced. The second subscript gives the
direction ofmechanical stress or the strain.
2.Piezoelectric Voltage coefficient ( g constant)The g coefficient is a measure of the field per unit stress or strain per unit charge density.
The first subscript indicates the direction of the generated voltage and the second indicates
the direction of the force. A "33" subscript signifies that theelectrical field generated and
themechanical stress are both along the polarization.
3.Relative Dielectric Constant expresses the relative dielectric constant of thematerial relative to that ofvacuum inthe 3 direction. Multiplying this by o, the dielectric constant of free space yields the
absolute dielectric constant (o = 8.85 x 10-12 farads/meter). The superscript T applies tothemechanically free condition.
4.Relationship Between g and d coefficientsAt frequencies far from resonance effects, piezoelectric ceramic transducers are
fundamentally capacitors. Consequently, the voltage coefficient gik are related to the
charge coefficient dik by the dielectric constant Ki , as in a capacitor thevoltage V is related
to charge Q by the capacitance C.
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5.Coupling Coefficients
Sometimes also referred as electromechanical coupling coefficients, these describe the
conversion ofenergy by the ceramic element fromelectrical to mechanical form or vice-
versa.
6.Young's ModulusTheeffectiveYoung's modulus with electrodes short circuited is lower than theelectrodesopen circuited. Furthermore, the stiffness is different in the3 - direction from that in the 1
or 2 direction. Therefore, in expressing mechanical quantities both direction and electrical
conditions must be specified. is theequivalent with theelectrodes open circuited inthe3 direction whereas isthemodulus in the 1 or 2 direction. The superscript D pointsout the open circuit condition. The inverse ofYoung's modulus Y is theelastic compliance
's'.
7.Dissipation Factor or tanThis is also frequently called loss tangent and is a measure of the dielectric losses in thematerial, expressed as the tangent of the loss angle or the ratio of resistance to reactance
of a parallelequivalent circuit of the ceramic element. It is measured directly at 1 kHz using
LCR Bridge.
8.Curie PointIt is the temperature at which the crystal structure of the material changes from a
piezoelectric to non-piezoelectric state. It is also the temperature at which the dielectric
constant peaks.
9.Mechanical Quality Factor (Qm)Themechanical Q is a dimensionless number which gives thequality of the ceramic as a
harmonic oscillator. It is the reciprocal of the damping factor.
10. Frequency Constants
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The frequency constant, N, is the product of the resonance frequency and the linear
dimension governing the resonance. It is also equal to half the sound velocity in the same
direction. The constant can be used to calculate the resonant frequency at which an
element would operate.Np = Planar mode of thin disc
Nt := Thickness mode of thin plate.
Fig1: piezoelectric properties(Sparkler Ceramics)
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Block Schematic
The principle of operation of this unit is similar to echometers used to measure the depth
of wells. The continuous ultrasonic level detector measures the time required for an
ultrasonic pulse to travel to the process surface and back. InstallationB shows the
transmitter and receiver are packaged as a single unit. When transmitter generate short
bust of ultrasonic energy and acoustic energy being produced then receiver is blanked off.When ultrasonic waves are on their way the receiver gate is open to detect theecho. It has
some disadvantage that someenergy is lost in travelling through vapor space.
Fig 2: various Installation
Theworking principle can beexplained as following-
1. A piezoelectric crystal acts as both a transmitter and Receiver2. When the crystal act as a transmitter, a pulse input of sufficient width is given (let
time T) for excitation. A timer is also started simultaneously.
3. Therefore a pulse of ultrasound generated at the resonant frequency of crystal thatenters themedium 1 and is reflected at the boundary ofmedium 1 and medium2.
4. The reflected ultrasound returns to Crystal at T1.5. The T1 is the time of flight for round trip and
6. The Crystal now acts as a receiver and converted theecho into a voltage pulse.7. Theecho is then amplified and rectified for triggering the timer to stop.
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Fig 3 : the firing logic for transmitter
Conditions:
y T must be sufficient enough to excite transducer.y
Fig 4 : block Diagram for complete Unit
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Fig 5: Steps to find time of flight
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Front end
Configurable parameters
y designa. Measurement range(operating frequency)b. Accuracyc. Resolutiond. Excitation method
y Manufacturinga. XTAL properties (in manufacturer hands)
y Usagee. Environmental conditionsf. Indication type
Installation related issues
a. Near field distanceb. Open/close tank
Environmental parameters
a. Temperatureb. Humidityc. Agitation in tank
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Design
Let us assume some properties of the system(for SP-5H)-
Piezoelectric charge constant d33 = 550*10^ -12 C/N Piezoelectric voltage constant 33 = 19*10^-3 Vm/N Density = 7500 Kg / m^3 Curie temperature = 190 'C Mechanical quality factor Qm = 65 Frequency constant = 1950 Hz-m Relative dielectric constant r = 3100 Dissipation factor tan = 0.020
1. Frequency selection(Operating range)As frequency of sound increases the amount of absorption increases significantly . For the
application of 1 m tank depth100 KHz is found reasonable good frequency.
40 KHz Upto 8m
80 KHz Upto 4m
100kHz Upto 2m
200 KHz Upto 1m
Above 1 MHz Biomedical applications
2. Dimension calculationIf the frequency of the AC field corresponds with the frequency where the thickness of the
crystal represents half a wavelength, the amplitude of the crystal vibration will bemuch
greater. This is called the crystal's fundamental resonance frequency. Thus --
t = /2 = (c/2) / f
as frequency constant Np = c/2 = 1950 Hz-m, and frequency of operation is 200 KHz then
t = 1950/(200 * 10^3)=0.0098m = ~ 10 mm .
3. Power calculationPmax =~ O.f
Where O = volume of transducer
f = operating frequency
For a max power of200 mW the area ia calculated as follows
Area*t = 200*10^-3/200*10^3 area = 0.0001 m*m
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Then for a circular disc diameter = 0.011m = 11mm
4. Stiffness calculationF = kx
K - stiffness
x - deformtion
Youngs modulus Y = (F/A) / (x/t) = FL/Ax =kt/A
Thus k = A/(L*s)
Where s - elastic compliance (inverse of young's modulus)
K = 0.0001/((15*10^-12)*(0.010)) = 6.6666E8 = 6.67*10^8 N/m
5. Mechanical natural frequencyWn = (k/m)
Wherem - mass of transducer
m = *O = 7500*((1*10^-4)*0.010) = 7.13*10^-3 kg.
Wn = (6.67*10^8/7.13*10^-3) =3.1*10^5 rad/s = 48.678 KHz
6. Damping factor = 1/(2*Q)
= 1/(2*65) = 0.0077
7. Damping resistance = (mk/Q)
= (7.13*10^-3 * 6.67*10^8) / 65 = 33.55 Ns/m
8. Crystal capacitanceC = or A/t = (8.85*10^-12)(3100)(9.5*10^-5)/(0.010) = 29 nF
9. Near fieldN = d^2 *f/4*c = 0.011^2 * 200000 / 4*350 = 0.0173m = 17.3mm
d - element diameterf - operating frequency
C - velocity of sound in medium
Max time to travel = 2/ 350 = 0.0057s = 5.7ms
10.Matching layer
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The acoustic impedance impedance ofmatching layer
Z = (Zair * ZXtal)
Zair = 406.5 Nsm3 (at 30'c)
Zxtal = 3.0*10^7
Z = (406.5*3.0*10^7) = 110430.97 rayls
Matching layer lenth l = /4Where, is wavelength
l = c/4f =350/(4*200000) = 0.0004m = 0.4 cm.
11.Backing layer
backing layerimpedance = XTAL impedance = 3.0*10^7 rayls
And must possess high attenuation coefficient(w)
12.Effective pressure
P = WAC
= 2200*103
(d33 E) 406.7
Let suitablevoltage = 50 V the P = 14.05 pas
Peff= P / 2 = 9.9 pascal
dB = 20 * log (9.9 / 20*10-6
) = 113.9 114 dbspl
13.Effective velocity
V = A
= 2f * d33E
= 0.0345 m/s
Veff = V / 2 =0.024m/s (particlevelocity)
14.Heat generated in piezo crystal during dynamic operation
P = power converted to heat(Watt)
tan = power factor
F = operating frequency(Hz)
C = piezo capacitance(F)
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V= peak to peakvoltage
P =0 227 watts
n i on nt l conditions
Thereare mainly two parameters that affect the performance ofsensor-
1 temperature2. Humidity3. Agitation
Both temperatureand humidityaffects thesound ofvelocityas well as the theattenuation
of sound. The changes are shown in the following graphs. The details are given in the
compensationsection.
Fig6
speed ofsoundversus temperature
Fig 7 : relative humidityversus percentagechangeinspeed ofsound
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Attenuation of sound is complex part, and affected due to several reasons. Some of them
are thermal relaxation, inverse square law, refraction absorption etc. the introduction to all
is given in compensation section.
Agitation (may be a stirrer) in the tank causes turbulence in the upper surface of
liquid level. The turbulence results in reflection of ultrasound in different directions. It may
happen that echo is not reflected to probe. Therefore it is better not to use ultrasonic
sensor for such application or constrained can be placed. Still work has to be done for this
unit.
Installation related issues
1. Near field: as the source is not a point source of ultrasound wave it follows sphericalwaves. Near field consists of interference region of these spherical waves.
2. Open/close tank: if the tank is close then provision has to bemade for mounting theprobe. Moreover if tank or atmosphere contains corrosive liquid, the probe has to be
provided with wear plate.
Governing equations:
y Time of flightthe distance covered is calculated as the speed of sound is calculated as following
y Speed of soundThe theoretical expression for the speed of sound c in an ideal gas is
Where P is the ambient pressure, p the gas density and the ratio of the specific heat of
gas at constant pressure to that at constant volume. The term is dependent upon the
number of degrees of freedom of the gaseous molecule. The number of
= 1.40 for diatomic molecules Thevelocity of sound c in dry air has the following experimentally verified values:
c = 331.45 0.05 m/s
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y Reflection coefficient Transmission coefficient
Fig 8 : reflection and transmission at boundry of two medium
For air water interface, Ro = 0.998 , Therefore it can be seen that all of incident energy is reflected back and negligible amount
is transmitted
y Attenuation of ultrasonic wave motionThere are several important factors which affect the propagation of sound: geometric
spreading, atmospheric effects surface effects and scattering. These are discussed
separately below.
y Attenuation due to scatteringThe propagation of ultrasonic wavemotion can be described as elastic deformation
of individual particles of medium. Due to internal friction and thermal conductivity, this
deformation is accompanied by the losses of oscillation energy, which are converted to
thermal energy.
If the avgerage size of crystalline grains is D then for condition
>= 20D, attenuation coefficient = af
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length ofmatching layer and backing layer should accept and absorbs sound wave at rear of XTAL thus
attenuation coefficient
Extended equation
Speed of sound :
The relation for speed of sound can beextended as following for several parameters
For temperature : For Humidity :
All previous discussion assumed dry air. Attention turns now to theeffects ofmoisture on
the speed of sound. Moisture affects the density of air and hence, the speed of sound in air.
Moist air is less dense than dry air (not particularly obvious). This causes an increase in the
speed of sound. Moisture also causes the specific-heat ratio to decrease, which would
cause the speed of sound to decrease. However, the decrease in density dominates, so the
speed of sound increases with increasing moisture.
The aboveequation is exact for dry air. Two terms must bemodified to include accurately
theeffects ofmoisture (water vapor) on the speed of sound. These are the specific heat
ratio (1.4 for dry air) and M, the average molecular weight of the different types of
molecules in the air. Development ofeach of these terms follows. The terms R (universalgas constant) and T (absolute temperature ) remain unchanged.
The specific-heat ratio can be expressed as an exact fraction by letting d equal the
number ofexcited degrees of freedom for the air molecules. This gives
The specific-heat ratio can be expressed as an exact fraction by letting d equal the
number ofexcited degrees of freedom for the air molecules. This gives
= d +2 / d
Since the composition of dry air is mostly two atommolecules, it is said to be a diatomic
gas. Diatomic gases have 5 degrees of freedom, three translational and two rotational; thus
d = 5 and or = 1.40,for dry air.
If h is defined to heequal to the fraction ofmolecules that arewater, then the presence of
water (with 6 degrees of freedom) causes the average number of degrees of freedom per
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molecule to increase to 5+ h cannow berewritten to include theeffects ofmoistureforair
as
w = (7+h)/(5+h)
Theaverage molecular weight ofairdecreases with added moisture. To see this M is
calculatedfirst fordryair. Dryaircompositionis
78
nitrogen (molecular weight =28)
21
oxygen (molecular weight = 32)
1
argon (molecular weight = 40)
fora total molecular weight e ual to
M = (0.78)(28) + (0.21)(32) + (0.01)(40) =29
The presence ofwater (with a molecular weight of18) causes the total average molecular
weight to decreaseto 29 (29 18) h, or
MW = 29 11h
affected by theaddition ofwatervapor to air.Both areafunction oftheintroduced water
moleculefraction h. Relative humidity RH (expressedasa percentage) isdefinedsuch that
where p e ualsambient pressure (1.013 l0^5 Pafor1atm reference pressure) and e(t) is
thevapor pressure ofwaterat temperature t.For temperaturevalues indegrees Celsius,
representativevalues ofe(t) are-
e (5) =5872 Pa e (20) =2338 Pa
e (l0) =1228 Pa e (30) = 4243 Pa
e (l5) =1705 Pa e (40) = 7376 Pa
To express the percentageincreasein thespeed ofsounddue to relative humidityall that
remains is to take the ratio of the wet anddry speeds, subtract 1, and multiply by 100.
Since both wet anddryspeed terms involve thesameconstant terms (R and T), theirratio
will cause these to cancel, leaving
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Subtracting 1 and multiplying by 100 yields
Attenuation of Sound
Attenuation of sound is a complex phenomena, and difficult to derive correct formulas.
Instead empirical relations can be used to get an approximate solution. The observation is
taken from a calculator available on the internet. The graphs are plots for
attenuation(db/m),temp (C),relative humidity(percentage), freq(KHz)
Fig 9 : temp vs attenuation (constant humidity)
Series 1(relative humidity=20%,freq =200KHZ)
Series 2(relative humidity=50%,freq =200KHZ)
Series 3(relative humidity=70%,freq =200KHZ)
0
5
10
15
20
25
30
35
40
45
0 50 100 150 200 250 300
Series1
Series2
Series3
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Fig 10 : humidity vs attenuation(constant temperature)
Series 1(temperature =30C,freq =200KHz)Series 1(temperature =60C,freq =200KHz)
Series 1(temperature =90C,freq =200KHz)
Fig 11 : frequency vs attenuation
Series 1(relative humidity =50% ,temp = 30C)
Series 1(relative humidity =50% ,temp = 90C)
Series 1(relative humidity =80% ,temp = 90C)
From the above graph it can be seen that themax attenuation occurs in usable range
is 40 dB/meter (at temp =240C and relative humidity = 20%). It means for flight of2m the
SPL will be reduced to 80dB.
due to geometric spreading(inverse square law) there is also decreased intensity
0
5
10
15
20
25
0 20 40 60 80 100 120
t =30
t =60
t = 90
0
2
4
6
8
10
12
14
0 50 100 150 200 250
Series1Series2
Series3
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Thus for r =2meter, I = 0.00398 watt/m2
For crystal of surface area A = 0.0001m2 , power = 0.00398*0.0001 = 3.98*10^ -7watt
Sound power = 10 * log(P/10^-12)dB = 56dB
Therefore, for proper detection ofecho it should be used at conditions where attenuationis low. At higher temperatures and lower relative humidity the attenuation increase
significantly.
Compensation
Compensation should be done for followingy change in environmental conditions that make changes in sound velocityy power supply compensationSince the change in ultrasound propagation velocity is not always linear, this could be
considered in measuring method. A method of comparing time can be used ,in which pulse
allowed to passed through measuring channel(transducer to level) as well as reference
channel(transducer to reflector), whose distance is known to us.
Where constant, and time interval corresponds to propagation time ofultrasonic wave in measuring and reference channel respectively.
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Mathematical modeling ofCrystal Tx/Rx
Effortvariables - voltage and force
Flowvariables - current and velocity
Two part network provides good representations of piezoelectric transducer transmitters
and receivers because piezoelectric effect is reversible.
Fig 12 : two port network for piezo transducers
Direct effect q = df
Inverseeffect x = dv
F = kx
Where, x = deformation of crystals
k = stiffness
Thus q = dkx
Hence, gives theRelation between flowvariables for piezo-crystal
Similarly ,
Hence
Hence , F = kdv gives the Relation between effort variables for
piezo crystal
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Fig 13 : two port network for piezo transducers(Detailed form)
The crystal has electric capacitance C and corresponding electric impedance is purely
capacitive.
Mechanics of crystal-
y Mass (Resistance )y spring(Stiffness k)y Damping(inertance M)
Themechanical circuit consists ofm,1/k and in series and has mechanical impedance
Zm(s) = ms + + k/s
Mechanical system -
Inertia force = Damping force + Spring restoration force + Applied force
md2x / dt
2= -fdx / dt - kx + f(t)
f(t) = md2x / dt
2+ fdx/dt + kx
Taking Laplace transform, assuming initial condition for
ms2
x(s) + fs x(s) + k x(s) = f(s)
x(s) / f(s) = 1 / ms2
+ fS + k
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Electrical sytem -
e(t) = Ri(t) + L di(t) / dt + 1/c i(t)dt
Taking Laplace transform
E(s) = RI(S) + SLI(s) + 1/SC I(S)
I(s) / E(s) = 1/ ((1/SC) + SL + R)f(t) = m (d
2x / dt
2) + dx/dt + kx
e(t) = L (d2q/dt
2) + R dq/dt + q/c
Mechanical term to Electric terms-
Force(f) Voltage (V)
Inertia / mass (m) inductance(L)
Variable friction / damping() Resistance (R)
Spring stiffness (K) Reciprocal of capacitance (1/C)
Displacement (x) Charge (q)
Velocity(x) current (i)
VS (s) = ZG i(S) +( 1 / CS )* i(S)
i(S) / Vs(S) = 1 /( ZG + 1/ CS )
If ZG = 0 then VS(S) / i(S) = 1/CS
Fin(S) = (ms + + k/s) * x(s)
Fin(s) = dKV(s)
dkV(S) = (mS + + k/s) * x(s)
x(s) / V(s) = dk / mS + + (k/s) + ZM -------------(1)
x = dx/dt
x(s) = Sx(s)
x(s) / x(s) = 1/CS ----------(2)
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Multiplying equation 1 and 2
(x(s) / v(s)) * (x(s) / x(s)) = dk / S(ms + + k/s z Min)
V(S) = (1/CS) / (ZG + 1/ CS) * Vs(S)
Vs(S) = (ZG + 1/CS) / (1 / CS) * V(S)
X(S) / Vs(S) = dk / S(ms + + k/s + zMin) * 1/CS / ZG + 1/CS ---------( 3)
Special case ZG = 0
ZMin = 0
X(S) / Vs(S) = dk / S(ms + +k + Z Min)
= d / ( (ms2 / k) + (S/k) +1 )
Thus transmitter transfer function is presented as
=
n = mechanical natural frequency
= damping ratio
Q = Quality factor = 1 / 2
Fout(S) = (ms + + k/s + ZMout) * x(s)
x(s) / Fout(S) = 1/ (ms + + k/s ++ Z Mout) --------------(4)
Vout(S) = (ZL / CS) / (ZL + 1/CS) * dk x(s)
Vout(S) / x(S) = dk (Z L / ) --------------5
multyplying 4 and 5
Vout(S) / Fout(S) = dk / (ms + + k/s + ZMout) * (ZL / ZLCS + 1)
special case ZMout 0 and ZL = RL
Vout(S) / Fout(S) = d / ( ms/k + /k + 1/S)SC * R LSC / RLSC + 1
= d / ( C(ms2/k + s/k) ) * S / S+ 1
let RLC = s
then receiver transmitter function can be presented as
=
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Transmitter responseThe transmitter transfer function is found to be
=
Threeessential requirements for transmitter system-
The sound wave launched from the rear face of the transducer must not enter thegas ,otherwise pulse transit timewill be confused
The overall Q should be ideally around 0.7 to minimize pulse distortion Maximum power should be transmitted from the front face of the transducer into
the gas
Programming in MATLAB to find response
d = input('enter piezoelectric charge constant');
%piezo charge constant d= 550*10^-12 C/N
wn = input('enter mechanical resonance frequency');
%mechanical resonance freq.
zi1 = input('enter damping factor');
zi2 = input('enter corrected damping factor');
numt = [d*50]; % for a 50 V excitation
a = 1/wn^2;b = (2*zi1)/wn;
c = (2*zi2)/wn;
% zi = damping coefficient = 0.7(as optimum for pulse system)[we have to get zi = 0.7 by
backing layer of
% absorbing material, currently let zi = 0.0077 ,Q(mechanical quality
% factor) = 65 and Q = 1/2? ]
dent1 = [a b 1]; % [1\wn^2 2*zi1\wn 1]
dent2 = [a c 1]; % [1\wn^2 2*zi2\wn 1]
%syst indicates transmitter transfer functionsyst = tf(numt,dent1)
syst1 = tf(numt,dent2)
step(syst,'r',syst,'b')
Result
enter piezoelectric charge constant 550*10^-12
enter mechanical resonance frequency 3.1*10^5
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enter damping factor 0.0077
enter corrected damping factor0.7
Transfer function:
2.75e-008
---------------------------------
1.041e-011 s^2 + 4.968e-008 s + 1
Transfer function:2.75e-008
---------------------------------
1.041e-011 s^2 + 4.516e-006 s + 1
Fig 14:response of Tx with
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Fig 15:response of Tx with
(a)
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(b)
(c)
Fig16:response of Tx with (blue)Receiver responseThe receiver transfer function is found to be
= % ultrasonic receiver function
clear all;
d = input('enter piezoelectric charge constant');
%piezo charge constant d= 550*10^-12 C/N
wn = input('enter mechanical resonance frequency');
C = input('input XTAL capacitance');
R = input('enter the load resistance');
%mechanical resonance freq.
zi1 = input('enter damping factor');
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zi2 = input('enter corrected damping factor');
tau = R*C;
temp = d/C
numr = temp * [tau 0];
a = 1/wn^2
b = (2*zi1)/wn
c = (2*zi2)/wn
% zi = damping coefficient = 0.7(as optimum for pulse system)[we have to get zi = 0.7 bybacking layer of
% absorbing material, currently zi = 0.0077 ,Q(mechanical quality
% factor) = 65 and Q = 1/2zi ]
denra = [a b 1]% [1\wn^2 2*zi1\wn 1]
denrb = [a c 1] % [1\wn^2 2*zi2\wn 1]
dentemp = [tau 1];
denr = conv (dentemp,denra)
denr1 = conv (dentemp,denrb)
%sysr indicates receiver transfer functionsysr = tf(numr,denr)
sysr1 = tf(numr,denr1)
step(sysr,'r',sysr1,'b')
legend('with zi =0.77','with zi =0.7');
Result
enter piezoelectric charge constant 550*10^-12
enter mechanical resonance frequency 3.1*10^5input XTAL capacitance29*10^-9
enter the load resistance47000
enter damping factor 0.0077
enter corrected damping factor0.7
temp =
0.0190
a =
1.0406e-011
b =
4.9677e-008
c =
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32
4.5161e-006
denra =
0.0000 0.0000 1.0000
denrb =
0.0000 0.0000 1.0000
denr =
0.0000 0.0000 0.0014 1.0000
denr1 =
0.0000 0.0000 0.0014 1.0000
Transfer function:
2.585e-005 s
------------------------------------------------
1.418e-014 s^3 + 7.812e-011 s^2 + 0.001363 s + 1
Transfer function:
2.585e-005 s
------------------------------------------------1.418e-014 s^3 + 6.166e-009 s^2 + 0.001368 s + 1
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Fig 17 : response ofRx with (blue)
Fi 18 i f f T (l f ) d R ( i h )