Ultrasonic Sensor for Level Measurement

8/8/2019 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)



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)



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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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 )

Ultrasonic Sensor for Level Measurement

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Transcript of Ultrasonic Sensor for Level Measurement