let’s compare the far-field response with the on axis ...the plots for the two cases are shown...

Oelze ECE/TAM 373 Notes - Chapter 7 pg 27

Let’s compare the far-field response with the on axis response. From the near-field on axis response

( )2

2,0 2 sin 1 12o okr a

p r cUr

ρ = + −

The axial pressure amplitude for the far field (Fraunhofer zone) is given by (for θ = 0)

( ) ( ) ( )212 sin

, ,2 sin

j t kro o J kaj a Up r t e

r kaω θωρ

θθ

− =

so

( )2

,0,02o oa U

p rr

ωρ=

( ) 2

2o okc a Ur

ρ=

12 o o

acU ka

rρ =

(Eq 7.4.7)

Normalizing axial-only ( ),0p r and far-field axial ( ),0 ,0p r to 2 o ocUρ gives:

( ) 2

2

, 0sin 1 1

2 2o o

p r kr acU rρ

= + − and

( ),0,0 12 4o o

p r aka

cU rρ =

Plotting these two values

where the solid line represents 3aλ = , the dashed line represents the 7a

λ = . Did one reduce to the

other? Why not?So,

( ) ( ) ( )121, 2 sin

2

kj R rj t

o ok

p r t j cU e e R rωρ− + = −

where 2 2

1R r a= +

and

( ) ( ) ( )212 sin

, ,2 sin

j t kro o J kaj a Up r t e

r kaω θωρ

θθ

− =

Homework EXERCISE: Show that these two expressions are the same, on-axis, in the FraunhoferZone.


*************************** Example 7.4 ***************************Compare and contrast the directivity and number of side lobes for the case of a continuous-linesource of length L, and a circular transducer of radius L excited at the same frequency and into thesame medium.

ANSWER: The equations for the line source and circular transducer are listed below:

( ) sinsinc

2LkL

Hθ

θ =

(Line source)

( ) ( )12 sinsinC

J kaH

kaθ

θθ

=

(Circular transducer)

The plots for the two cases are shown below. Note that the continuous-line source has the x-axis

“expanded out” by a factor of two because of the factor of 0.5 in the expression ( )1sin

2kL θ . Thus,

one can immediately see that the circular transducer (i.e. the circular piston source) has greaterdirectivity, but at the same time has more side lobes than the continuous-line source. Also note thatL for the line source is the “full length” whereas L (a) for the piston source is “1/2 the length.”

1510500-80

-70

-60

-50

-40

-30

-20

-10

0

10Continuous-line SourceCircular Piston Source

Log Comparison of Directivity

kL*sin(theta)

20 lo

g |H

(the

ta)|

15105000.00

0.20

0.40

0.60

0.80

1.00 Continuous-line SourceCircular Piston Source

Direct Comparison of No. of Sidelobes

kL*sin(theta)

|H(t

heta

)|

How would the plots be different if L D= , the diameter of the circular transducer?

******************************************************************


The far field pattern can be modified by changing U0 over the face of the source. For example, if U0

decreases toward the edges of the circular piston then the side lobe levels will decrease and the mainlobe will be broader.

Ex 1 – High frequencies in a loudspeaker move the center more than outer parts.

Ex 2 – “Shading” of a piezoelectric transducer.

Transducer dimension (thickness) changes in response to electric field. “Shading” lowers the fieldin the outer portions of the disk.

Beamwidth

Defined as angular width where the intensity is down a specified amount from that on the axis.

Ex 0.5 → 3 dB down0.25 → 6 dB down0.1 → 10 dB down

Thus if θ = 13° for the acoustic amplitude down by 10 dB then the –10 dB beam width is 2θ = 26°.


*************************** Example 7.5 ***************************Computation of –10 dB beamwidth

For –10 dB the intensity directivity function will have a value of 0.1

Thus, 2

12 ( )0.1

J xx

=

, where x = ka sin(θ). This occurs at approximately x = 2.74.

Consider two different values of ka

For ka = 7, θ = sin -1 2.74

7

= 23o , so the beamwidth is 2θ = 46°

For ka = 21, θ = sin -1 2.74

21

= 7.5o , so the beamwidth is 2θ = 15°

******************************************************************

Alternatively, at a specific distance from the source the beam width might be specified in termsof a linear distance as opposed to an angular width.

I

I0

0.5 I0

1/2 power beam width

Beam patterns for circular sources with uniform velocity amplitude distributions.

6 z03 z0z01/2 z00

p

2θ1

SphericalWaves-Field

y

Sound field of a circular piston. 2a/λ = 16.

Transverse pattern in near field is very irregular.


(7.5) Radiation Impedance

An electromechanical transducer has an electrical impedance which is, through someelectromechanical coupling parameters (depends on type of transducer), directly dependent uponthe mechanical parameters of the transducer and its load.

∴ Total mechanical impedance = mechanical impedance of transducer (radiating into a vacuum)plus the radiation impedance of the fluid load

dfs

comp. ⊥ surface

zr =

df→

sr u ∫

The mechanical impedance of the transducer is analogous to discussion from Chapter 1 for the massloaded, damped driven harmonic spring system.

Piston → mass mmechanical resistance Rm

stiffness s

Driven by external force f = Fe jωt to give 0 0 0 for simple harmonic time dep.j tu U e jω ωξ= =

r

r 0 force of fluid = usf z= % where r z% is the radiation impedance

We can plug all of this into a second order differential equation (just like the damped, drivenharmonic oscillator)

20 0

0 2– – –s md d

f f R s mdt dtξ ξ

ξ =r r

or

00 0s m

duf f R u s u dt m

dt= + + +∫

We can assume a simple harmonic time dependence like the driving force so that

( )

0

0

s m

r m

sf f R j m u

j

z z u

ωω

= + + +

= + %% %

where


m ms

z R j mj

ωω

= + +%is the mechanical impedance. Hence,

0m r

fu

z z=

+

%% % %

dividing by e jωt gives

0m r

FU

z z=

+% %where

( ) N–s

mj

r r r rz z e R jXθ= = +%↑ ↑

radiation radiationresistance reactance

Radiation resistance – loss term for the systemtwo loss terms – Rm → loss in transducer

Rr → radiated energy

Power radiated 20

12r rU RΠ =

Let’s look at the Simple Source

( )

20

20

( ) unbaffled, where S is the surface area

4baffled simple source( )

twice and 2

r

rr r

c kSR

c kSR

R

ρπ

ρπ

=

=Π

_________________

Looking at

–

total m r

m r r

Z Z Z

sR j m R jXω

ω

= +

= + + +

a positive Xr means that we are loading the system with mass from the surrounding fluid so that theresonance frequency is decreased for the oscillator.

Thus, we define a radiation mass rr

Xm

ω= .

Since the resonance frequency of the system is


0 0r

s smass m m

ω ω= ⇒ =+

The radiation mass reduces the resonance frequency. mr is related to mass of material moved infront of source. Less mass at higher frequency.What does this mean for a transducer operating in air compared to a transducer operating in water?

Let’s look briefly at one particular case: The Circular Piston

( )( )

0 1

0 1

2

2r

r

R cSR ka

X cSX ka

ρ

ρ

=

=

where

( ) ( )

( ) ( )

11

11

2 22 1

22 2

22

J kaR ka

kaH ka

X kaka

= −

= (To be shown in homework)

and 2S aπ= (the area of the piston face).

1.0

0.5

00 4 8 12

x

X1(x)

R1(x)

Figure 7.5.2 Radiation resistance andreactance for a plane circular piston ofradius a radiating sound of wavenumber k(x = 2ka).

For ka << 1 (a small compared to λ)

We use the series expansion for J and H to get2

0

0

0

1( )

28

( )383

r

r

r

R cS ka

X cS ka

am S

ρ

ρπ

ρπ

≈

≈

≈

Physically, this means that the piston appears to be loaded with a cylindrical volume of fluid masswith cross sectional area S (the same as the piston surface) and a depth of the fluid mass cylinder

reaching to 8

0.853

az a

π= ; .We see how the fluid changes the resonant frequency.


For ka >> 1 (a large compared to λ)Again we use the series expansion for J and H at large ka.

0

0

02

(1)2

0 for very large

2 0 for very large

r

r

r

R cS

X cS kaka

am ka

k

ρ

ρπ

ρ

≅

≅ ⇒

≅ ⇒

Thus, at large ka oraλ

the radiation impedance is almost real with no mass loading

So, the power is given by2 2

00 0

1 12 2r rU R cSUρΠ = ≅ .

This is the same power as a plane wave propagating through area S with characteristic impedance0cρ .

*************************** Example 7.6 ***************************Calculate Rr, Xr and mr in air and water for a circular piston source

Take a = 3 cm, f = 10 kHz, and U0 = 1 m/s

Air44 10

2 0.03 11343

kaπ

= =

Water44 10

2 0.03 2.551481

kaπ

= =

Parameter Air Waterπ a

2ρ0c 1.173 4,128

2ka 11 2.55R1 1.0321 0.6375Rr 1.21 2,632 N–s

m

Πr = 20

12 rU R

0.61 1,316 (W)

X1 0.1464 0.693Xr 0.172 2,861mr 2.73 × 10–6 4.56 × 10–2 (kg)

Physically it takes a lot more work done against Rr in water than in air. Hence, the more powerexpended in water than in air by the transducer.

******************************************************************


(7.6) Fundamental Properties of Transducers

(a) Directional Factor and Beam Pattern

The general expression for beam pattern, ( , )b θ φ , represents the variation of intensity level withangle. The expression is given by

( )),(log20

)(),,(

log20

)(),,(

log10),(

φθ

φθ

φθφθ

H

rprP

rIrI

b

ax

ax

=

=

=

Note: is generally a function of both θ and φsince it may not have circular symmetry.

(b) Beam WidthDiscussed previously – must specify criterion: half-power, quarter power, 0.1 etc. Rememberin some cases, for a specified distance from source, can use a linear measure.

(c) Source Level

Source Level = SL(Pref) = 20 log (1)e

ref

PP

where Pe(1) is effective axial pressure at r = 1 m,

extrapolated back from far field. Must specify reference pressure

Pref = 1 µPa, 20 µPa, 1 µbar

(d) Directivity

directivity = D = 2

2

nonspherical

( ) directional( )source

( )(spherical) ( )

ax

ax

SS

I rP r

I r P r

=

for a simple source having the same acoustic power as the directive source. For the simplesource

( )2 2

0

14

2s Sr P r

cπ

ρΠ =

and there is no preferred direction. For the directional source

2 2

40

1( , , )

2directional P r r dc π

θ φρ

Π = Ω∫ where dΩ = sin(θ) dθ dφ and the integration is

over 4π steradians (φ varies from 0 to 2π and θ from 0 to π)


( , , ) ( ) ( , )axP r P r Hθ φ θ φ=

2 22

40

1( ) ( , )

2directional axr P r H d

c πθ φ

ρΠ = Ω∫

Since the acoustic powers are the same, we can solve for 2 ( )ax

P r and 2 ( )S

P r in terms of the

acoustic power giving

02 2 2( )

422 0

42

2( , ) 4

2 ( , )4

rax

S

cP r H d

Dc H dP

r

π

π

ρθ φ πρ θ φ

π

ΠΩ

∴ = = =Π Ω

∫∫

The larger the ratio a/λ the more directional the beam since 2

4( , )H d

πθ φ Ω∫ is smaller.

(e) Directivity Index

Directivity Index = DI = 10 log D

Directivity of baffled piston source

222

140

4 4( , ) 2 ( sin )

2 sinsin

DH d J ka

dka

ππ

π πθ φ θ

π θ θθ

= =Ω

∫ ∫

2

1

( ) again Bessel functions are tabulated in appendix A6

(2 )1–

kaD

J kaka

=

for ka >> 1

D ≅ (ka)2

which is greater for greater a/λ, or the beam is more highly directive when the frequency isincreased


*************************** Example 7.7 ***************************Examples of directivity calculations.

10Take 1 10 100

215

2 15 2252

a ka D

a ka D

λπ

λπ

= > = ≅

= > = ≅

Thus the source starts to become quite directive even for a/λ not much greater than 1.

******************************************************************Reversible Transducers

Ex. speakers ⇒ intercomultrasonic transducers

Transmit and receiving patterns are the same.

H(θ,φ)receive = H(θ,φ)transmit

(f) Rough Estimates of Radiation Patterns

Extent of Near Field

The extent of the near field for a source is defined by the two distances rmin (the distance from thefield point to the closest element of the source) and rmax (the distance from the field point to thefurthest element of the source).When the field point is moved towards the source such that rmax – rmin ≈ λ 2 then the axialpressure is sufficiently shifted from the far field axial response to alter the far field axial pressureamplitude. The above relation can be shown through geometry to imply the extent of the near fieldresponse is quantified by

min 14

r LL λ

≅ or

2

2

min2

4

LL

rλ λ

= =

where L is the greatest extent of the source.


For a rectangular source:

L1

L2

L2 > L122

min 4

Lr

λ=

First angular null

1sinLλ

θ ≅

θ1 decreases with L decreases with frequency

L

λ

θ

λ2

phase cancellation(ON Transmit and ON Receive)

Estimate of Directivity

4 determine solid angle for main lobe–for side lobes low

effective

Dπ

≈ ∴Ω

Text pressure down 0.5 (–6 dB) as a good estimate

Piston like source

21 2

14

D k L L≅

increases with dimensions andfrequency L1

L2

let’s compare the far-field response with the on axis ...the plots for the two cases are shown...

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