Ultrasound lecture 1 post

Advanced Imaging 1024

Jan. 7, 2009 Ultrasound Lectures

History and tour Wave equation Diffraction theory Rayleigh-Sommerfeld Impulse response Beams

Lecture 1:Fundamental acousticsDG: Jan 7

Absorption

Reflection

Scatter

Speed of sound

Image formation:

- signal modeling- signal processing- statistics

Lecture 2:Interactions of ultrasound with tissue and image formationDG: Jan 14

The Doppler Effect

Scattering from Blood

CW, Pulsed, Colour Doppler

Lecture 3: Doppler Ultrasound IDG: Jan 21

Velocity Estimators

Hemodynamics

Clinical Applications

Lecture 4: Doppler US IIDG: Jan 28

Lecture 5: Special TopicsMystery guest: Feb 4 or 11

ULTRASOUND LECTURE 1

Physics of Ultrasound Waves: The Simple View


Physics of Ultrasound: Longitudinal and Shear Waves


Physics of Ultrasound Waves: Surface waves


Physics of Ultrasound Waves1) The wave equation

y

z

u

t

ux

u+Δu

Particle Displacement =

Particle Velocity =

Particle Acceleration =

u

x

vtu =

∂∂

t

v

t

u

∂∂=

∂∂

2

2

Equation of Motion

dV

dxdydzxp

∂∂+

Net force = ma p = pressure

= P – P0

=ρ densityt

vdVdV

x

p

∂∂=

∂∂ ρ

ort

vp

∂∂=∇ ρ (1)

dydzpdydzp

Definition of Strain

,xSxxu

u ∆=∆∂∂=∆ S = strain

Also

TermsNonlinear

SC

SB

SAp ...!3

32 +++=

bulk modulusxu

∂∂=

Taking the derivative wrt time of (2)

tS

∂∂

=xv

∂∂

(4)

(2)

(3)

Substituting for from Eq (1) (in one dimension)v∂

txp

xtS ∂

∂∂

∂∂=

∂∂

ρ1

2

2

2

2 1

x

p

t

S

∂∂=

∂∂

ρ

Substituting from

2

2

2

2

x

pA

t

p

∂∂=

∂∂

ρ

or pct

p 2202

2

∇=∂∂

; ==ρA

c0 wave velocity

3 dimensions

(3)

(5)

For the one dimensional case solutions are of the form

( ) )(, kxtjetxp −= ω; fk πω

λπ

22 == (6)

for the forward propagating wave.

A closer look at the equation of state and non-linear propagation

Assume adiabatic conditions (no heat transfer)

P = P0=

γ

ρρ

0

P0

γ

ρρρ

−

+0

01

Condensation = S'Gamma=ratio of specific heats

P = P0[ ] γ'1 S+ ,

v

p

c

c=γ

Expand as a power series

P = P ( ) 30

'0

'0 '

!3

11

2

1 2

SCPSPSP

BA

+−++ γγγ0

. . . .

B/A = ( )

11 −=− γ

γγγ Depends solely on

thermodynamic factors

Material B/A

Water 5

Soft Tissues 7.5

Fatty Tissues 11

Champagne

(Bubbly liquid)

( )

+−≅

00

),(,

v

txv

l

xkxwtSin

v

txv

Additional phase term small for small xand increasingly significant as

= shock distance

(8)

In terms of particle velocity, v, Fubini developed a non linearsolution given by:

lx⇒l

Nonlinear Wave Equation

202

2

)( vct

v β+=∂∂

2

2

x

v

∂∂

(7)

AB

21+=β

kk

c

vl

βεβ

11

0

0

=

=

Mach #

(9)

ε=

• At high frequencies the plane wave shockdistance can be small.

• So for example in water:

5.3=β MHzf 5.30 = MPap 10 =

Shock distance = 43 mm

Shock Distance, l

Where

=lnx

Jnxl

B nn2

(11)

Thus the explicit solution is given by

( ) ( )∑∞

=

−=10

2n

n kxtnSinlnx

lnxJ

v

v ω (12)

We can now expand (Eq. 8) in a Fourier series

( )[ ]∑∞

=−=

10

nn kxtnSinBvv ω

0/ vv

(10)

Hamilton and Blackstock Nonlinear Acoustics 1998

Aging of an Ultrasound Wave


lx /

Re

lativ

e A

mpl

itud

eHarmonic Amplitude vs Distance

(narrow band, plane wave)

Focused Circular Piston

2.25 MHz, f/4.2, Aperture = 3.8 cm, focus = 16 cm



Propagation Through the Focus

Nonlinear Propagation: Consequences

_______________________

• Generation of shock fronts

• Generation of harmonics

• Transfer of energy out of fundamental

RADIATION OF ULTRASOUND FROM AN APERTURE

We want to consider how the ultrasound propagates in the field of the transducer. This problem is similar to that of light (laser) in which the energy is coherent but has theadded complexity of a short pulse duration i.e. a broad bandwidth.

Start by considering CW diffraction theory based on the linear equation in 1 dimension

pct

p ∇=∂∂ 2

02

2

Laplacianyx

=∂∂+

∂∂+

∂∂=∇

2

2

2

2

2

22

δ

The acoustic pressure field of the harmonic radiator can be written as:

( ) ( ){ }tjerPtrp ω⋅=Re,

Where is a complex phasor function satisfying the Helmholtz Equation

( )rP

(13)

( ) ( ) 022 =+∇ rPk (14)

To solve this equation we make use of Green’s functions

( )∏

=41

rP dsnG

PGnP

s aa

∂∂−⋅

∂∂∫ (15)

s = surface area n∂∂ / = normal derivative

( )11, yxPPa = = pressure at the aperture

Rayleigh Sommerfeld Theory

Assume a planar radiating surface in an infinite “soft” baffel

Aperture

Field PointConjugate field point

n

'r

use ( )'~'

'~'

re

re

rGrjkjkr −

−=

'~r

as the Greens function

r

= 0 in the aperture

θ

( ) dsn

GPrP as

∂∂−∫=

π4

1

'

'

2r

ejkCos

rG

nG jkr

=∂∂=

∂∂ θ

( ) dsCosr

eP

jkrP

jkr

as θπ '

'

2∫−=

dsr

eP

j

jkr

a '

'1 ∫≅λ

(16)

Equation 15 can now be written:

Example

Consider the distribution of pressure along the axis of aplane circular source:

radius = a

22 σ+=′ zr

σd

σz

From Equation (16) in r, , z coordinates

∫+

+−=a

az d

z

zjkejkPP

0)( 2

22

22

2σπσ

σ

σ

π

The integrand here is an exact differential so that

azjk

az jke

jkPP

0

)(

22

−=

+σ

θ

[ ]22

)( azjkjkza eePzP +−=

The pressure amplitude is given by the magnitude of this expression

[ ] tja ezaz

kSinPzP ω⋅

−+= 22

22)(

[ ]

−+= zazk

SinZ

PzI a 222

2

24)(

(17)

To look at the form of (17) find an approximation for:

zz

azzaz −

+=−+

2

2222 1

zz

az −+=

2

2

1

za2

2

≅

zz

az −

+≅

2

2

21

( )

=∴

z

kaSin

z

PzI a

4

4 22

2

Maxima24

2 ∏= mz

ka m = 1, 3, 5, . . .

242 2 ∏∏ = m

za

λ

ma

zormza

λλ

22

==

Minimam

az

λ

2

= ; m = 0, 2, 4, 6

∞=z

0 50 100 150 2000.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

ka2/(4z)

a = 5 mmfrequency = 5 MHz

Inte

nsity

* 4

p a2 /Z

Axial Distance (mm)

Eq. 17

mma

z 3.832

==λ

M: 54 3 2 1 0

THE NEAR FIELD (off axis)

( ) ( ) 210

210

2' yyxxzr −+−+=

2

210

2

210 )()(

1z

yy

z

xxz

−+−+=

1y

1x 0y

0x

z'r

Fresnel approximation (Binomial Expansion)

++

++≅

2

10

2

10'

2

1

2

11

z

yy

z

xxzr (19)

From (16) we have

( ) 2110

20

210 2 xxxxxx +−=− 2

1102

02

10 2 yyyyyy +−=−

( ) ( ) ( ) ( )[ ]11

21100

210

210

,1

, dydxeeyxPzj

yxPyyxx

z

jkjkz

a

−+−∞

∞−∫∫=

λ

Note that the r’ in the denominator is slowly varying and is therefore ~ equal to z

Grouping terms we have

z

jk

e−

⋅[ ]0101 yyxx +

[ ]

)(

200

20

20

),(

zK

yxz

jkjkz

ezj

eyxP

+=

λ( ) [ ]2

12

121,1

yxz

jk

eyxP+

∫∫

11dydx

( ) ( ) ( ) ( ) ( )11

221100

11

21

21

,, dydxeeyxPKyxP yxjyxz

jk

zyx υυπ +−+

∫∫=

Wherez

xx λ

υ 0=z

yy λ

υ 0= (20)

We can eliminate the quadratic term by “focusing” the transducer

Thus the diffraction limit of the beam is given by:

( ) ( ){ }1100 ,, yxPyxP zℑΚ=

Consider a plane circular focused radiator in cylindrical coordinates

( )σ,zPσz

radius a

Circular Aperture

( )

ℑΚ=

ar

CirczP zσ,

( )xxJ

zk ππ1'= Where

za

xλ

σ2=

FWHM a

z

241.1

λ=

x

xJ

ππ )(

2 1az 222.1 λ

(21)

Square Aperture

Need to consider the wideband case. Returning to Eq. 16we have:

( ) ( ) 11'1100

'

,2

, dydxr

eyxP

jkyxP

jkr

∫∫= −

π

cck

v ωλ

=== ∏∏ 22

( ) ( ) ωπω

ω

ddydxr

eyxP

c

jyxP

rcj

11

'

1100 ',

2, ∫∫−∫=

( ) ωω

ω

πddydx

r

eyxP

jrcj

c 11'11

'

,2

∫∫∫= −

This is a tedious integration over 3 variables even aftersignificant approximations have been made

There must be a better way!

Impulse Response Approach to Field Computations

Begin by considering the equation of motion for an elemental fluid volume i.e. Eq. 1

tv

p∂∂=∇− ρ (22)

Now let us represent the particle velocity as the gradientof a scalar function. We can write

ϕ⋅∇=v

Where is defined as the velocity potential we are assuming here that the particle velocity is irrotational

i.e. 0=×∇ v

~ no turbulence

~ no shear waves

~ no viscosity

Rewrite as(22)

tp

∂∂∇=∇− ϕρ

∂∂+∇=t

pϕρ0

tp

∂∂−= ϕρ (23)

ϕ

The better way: Impulse response method

r

'r

( )tV0

( ) ( )t

trtrp

∂∂−= ,

,ϕρ

•

ds

( ) ( )ds

r

crtVtr

'

'

2

1, 0 −∫∫= πϕ

s

( ) ( )trhtV ,0 ∗=

Impulse Response

where

( ) ( )ds

r

crttrh

'

'

2

1,

−∫= δπ (24)

Thus

( ) ( ) ( )trht

tVtrp ,, 0 ∗

∂∂−= ρ (25)

Useful because is short!convolution easy

Also is an analytic functionNo approximations!

Can be used in calculations

You will show that for the CW situation

( )trh ,

( ) ( ){ }0

,, 00 ωωωρ =ℑ−= trhvjtrp (26)

( )trh ,

IMPULSE RESPONSE THEORY EXAMPLE

Consider a plane circular radiator

z

2r

0r

'rσ

1r

1σ

is the shortest path to the transducer

near edge of radiating surface

far edge of radiating surface

0r

1r

2r

σdrlds ⋅= )( '

s( ) ( )sd

r

crttrh

'

'

2

1,

−∫= δπ

σd

'

'

'rd

drSin r

σσ

θ ==

'rd

σd

σ'r

'rθσ

σ''rdr

d =

( )σ

''' drrrlds =

Also letcr '

=τSo that

( ) ( ) ( )σ

τδπ

''

'2

1,

rrl

r

ttrh

−∫=⋅

cdτ

( ) ( )σπ2

,cctl

trh = (27)*

* a very powerful formula

( ) σπ2' =rl while the wavefront lies between

21 rr and

Thus we have: (next page)

( ) cc

trh ==πσσπ

2

2,ie

( ) =trh ,

c

rt 00 <<0

( )

−

−+−−21

02

221

20

21

))((2 rctr

arctCosc ρπ

cc =

σσ

ππ2

2

0

c

rt

c

r 10 <<

c

rt

c

r 21 <<

c

rt 2>

Planar Circular Aperture

( )caz

tcz

ctrh22

,+<<=

0= otherwise

hz small z large

cz

caz 22 +

t

Consider the on axis case:

01 =σ zr =0 21 rr =

Recall Equ 26

( ) ( )trht

Vtrp ,, 0 ∗

∂∂−= ρ

so that the pressure wave form is given by

( )tzP ,

t

),( trh z small z large

Off axis case

( )tzh ,, 1σ

( )tzp ,, 1σ

c

r0c

r1c

r2

Off axis case5 mm radius disk, z = 80 mm

Spherically focused aperture- relevant to real imaging devices

Spherically focused aperture impulse response

Spherically focused aperture impulse responses

Spherically focused aperture pressure distribution

a. f/2b. f/2.4c. f/3

Frequency = 3.75 MHz

Health & Medicine

Ultrasound lecture 1 post