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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
ULTRASOUND LECTURE 1
Physics of Ultrasound: Longitudinal and Shear Waves
ULTRASOUND LECTURE 1
Physics of Ultrasound Waves: Surface waves
ULTRASOUND LECTURE 1
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
Hamilton and Blackstock Nonlinear Acoustics 1998
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
Hamilton and Blackstock Nonlinear Acoustics 1998
Hamilton and Blackstock Nonlinear Acoustics 1998
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