Advanced Imaging 1024
Jan. 7, 2009
Ultrasound Lectures
Lecture 1:Fundamental acoustics
DG: Jan 7
History and tour
Wave equation
Diffraction theory
Rayleigh-Sommerfeld
Impulse response
Beams
Lecture 2:Interactions of ultrasound
with tissue and image formation
DG: Jan 14
Absorption
Reflection
Scatter
Speed of sound
Image formation:
- signal modeling
- signal processing
- statistics
Lecture 3: Doppler Ultrasound I
DG: Jan 21
The Doppler Effect
Scattering from Blood
CW, Pulsed, Colour Doppler
Lecture 4: Doppler US II
DG: Jan 28
Velocity Estimators
Hemodynamics
Clinical Applications
Lecture 5: Special Topics
Mystery 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 Waves
1)
The wave equation
x
u
z
u
y
x
t
Particle Displacement
Particle Velocity
Particle Acceleration
u+Δ
u
= u
u
v
=
t
 2u v
=

2
t
t
Equation of Motion
p
p dydz +
x
p dydz
Net force
= ma
or p  
v
t
(1)
dxdydz
p = pressure
= P – P0
  density
p
v
dV  dV
x
t
dV
Definition of Strain
u
u 
x  Sx ,
x
S = strain
u

x
bulk modulus
Also
p AS +
B
1
S2 +
C 3
S + ...
3!
(2)
(3)
Nonlinear Terms
Taking the derivative wrt time of
S v
=
t x
(2)
(4)
Substituting for v from Eq (1) (in one dimension)
S
 1 p

t
t
x  x
2S
1 2 p

2
t
 x 2
Substituting from
(3)
2 p
A 2 p

2
 x 2
t
2 p
A
2 2

c

p
 wave velocity
or
; c0 
0
2

t
3 dimensions
(5)
For the one dimensional case solutions are of the form
2
j ( t kx )
(6)
k

  2 f
;
p x, t  e

 
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
 

 0


  P0

Gamma=ratio of specific heats

 
0
1 +
0








Condensation = S'

P = P0 1 + S
'


cp
, c
v
Expand as a power series
1
1
'2


+

+



+
P
S
1
P
S
CP0 S '3
P = P0
0
0
2
3!
'
A
B/A =
   1
  1

....
B
Depends solely on
thermodynamic factors
Material
Water
B/A
5
Soft Tissues
7.5
Fatty Tissues
11
Champagne
(Bubbly liquid)
Nonlinear Wave Equation
2
 2v
2  v
 (c0 + v)
2
t
x 2
(7)
B
  1+
2A
In terms of particle velocity, v, Fubini developed a non linear
solution given by:

vx, t 
x v( x, t ) 

 Sin wt  kx +
v0
l v0 

(8)
Additional phase term small for small x
and increasingly significant as x  l
l = shock distance
Shock Distance, l
1
1
l 

k
 v0 
  k
 c0 
Mach #
(9)

•
At high frequencies the plane wave shock
distance can be small.
•
So for example in water:
  3.5
f 0  3.5 MHz
Shock distance = 43 mm
p0  1 MPa
We can now expand v / v0 (Eq. 8) in a Fourier series
v v0 


n 1
Bn Sin n  t  kx 
2l
nx 

Jn  
Where Bn 
nx
 l 
(10)
(11)
Thus the explicit solution is given by

J n nx l 
v
 2
Sin n t  kx
v0
nx l
n 1
(12)
Aging of an Ultrasound Wave
Hamilton and Blackstock Nonlinear Acoustics 1998
Relative Amplitude
Harmonic Amplitude vs Distance
(narrow band, plane wave)
x/l
Hamilton and Blackstock Nonlinear Acoustics 1998
Focused Circular Piston
2.25 MHz, f/4.2, Aperture = 3.8 cm, focus = 16 cm
Hamilton and Blackstock Nonlinear Acoustics 1998
Propagation Through the Focus
Hamilton and Blackstock Nonlinear Acoustics 1998
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 the
added 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
2 p
2

c
0 p
2
t
2
2
2



2 
+ 2+
 Laplacian
2
2
x
y

The acoustic pressure field of the harmonic radiator
can be written as:

pr , t   Re Pr   e jt

(13)
Where Pr  is a complex phasor function satisfying
the Helmholtz Equation
2 + k 2  Pr   0
(14)
To solve this equation we make use of Green’s functions
P r  
1
Pa
G
4   s n  G  Pa n ds
s = surface area
(15)
 / n = normal derivative
Pa  Px1 , y1  = pressure at the aperture
Rayleigh Sommerfeld Theory
Assume a planar radiating surface in an infinite “soft” baffel
~
r'
Conjugate
field point
r'

r
Aperture
use
G r  
e
jkr'
e
 jk~
r'
 ~
r'
r'
as the Greens function
= 0 in the aperture
Equation 15 can now be written:
P r  
1
 G 
 s  Pa 
ds
4
 n 
jkr '
G G
e

Cos  2 jk
n
r
r'
 jk
e
Pr  
 s Pa ' Cos  ds
2
r
jkr '
jkr'
1
e

 Pa
ds
'
j
r
(16)
Example
Consider the distribution of pressure along the axis of a
plane circular source:
radius = a
d

r  z 2 +  2
z
From Equation (16) in r,  , z coordinates
P( z )
 jkPa a e


2 0
jk z 2 +  2
z2 +  2
2 d
The integrand here is an exact differential so that
 e jk z +
  jkPa 
jk


2
P( z )
2
a



0

P( z )  Pa e
jkz
e
jk z 2 + a 2

The pressure amplitude is given by the magnitude of this
expression
k
P( z )  2 Pa Sin 
2
 z + a  z  e
2
Pa
2 k
I ( z)  4
Sin 
Z
2
2

2
j t


z +a z 

2
2
(17)
To look at the form of (17) find an approximation for:
z2 + a2  z 
2

a
z 2 1 + 2   z
 z 
a2
 z 1+ 2  z
z

a2 
 z 1 + 2   z
 2z 
a2

2z
2


4 Pa
ka
2
 I z  
Sin 

z
 4z 
2
Maxima
ka 2

 m
4z
2
2  a2

 m
 4z
2
a2
a2
 m or z 
z
m
m = 1, 3, 5, . . .
a2
; m = 0, 2, 4, 6
Minima z 
m
z  
2.0
frequency = 5 MHz
a = 5 mm
1.8
1.6
2
Intensity * 4pa /Z
1.4
z 
1.2
1.0
M:
54 3
a2

2
 83.3mm
2
ka /(4z)
1
0
0.8
0.6
Eq. 17
0.4
0.2
0.0
0
50
100
150
Axial Distance (mm)
200
THE NEAR FIELD (off axis)
r '  z 2 +  x0  x1 2 +  y0  y1 2
y1
( x0  x1 ) 2 ( y0  y1 ) 2
 z 1+
+
2
z
z2
x1
r'
y0
x0
z
Fresnel approximation (Binomial Expansion)
2
2

x
+
x
y
+
y
1
1
 0 1
 0 1 
'
r  z 1 + 
 + 
 
 2  z  2  z  
(19)
From (16) we have
Px0 , y0 

1

  Pa x1 , y1  e e
jz 
x0  x1 2

jk

x0  x1 2 +  y0  y1 2
jkz 2 z
 x0  2 x0 x1 + x1
2
2

dx1dy1
y0  y1  y0  2 y0 y1 + y1
2
2
2
Note that the r’ in the denominator is slowly varying and is
therefore ~ equal to z
Grouping terms we have
jkz
e
P( x0 , y0 ) 
e
j z

jk 2
x0 + y0 2
2z

  Px1, y1  e

jk 2
x1 + y12
2z

K (z)
e

 jk x x + y y
1 0
1 0
z

dx1dy1
Px0 , y0   K  z    Px1 , y1  e
Where
x
x0

z

jk 2 2
x1 + y1
2z
y 

e

 j 2  x x1 + y y1
y0
z
We can eliminate the quadratic term by
“focusing” the transducer
Thus the diffraction limit of the beam is given by:
P x0 , y0    z P x1, y1 

dx1dy1
(20)
Circular Aperture
Consider a plane circular focused radiator in cylindrical
coordinates
radius a

P z, 
z
r

P  z,    z  Circ 
a

2a
' J 1 x 
 kz
Where x 
x
FWH
M
J1 (x)
2
x
z
 1.41
z
2a
1.22 z 2a
(21)
Square Aperture
Need to consider the wideband case. Returning to Eq. 16
we have:
P x0 , y0  
k 
Px0 , y0 
2



jkr '
jk
e
  Px1 , y1  ' dx1dy1
2
r
2v


c
c

j r'
c
j
e
 
  Px1 , y1 
dx1dy1d
2 c
r'


j
2 c
    Px1 , y1 
e

j r'
c
r
'
dx1dy1d
This is a tedious integration over 3 variables even after
significant 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
v
 p  
t
(22)
Now let us represent the particle velocity as the gradient
of a scalar function. We can write
v   
Where  is defined as the velocity potential we are
assuming here that the particle velocity is irrotational
i.e.
v  0
~ no turbulence
~ no shear waves
~ no viscosity
Rewrite (22) as

 p  
t
 


0   p +
t 


p  
t
(23)
The better way: Impulse response method
 
 r , t
pr , t    
t
V0 t 

r'
ds
r
 r , t 
V0 t  r ' c 
1


ds
2 s
r'
 
 V0 t  h r, t
Impulse Response
where
1  t  r ' c 
hr , t  

ds
2
r'
Thus
V0 t 
pr , t    
 h r, t
t
 
(24)
(25)
Useful because hr , t  is short!
convolution easy
Also hr , t is an analytic function
No approximations!
Can be used in calculations
You will show that for the CW situation
pr , t     j 0v0h r , t   0
(26)
IMPULSE RESPONSE THEORY EXAMPLE
Consider a plane circular radiator

d
r1
r'
r0
1
r2
r0 is the shortest path to the transducer
r1 near edge of radiating surface
r2 far edge of radiating surface
z


1  t  r' c
hr , t  
s
ds
'
2
r
ds  l ( r ' )  d
d r'
dr '


 '
d
r
Sin r '
d 
ds 
So that
dr ' r '
d
r'


r
 
l r ' r ' dr '
'

Also let

1  t    l r ' r 'd c
hr , t  

2  r '

r
 
c
'
l ct  c
hr , t  
2 
*
(27)
 
* a very powerful formula
l r '  2 while the wavefront lies between
r1 and r2 ie hr , t   2 c  c
2
Thus we have: (next page)
hr, t  
Planar Circular Aperture
0
2 c
c
2
2
2
2
2




ct

r
+


a
1
0
1
Cos 

2
1 2

2
r
((
ct
)

r
)
0


c
0
0t 
r0
c
r0
r
t  1
c
c
r1
r
t  2
c
c
t
r2
c
Consider the on axis case:
r1  r2
r0  z
1  0
h r, t   c
 0
z
z2 + a2
t 
c
c
otherwise
z small
h
z large
t
zc
z2 + a2
c
Recall Equ 26
V0
p r , t    
 h r , t 
t
h( r , t )
z small
z large
so that the pressure wave form is given by
P  z, t 
t
Off axis case
h z,  1 , t 
p z,  1 , t 
r0
r1
r2
c
c
c
Off axis case
5 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
Frequency = 3.75 MHz
a. f/2
b. f/2.4
c. f/3