basics of ultrasound

1. Advanced Imaging 1024
Jan. 7, 2009 Ultrasound Lectures
History and tour
Wave equation
Diffraction theory
Rayleigh-Sommerfeld
Impulse response
Beams
Lecture 1:Fundamental acoustics
DG: Jan 7
Absorption
Reflection
Scatter
Speed of sound
Image formation:
- signal modeling
- signal processing
- statistics
Lecture 2:Interactions of ultrasound
with tissue and image formation
DG: Jan 14

2. The Doppler Effect
Scattering from Blood
CW, Pulsed, Colour Doppler
Lecture 3: Doppler Ultrasound I
DG: Jan 21
Velocity Estimators
Hemodynamics
Clinical Applications
Lecture 4: Doppler US II
DG: Jan 28
Lecture 5: Special Topics
Mystery guest: Feb 4 or 11

3. ULTRASOUND LECTURE 1
Physics of Ultrasound Waves:
The Simple View

4. ULTRASOUND LECTURE 1
Physics of Ultrasound: Longitudinal and Shear Waves

5. ULTRASOUND LECTURE 1
Physics of Ultrasound Waves: Surface waves

6. ULTRASOUND LECTURE 1
Physics of Ultrasound Waves
1) The wave equation
y
z
u
t
u
x
u+Δu
Particle Displacement =
Particle Velocity =
Particle Acceleration =
u
x
v
t
u
=
∂
∂
t
v
t
u
∂
∂
=
∂
∂
2
2

7. Equation of Motion
dV
dxdydz
x
p
∂
∂
+
Net force = ma p = pressure
= P – P0
=ρ density
t
v
dVdV
x
p
∂
∂
=
∂
∂
ρ
or
t
v
p
∂
∂
=∇ ρ (1)
dydzpdydzp

8. Definition of Strain
,xSx
x
u
u ∆=∆
∂
∂
=∆ S = strain
Also

TermsNonlinear
S
C
S
B
SAp ...
!3
32 +++=
bulk modulus
x
u
∂
∂
=
Taking the derivative wrt time of (2)
t
S
∂
∂
=
x
v
∂
∂
(4)
(2)
(3)

9. Substituting for from Eq (1) (in one dimension)v∂
t
x
p
xt
S
∂
∂
∂
∂
∂
=
∂
∂
ρ
1
2
2
2
2
1
x
p
t
S
∂
∂
=
∂
∂
ρ
Substituting from
2
2
2
2
x
pA
t
p
∂
∂
=
∂
∂
ρ
or pc
t
p 22
02
2
∇=
∂
∂
; ==
ρ
A
c0 wave velocity
3 dimensions
(3)
(5)

10. For the one dimensional case solutions are of the form
( ) )(
, kxtj
etxp −
= ω
; fk πω
λ
π
2
2
== (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
ρ
ρ
0
P0
γ
ρ
ρρ












−
+
0
0
1
Condensation = S'Gamma=ratio of specific heats

11. P = P0[ ]
γ'
1 S+ ,
v
p
c
c
=γ
Expand as a power series
P = P ( ) 3
0
'
0
'
0 '
!3
1
1
2
1 2
SCPSPSP
BA
+−++ γγγ0
. . . .
B/A =
( )
1
1
−=
−
γ
γ
γγ Depends solely on
thermodynamic factors

12. Material B/A
Water 5
Soft Tissues 7.5
Fatty Tissues 11
Champagne
(Bubbly liquid)

13. ( )






+−≅
00
),(,
v
txv
l
x
kxwtSin
v
txv
Additional phase term small for small x
and increasingly significant as
= shock distance
(8)
In terms of particle velocity, v, Fubini developed a non linear
solution given by:
lx ⇒
l
Nonlinear Wave Equation
2
02
2
)( vc
t
v
β+=
∂
∂
2
2
x
v
∂
∂
(7)
A
B
2
1+=β

14. k
k
c
v
l
βε
β
11
0
0
=






=
Mach #
(9)
ε=
• At high frequencies the plane wave shock
distance can be small.
• So for example in water:
5.3=β MHzf 5.30 = MPap 10 =
Shock distance = 43 mm
Shock Distance, l

15. Where 




=
l
nx
J
nx
l
B nn
2 (11)
Thus the explicit solution is given by
( ) ( )∑
∞
=
−=
10
2
n
n
kxtnSin
lnx
lnxJ
v
v
ω (12)
We can now expand (Eq. 8) in a Fourier series
( )[ ]∑
∞
=
−=
1
0
n
n kxtnSinBvv ω
0/vv
(10)

16. Hamilton and Blackstock Nonlinear Acoustics 1998
Aging of an Ultrasound Wave

17. Hamilton and Blackstock Nonlinear Acoustics 1998
lx /
RelativeAmplitude Harmonic Amplitude vs Distance
(narrow band, plane wave)

18. Focused Circular Piston
2.25 MHz, f/4.2, Aperture = 3.8 cm, focus = 16 cm
Hamilton and Blackstock Nonlinear Acoustics 1998

19. Hamilton and Blackstock Nonlinear Acoustics 1998
Propagation Through the Focus

20. Nonlinear Propagation: Consequences
_______________________
• Generation of shock fronts
• Generation of harmonics
• Transfer of energy out of fundamental

21. 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
pc
t
p
∇=
∂
∂ 2
02
2
Laplacian
yx
=
∂
∂
+
∂
∂
+
∂
∂
=∇ 2
2
2
2
2
2
2
δ

22. The acoustic pressure field of the harmonic radiator
can be written as:
( ) ( ){ }tj
erPtrp ω
⋅=Re,
Where is a complex phasor function satisfying
the Helmholtz Equation
( )rP
(13)
( ) ( ) 022
=+∇ rPk (14)

23. To solve this equation we make use of Green’s functions
( )
∏
=
4
1
rP ds
n
G
PG
n
P
s a
a
∂
∂
−⋅
∂
∂
∫ (15)
s = surface area n∂∂ / = normal derivative
( )11, yxPPa = = pressure at the aperture

24. Rayleigh Sommerfeld Theory
Assume a planar radiating surface in an infinite “soft” baffel
Aperture
Field Point
Conjugate
field point
n
'
r
use ( )
'~'
'~'
r
e
r
e
rG
rjkjkr −
−=
'~r
as the Greens function
r
= 0 in the aperture
θ

25. ( ) ds
n
G
PrP as 





∂
∂
−∫=
π4
1
'
'
2
r
e
jkCos
r
G
n
G jkr
=
∂
∂
=
∂
∂
θ
( ) dsCos
r
e
P
jk
rP
jkr
as θ
π '
'
2
∫
−
=
ds
r
e
P
j
jkr
a '
'
1
∫≅
λ
(16)
Equation 15 can now be written:

26. Example
Consider the distribution of pressure along the axis of a
plane circular source:
radius = a
22
σ+=′ zr
σd
σ
z

27. From Equation (16) in r, , z coordinates
∫
+
+
−
=
a
a
z d
z
zjk
ejkP
P
0
)( 2
22
22
2
σπσ
σ
σ
π
The integrand here is an exact differential so that
a
zjk
az
jk
e
jkPP
0
)(
22








−=
+σ
θ

28. [ ]22
)( azjkjkz
a eePzP +
−=
The pressure amplitude is given by the magnitude of this
expression
[ ] tj
a ezaz
k
SinPzP ω
⋅






−+= 22
2
2)(
[ ]





−+= zaz
k
Sin
Z
P
zI a 222
2
2
4)( (17)

29. To look at the form of (17) find an approximation for:
z
z
a
zzaz −







+=−+ 2
2
222
1
z
z
a
z −+= 2
2
1
z
a
2
2
≅
z
z
a
z −





+≅ 2
2
2
1

30. ( ) 





=∴
z
ka
Sin
z
P
zI a
4
4 2
2
2
Maxima
24
2 ∏
= m
z
ka m = 1, 3, 5, . . .
24
2 2 ∏∏
= m
z
a
λ
m
a
zorm
z
a
λλ
22
==

31. Minima
m
a
z
λ
2
= ; m = 0, 2, 4, 6
∞=z

32. 0 50 100 150 200
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
ka
2
/(4z)
a = 5 mm
frequency = 5 MHz
Intensity*4pa
2
/Z
Axial Distance (mm)
Eq. 17
mm
a
z 3.83
2
==
λ
M: 54 3 2 1 0

33. THE NEAR FIELD (off axis)
( ) ( )2
10
2
10
2'
yyxxzr −+−+=
2
2
10
2
2
10 )()(
1
z
yy
z
xx
z
−
+
−
+=
1y
1x 0y
0x
z
'r

34. Fresnel approximation (Binomial Expansion)













 +
+




 +
+≅
2
10
2
10'
2
1
2
1
1
z
yy
z
xx
zr (19)
From (16) we have
( ) 2
110
2
0
2
10 2 xxxxxx +−=− 2
110
2
0
2
10 2 yyyyyy +−=−
( ) ( )
( ) ( )[ ]
11
2
1100
2
10
2
10
,
1
, dydxeeyxP
zj
yxP
yyxx
z
jk
jkz
a
−+−∞
∞−
∫∫=
λ

35. 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 +
[ ]
)(
2
00
2
0
2
0
),(
zK
yx
z
jkjkz
e
zj
e
yxP
+
=
λ
( )
[ ]2
1
2
1
2
1,1
yx
z
jk
eyxP
+
∫∫
11dydx

36. ( ) ( ) ( )
( ) ( )
11
22
1100
11
2
1
2
1
,, dydxeeyxPKyxP
yxjyx
z
jk
z
yx υυπ +−+
∫∫=
Where
z
x
x
λ
υ 0
=
z
y
y
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=

37. Consider a plane circular focused radiator in cylindrical
coordinates
( )σ,zP
σ
z
radius a
Circular Aperture

38. ( )






ℑΚ=
a
r
CirczP zσ,
( )
x
xJ
zk
π
π1'
= Where
z
a
x
λ
σ2
=
FWHM a
z
2
41.1
λ
=
x
xJ
π
π )(
2 1
az 222.1 λ
(21)

39. Square Aperture

40. Need to consider the wideband case. Returning to Eq. 16
we have:
( ) ( ) 11'1100
'
,
2
, dydx
r
e
yxP
jk
yxP
jkr
∫∫= −
π
cc
k
v ω
λ
===
∏∏ 22
( ) ( ) ω
π
ω
ω
ddydx
r
e
yxP
c
j
yxP
r
c
j
11
'
1100
'
,
2
, ∫∫−∫=
( ) ωω
ω
π
ddydx
r
e
yxP
j
r
c
j
c 11'11
'
,
2
∫∫∫= −

41. This is a tedious integration over 3 variables even after
significant approximations have been made
There must be a better way!

42. Impulse Response Approach to Field Computations
Begin by considering the equation of motion for an
elemental fluid volume i.e. Eq. 1
t
v
p
∂
∂
=∇− ρ (22)
Now let us represent the particle velocity as the gradient
of a scalar function. We can write
ϕ⋅∇=v

43. 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)
t
p
∂
∂
∇=∇−
ϕ
ρ




∂
∂
+∇=
t
p
ϕρ0
t
p
∂
∂
−=
ϕ
ρ (23)
ϕ

44. The better way: Impulse response method
r
'r
( )tV0
( ) ( )
t
tr
trp
∂
∂
−=
,
,
ϕ
ρ
•
ds

45. ( ) ( ) ds
r
crtV
tr
'
'
2
1
, 0 −
∫∫=
π
ϕ
s
( ) ( )trhtV ,0 ∗=
Impulse Response
where
( ) ( ) ds
r
crt
trh
'
'
2
1
,
−
∫=
δ
π (24)
Thus
( ) ( ) ( )trh
t
tV
trp ,, 0
∗
∂
∂
−= ρ (25)

46. Useful because is short!
convolution easy
Also is an analytic function
No approximations!
Can be used in calculations
You will show that for the CW situation
( )trh ,
( ) ( ){ } 0
,, 00 ωωωρ =ℑ−= trhvjtrp (26)
( )trh ,

47. IMPULSE RESPONSE THEORY EXAMPLE
Consider a plane circular radiator
z
2r
0r
'r
σ
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
crt
trh '
'
2
1
,
−
∫=
δ
π
σd

48. '
'
'
rd
dr
Sin r
σ
σ
θ ==
'rd
σd
σ
'r
'
r
θσ
σ
''
rdr
d =
( )
σ
'''
drrrl
ds =
Also let
c
r'
=τ
So that
( ) ( ) ( )
σ
τδ
π
''
'
2
1
,
rrl
r
t
trh
−
∫=
⋅
cdτ

49. ( ) ( )
σπ2
,
cctl
trh = (27)
*
* a very powerful formula
( ) σπ2'
=rl while the wavefront lies between
21 rr and
Thus we have: (next page)
( ) c
c
trh ==
πσ
σπ
2
2
,ie

50. ( ) =trh ,
c
r
t 0
0 <<0
( )






−
−+−−
21
0
2
22
1
2
0
2
1
))((2 rctr
arct
Cos
c ρ
π
c
c
=
σ
σ
π
π
2
2
0
c
r
t
c
r 10
<<
c
r
t
c
r 21
<<
c
r
t 2
>
Planar Circular Aperture

51. ( )
c
az
t
c
z
ctrh
22
,
+
<<=
0= otherwise
h
z small
z large
cz
c
az 22
+
t
Consider the on axis case:
01 =σ zr =0 21 rr =

52. Recall Equ 26
( ) ( )trh
t
V
trp ,, 0
∗
∂
∂
−= ρ
so that the pressure wave form is given by
( )tzP ,
t
),( trh
z small
z large

53. Off axis case
( )tzh ,, 1σ
( )tzp ,, 1σ
c
r0
c
r1
c
r2

54. Off axis case
5 mm radius disk, z = 80 mm

55. Spherically focused aperture
- relevant to real imaging devices

56. Spherically focused aperture impulse response

57. Spherically focused aperture impulse responses

58. Spherically focused
aperture pressure
distribution
a. f/2
b. f/2.4
c. f/3
Frequency = 3.75 MHz