ultrasound imaging: lecture 2 absorption reflection scatter speed of sound signal modeling signal...
TRANSCRIPT
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Ultrasound Imaging: Lecture 2
• Absorption
• Reflection
• Scatter
• Speed of sound
• Signal modeling
• Signal Processing
• Statistics
Interactions of ultrasound with tissue
Image formation
Jan 14, 2009
• Steering• Focusing• Apodization• Design rules
Beams and Arrays
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Anatomy of an ultrasound beam• Near field or Fresnel zone
• Far field or Fraunhofer zone
• Near-to-far field transition, L
2aL
L
Lateral distance (mm)Depth (mm)
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Anatomy of an ultrasound beam
• Lateral Resolution (FWHM)
FWHM
Lateral distance (mm)Depth (mm)
numberFR
aFWHM 2
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Anatomy of an ultrasound beam
• Depth of Field (DOF)
DOF
Lateral distance (mm)Depth (mm)
2)(7 numberFDOF
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Array Geometries
• Schematic of a linear phased array
• Definition of azimuth, elevation
• Scanning angle shown, , in negative scan direction.
ya (elevation)
xa (azimuth)
za (depth)
array pitch
Acoustic beam
t
trtrp
,
,
N
ii ttrhWtrh1
),(,
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Some Basic Geometry
• Delay determination:
– simple path length difference
– reference point: phase center
– apply Law of Cosines
– approximate for ASIC implementation
• In some cases, split delay into 2 parts:
– beam steering
– dynamic focusing
x
z
x
r
r
0
rx
crr x
rrrxxc
22 cos21
fs
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Far field beam steering
• For beam steering:– far field calculation
particularly easy
– often implemented as a fixed delay
c
xs
sin
x
z
xr
0
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Beamformation: Focusing• Basic focusing type beamformation
• Symmetrical delays about phase center.
-40 -30 -20 -10 0 10 20 30 40
-20
-10
0
10
20
point source
wavefronts before correction
transducer elements
delay lines
wavefronts after correction
summing stage
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Beamformation: Beam steering
• Beam steering with linear phased arrays.• Asymmetrical delays, long delay lines
-40 -30 -20 -10 0 10 20 30 40
-20
-10
0
10
20
point source
wavefronts before correction
array elements
delay lines
wavefronts after beam steering and focusing
summing stage
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Anatomy of an ultrasound beam• Electronic Focusing
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Grating Lobes
• Linear array:
– 32 element array
– 3 MHz
– ‘pitch’ l = 0.4 mm
– = 0.51 mm
– L= N l = 13 mm
• How to avoid:
– design for horizon-to-horizon safety
275.14.
51.)(
)(
g
g
Sin
lSin
How many elements?
What Spacing?
gl
g
l
Main Lobe
Grating Lobe
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Array design
• Linear array:
– 32 element array
– 3 MHz
– pitch l = 0.4 mm
– = 0.51 mm
– Larray= N l = 13 mm
• How to avoid:
– design for horizon-to-horizon safety
2
l
How many elements?
What Spacing?
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Apodization
• Same array:– 32 element array– 3 MHz– pitch l = 0.4 mm– = 0.51 mm– Larray = N l = 13 mm
• With & w/o Hanning wting.• Sidelobes way down.• Mainlobe wider• No effect on grating lobes.
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Summary of Beam Processing
• Beam shape is improved by several processing steps:
– Transmit apodization
– Multiple transmit focal locations
– Dynamic focusing
– Dynamic receive apodization
– Post-beamsum processing
• Upper frame: fixed transmit focus
• Lower frame: the above steps.
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I INTERACTIONS OF ULTRASOUND WITH TISSUE
Some essentials of linear propagation
Recall the equation of motion
t
v
x
p
0 (1)
Assume a plane progressive wave in the +x direction that satisfies the wave equation
ie)(
0kxtepp
(2)
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Substituting 2 into 1 we have
t
vjkep kxtj
0
)(0
Z
p
c
pv
epf
ejk
j
pv
dtejkp
v
kxtj
kxtj
kxtj
0
00
)(
0
0
)(
0
0
2
2
Acoustic impedance
(3)
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cZ 0Where
= Characteristic Acoustic Impedance
Define a type of Ohm’s Law for acoustics
Electrical:Acoustical:
Extending this analogy to Intensity we have
vZp
IRV
20
20
2
1
2
1Zv
Z
pI
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Propagation at an interface between 2 media
111 cZ 222 cZ
iP
rP
tP
xktj
tt
xktjrr
xktjii
ePp
ePp
ePp
2
1
1
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Define Reflection/Transmission Coef
i
t
i
r
p
pT
p
pR ,
You will show:
21
2
12
12 2ZZ
ZT
ZZZZ
R
Example: Fat – Bone interface
38.16.7)6.7(2
38.16.738.16.7
TR
70.0 69.1
(4)
(5)
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THE DECIBEL (dB) SCALE
refsBd A
ALogA 10)( 20
Where A = measured amplitudeAref = reference amplitude
In the amplitude domain
6 dB is a factor of 2-6 dB is a factor of .5 (i.e. 6dB down)
20 dB is a factor of 10-20 dB is a factor of .1 (i.e. 20dB down)
(6)
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0
-10
-20
-30
-40
-50
Reflection Coefficients
Air/solid or liquidBrass/soft tissue or waterBone/soft tissue or water
Perspex/soft tissue or water
Tendon/fatLens/vitreous or aqueous humourFat/non-fatty soft tissuesWater/muscle
Fat/water
Muscle/blood
Muscle/liver
Kidney/liver, spleen/blood
Liver/spleen, blood/brain
Water/soft tissues
R = 1.0
R = .1
R = .01
Ref
lect
ion
Coe
f. dB
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3) ULTRASOUND IMAGING AND SIGNAL PROCESSING
Thus far we have been concerned with the ultrasound transducerand beamformer. Let’s now start considering the signalprocessing aspects of ultrasound imaging.
Begin by considering the sources of information in an ultrasound image
a) Large interfaces, let a = structure dimension
a- specular reflection-
- reflection coefficient12
12ZZZZ
R
where cZ
- strong angle dependance- refraction effects
density speed of sound
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b) Small interfaces
a-
- Rayleigh scattering
Cosak
D0
0
0
032
2
33
3
Compressibility Density
and Arp , Dr
eikr(7)
Morse and Ingard Theoretical Acousticsp. 427
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SCATTER FROM A RIGID SPHERE Cosr
ac
Ds 3134 32
*
*
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SCATTER FROM A RIGID SPHERE (Mie Scatter)
*
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ATTENUATION
= absorption component + reflectivity component
xepxp 0
The units of are cm-1 for this equation. However attenuationis usually expressed in dB/cm. A simple conversion is givenby
1686.8 cmcm
dB
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Attenuation in Various Tissues
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Speed of Sound in Various Tissues
0%
5%
10%
15%
-5%
-10%
Assumed speed of sound = 1540 m/s
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SUMMARY ULTRASONIC PROPERTIESTable 1
Material Speed of Sound Impedance Attenuation Frequency
ms-1 Kg m-2 s-1
X 106
At 1 MHz (dB cm-1) Dependency
water 1490 @ 23ºC 1.49 0.002 2
muscle 1585 @ 37ºC 1.70 1.3-3.3 1.2
fat 1420 @ 37ºC 1.38 0.63 1.5-2
liver 1560 @ 37ºC 1.65 0.70 1.2
breast 1500 + 80 @ 37ºC ------ 0.75 1.5
blood 1570 @ 37ºC 1.70 0.18 1.2
skull bone 4080 @ 37ºC 7.60 20.00 1.6
air 331 @ STP 0.0004 12.00 2
PZT 4300 @ STP 33.00 ------ --
smc /1540
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2.2 Modeling the signal from a point scatterer
Imagine that we have a transducer radiating into a medium and we wish to know the received signal due to a single point scatterer located at position
By modifying the impulse response equation (Lecture 1 Equ. 25 ) we can write:
r
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trhtrhtstgtgt
VktrV rtout ,*,***,*, 2
0
transmit + receiveelectromechanical IR’s
scatterer IR
transmitIR
receiveIR
pulse (t)
trHtpulse
trhtrhtpulsetrV rtout
,*)(
,*,*,
easily measured
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Now consider a complex distribution of scatterers
Isochronousvolume
rxri
(1)
(2)
(3)
(4)
At any point in the isochronous volume there exists a transmit –receive path length divided by c for a time, t, such that
c
zt
c
ll
21
zl1
l2
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If we look at the four field points shown on the previous pagewe would see the following impulse responses
(1)
(2)
(3)
(4)
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The total signal for a given ray position rx is given by
trHrWtpulsetrVout xiN
iiix ,
1*)(,
(9)
scattererstrength
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The resultant signal is the coherent sum of signals resultingfrom the group of randomly positioned scatterers that make up the isochronous volume as a function of time.
A useful model of the signal is:
ttCostatytVout 2
Envelope Modulatedcarrier
Phase
Grayscale informationfor B-scan Image
How do we calculate a(t) and (t)?
Velocity informationfor Doppler
(10)
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3.3 Hilbert Transform
The Hilbert transform is an unusual form of filtration in which thespectral magnitude of a signal is left unchanged but its phase is altered by for negative frequencies and for positive frequencies
2 2
Definition
)(*1
1
xfx
xdxx
xfxFH
(11)
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In the frequency domain
sH FsjxF )sgn(
Consider the Hilbert transform of Cos x
RE RE
(12)
IMIM
xCos sjSgn
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The application of two successive Hilbert transforms resultsin the inversion of the signal – we have 2 successive rotations in the negative frequency range and 2 rotations in the positive frequency range. Thus the total shift in each direction is .
2
2
1
II
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xfxx
xFx H
111
sFsjsj sgnsgn
xfF
sF
s
1
1
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The Hilbert transform is interesting but what good is it?
ANALYTIC SIGNAL THEORY
Consider a real function . Associate with this function another function called the analytic signal defined by:
where = Hilbert Transform
The real part of the analytic signal is the function itself whereas the imaginary part is the Hilbert transform of the function.
Note that the real and imaginary components of the analyticsignal are often called the “in phase”, I, and “quadrature”, Q,components.
tjztytf tz(13)
ty
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Just as complex phasors simplify many problems in AC circuit analysis the analytic signal simplifies many signal processing problems.
The Fourier transform of the analytic signal has an interestingproperty.
0,2
0,0
][
sY
s
YsSgnY
YsSgnjjYtjzty
s
s
ss
s sy
0
2
s
Y s
(14)
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Equation 14 gives us an easy way to calculate the analyticsignal of a function:
1) Fourier transform function2) Truncate negative frequencies to zero3) Multiply positive frequencies by 24) Inverse Fourier Transform
Recall that our resultant ultrasound signal can be expressedas:
ttCostaty 2
Its analytic signal is then
ttetatf
2 (15)
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which on the complex plane looks like:
IM
RE ty
t ta tz
Where
and the phase is given by
tztyta 22
tytz
Tant)(1
(16)
(17)
a(t) envelope
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Demodulation: estimate using
1) Analytic signal method using FFT (slow)2) Analytic signal using baseband quadrature approach3) Sampled quadrature
)(),( tta QI ,
Baseband Quadrature Demodulation
X
X
Low Pass
Low Pass
tCos 2
tQ
tSin
2
ty
tIt )Re(
Baseband Inphase Signal
)()Im( tQt
Baseband Quadrature Signal
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tCostCosa
tCostCosaI
tt
ttt
22
2
22
Use shift and convolution theorems to calculate spectra
ttnote :
2
2 tjt eAI
(slowly varying)
tjeA 2 2
21
tjt eAI
21
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tjtjt eAeAI
2
1
2
1
tt CostaI 21
tjt etas )(
2
1
Similarly
tt SintaQ )(21 Baseband
AnalyticSignalNo carrierPhase preserved
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ttt
t
ttt
QIa
a
CosSinaQI
22
2
22222
2
4
1
)(4
1
Thus
)()(
)()(
)(
tItQ
ArcTan
I
QTan
t
t
tt
and
Sampled Quadrature
Begin with the signal of the ultrasound waveform
ttt Cosay 2
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Sample with period 1T
* **
)(nTI
Tt
IIIy t
)(nTQ
Tt
IIIy t
Recall that the quadrature signal is the Hilbert Transform of theinphase component of the analytic signal i.e. for a cos wave it is a negative sine wave. Thus we see that . . .
nTIT
tIIIy t
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If the inphase and quadrature signals are slowly varyingwe can get the quadrature signal simply by sampling theinphase signal 90º or ¼ period later
Sampling t = nT for I samplest = nT+T/4 for Q sample
142)()(
)2()()(
T
nTTnTCosnTanTQ
nTrTCosnTanTI
let
nTSinntanTnCosnTanTQ
TnCosnTanTnCosnTanTI
)()22()()(
)()()2()()(
(18)
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1
Overall Imager Block Diagram
D opplerBeam form er
D ig ita lR eceive
Beam form er
Beam form erC entra lC ontro l
D ig ita lT ransm it
Beam form er
T ransm itD em ux
R eceiveM ux
TransducerC onnectors
SystemC ontro l
Im ageProces-
sing
23 4 5
6
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Imaging System Signals
D opplerBeam form er
D ig ita lR eceive
Beam form er
Beam form erC entra lC ontro l
D ig ita lT ransm it
Beam form er
T ransm itD em ux
R eceiveM ux
TransducerC onnectors
SystemC ontro l
Im ageProces-
sing
23 4 5 6
1
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Coarse and Fine Beamforming Delays
Ho()e-j/4
Ho()e-j/2
Ho()e-j3/4
Ho()
MUXFIFO
Input fromADC at 20to 40 MHz,8 to 12 bits
Output withdelay accuracyup to 160 MHz
To apodizationand furtherprocessing
CoarseDelay
ControlFine
DelayControl
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SIGNAL STATISTICS
Recall that the ultrasound signal is the sum of harmoniccomponents with random phase and amplitude. It can be shownthat the probability density function for such a situation isGaussian with zero mean i.e.
2
2
221
)(
y
eyp
(19)
The quadrature signal will also be Gaussian with the same standard deviation
22
221
)(
z
ezp
(20)
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Since p(y) and p(z) are independent random variables the jointprobability density function is given by
2
22
2
2
2
2
22
22
2
1
2
1
2
1),(
zy
zy
e
eezyp
(21)
The probability of a joint event (corresponding to a particularamplitude of the envelope) is the probability that:
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)(zp
)( yp
total area = ada2
The probability that a lies betweena and a + da is
222 zya
daea
daapa
22
222
2)(
adad
adad
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So that the probability density function for the radiofrequency signal is given by
22
22
a
ea
ap
Rayleigh Prob.Density function
aa
)(ap
few white pixels
many gray pixels
fewblackpixels
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The speckle in an ultrasound image is described by thisprobability density function. Let’s define the signal as and the noise as the deviation from this value
arms
21
2212 aaaN Thus
daea
daapaa
a
o
o
2
2
22
2
Recall
2a
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Thus:
21
21
22
22
2
22
2
Na
SNR
SNR = 1.91 and is invariant (25)
Note that the SNR in ultrasound imaging is independent of signal level. This is in contrast to x-ray imaging where thenoise is proportional to the square root of the number ofphotons.
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Speckle Noise in an Ultrasound Image
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a
ias
as
i
00
x
Let’s make several independent measurements of so and si
These measurements will form distributions
i 0
is 0s
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The parameter used to define image quality includes boththe observed contrast and the noise due to speckle in the following fashion:
Define Contrast:
Define Normalizedspeckle noise as:
and finally, define our quality factor as the contrast to speckle noise ratio (CSR)
0
0
s
ss i
0
21
220
si
220
0
i
issCSR
(26)
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Suggested Ultrasound Book References:
General Biomedical Ultrasound (and physical/mathematical foundations): “Foundations of Biomedical Ultrasound”, RSC Cobbold, Oxford Press 2007.
General Biomedical Ultrasound (bit more applied): “Diagnostic Ultrasound Imaging: inside out” TL Szabo Academic Press 2004.
Ultrasound Blood flow detection/imaging: “Estimation of blood velocities with ultrasound” JA Jensen Cambridge university press 1996
Basic acoustics: “Theoretical Acoustics” PM Morse and KU Ingard, Princeton University Press (many editions).
Bubble behaviour: “The Acoustic bubble” TG Leighton Academic Press 1997.
Nonlinear Acoustics: “Nonlinear Acoustics” Hamilton and Blackstock, Academic Press 1998.