Ultrasound Kim Butts Pauly 1

Ultrasound
Kim Butts Pauly
[email protected]
1
Ultrasound Cheat Sheet
Material
Air
Fat
Water Soft Tissue
Bone
Velocity (m/s)
330
1450
1540
1540
3300
% reflected = 100 * (Z2-Z1)2/(Z2+Z1)2
% transmitted = 100 - %reflected
Material
Acoustic Impedence (Rayl)
Air
0.0004
Fat
1.38
Soft Tissue 1.54
Bone
7.8
v = λf
1/2 -> -3 db
1/10 -> -10 db
SPL = n*λ
Axial resolution = n* λ/2
attenuation = α f distance
Material
Water Fat
Soft Tissue
Air
Bone
Attenuation(dB/cm/MHz)
0.0022
0.5 - 1.8
1
12
20
fshift = 2 * (v/c) * fo * cos(θ)
Z = ρv
2
Basic Concepts:
Sound in tissue
3
Transducer - Piezoelectric Crystal
Transducer
+ + +
+- - -+ --+ ++- -+-+ -+-+ ++ -++ -++ +- - -+ --+ ++- -+-+ -+-+ ++ -++ -++ - - -
1.An oscillating electric field is applied.
2. This realigns the dipoles in a piezoelectric crystal.
4
Basic Transducers
• Placed next to tissue, the transducer creates a pressure wave the ultrasound pulse.
• Common frequencies:1, 2.25, 3.5, 5, 7, 10 MHz.
5
Sound in Matter
Velocity - speed of the sound wave
Frequency - # cycles/s
Wavelength - spatial size of one cycle
Material
Air
Fat
Water Soft Tissue
Bone
Velocity (m/s)
330
1450
1540
1540
3300
6
The pulse moves through
tissue at a velocity
specific to the tissue: the
velocity depends on the
density and stiffness of
the tissue.
Sound in Matter
The pulse moves through
tissue at a velocity specific
to the tissue and
independent of the applied
frequency.
f=
3.0
MHz
2
1/
th
g
len
e
v
wa
λ = 0.5 mm
=
1.5
MHz
f
λ = 1 mm
• Velocity (v), frequency (f), and wavelength (λ) are related:
• v = λf
• v = distance/time
7
1. The speed with which a sound pulse travels through tissue depends on the _______.
A. Ultrasound frequency
B. Duration of the pulse
C. Tissue density and stiffness
D. Pulse amplitude and attenuation
2. The wavelength of a 2 MHz ultrasound beam is ____ mm.
A. 0.02
B. 0.55
C. 0.77
D. 2.0
E. 5.0
3. An ultrasound transducer that is used to examine tissue receives a reflection echo 64 μs after the signal is sent. The depth of the
structure that caused the echo is _____ cm.
A. 2
B. 5
C. 7
D. 10
E. 15
4. The ultrasound propagation velocity through ______ is at least twice the velocity of the propagation through the other human
tissues listed.
A. Brain
B. Fat
C. Muscle
D. Soft tissue
E. Skull (bone)
5. The wavelength (λ) of a 3 MHz sound beam is shortest in
A. Air
B. Castor Oil
C. Fat
D. Muscle
E. The same in all of the above
8
6. Sound waves
A. Have the same velocity in all media.
B. Are at the low energy end of the electromagnetic spectrum.
C. Can ionize only hydrogen atoms.
D. Cannot travel in a vacuum.
E. Transfer most of their energy to tissue in side lobes.
7. The resonant frequency of an ultrasonic transducer is determined by the _______.
A. Thickness of the piezoelectric crystal.
B. Frequency of the applied voltage pulse.
C. Material surrounding the transducer.
D. Transducer material.
E. Backing block.
8. What happens when a 5 MHz ultrasound beam travels from liver into fat?
A. Wavelength increases.
B. Wavelength decreases.
C. Frequency increases.
D. Frequency decreases.
E. A very large reflection at the boundary.
9
1. C
2. C
Wavelength = velocity / frequency.
The average velocity in tissue is 1540 m/s. Frequency is 2x106/s.
3. B
Half the time is required for the pulse to go forward, and half for it to return.
Distance = velocity x time
D = (1540 m/s) x (32x10-6 s) = 4.9 cm
4. E
Ultrasound velocity is inversely related to the compressibility of the conducting material. All body tissues have approximately the
same compressibility and velocity (1540 m/s) except bone, which is much less compressible than the other tissues. The velocity in
bone is about 4080 m/s.
5. A
The relationship between the velocity of a sound wave and the frequency and the wavelength is v = f λ.
The velocity of sound is constant in any given material. Since f is also fixed (3 MHz), λ must vary with the compressibility of the
material. The velocity is about the same in all materials except air. Velocity in air is very much slower, and therefore, the wavelength
is shortest.
6. D
Sound waves are propagated by the motion of particles in a medium; a vacuum has no particles and, hence, does not transmit sound.
Sound is not EM radiation, and no ionization takes place. Most of the energy is given to molecules in the medium in the form of
kinetic energy resulting in heat.
7. A
The transducer crystal resonates in the thickness mode, where the emitted wavelength is equal to twice the crystal thickness, or
crystal thickness = λ/2. The frequency is selected by choosing a crystal thickness: higher frequencies require thinner crystals.
8. B
Frequency is unchanged. Velocity is lower in fat than in liver, so λ=v/f must also decrease. Acoustic impedances are similar, so there
is a small reflection at the boundary.
10
Attenuation
11
Attenuation
• Tissue is made up of small
reflectors/scatterers
• Eg. tissue collagen
• Sound is scattered in all
directions
• Energy is absorbed
=> loss of signal
12
Attenuation
attenuation (dB) = α f z
where
α = attenuation coefficient
-1 dB/cm/MHz for most soft tissues
f = frequency in MHz
z = distance in cm
Material
Water Blood Fat
Soft Tissue
Air
Bone
Lung
Attenuation(dB/cm/MHz)
0.0022
0.02
0.5 - 1.8
1
12 (Huda)
20
40 (Bushberg)
13
Relative Sound Intensity
Relative Sound Intensity(dB) = 10 log10 (I/Io)
when I = 0.5 Io, relative sound intensity is -3 dB
when I = 0.1 Io, relative sound intensity is -10 dB
multiply
add
when I = 0.25 Io, relative sound intensity is -6 dB
when I = 0.001 Io, relative sound intensity is -30 dB
14
Attenuation Examples
attenuation (dB) = α f z
• A 5 MHz beam passes thru 6 cm of tissue. The intensity is
attenuated by how much?
attenuation = -1 dB/cm/MHz * 5MHz * 6cm
= -30dB
The received signal is 1/1,000 the transmitted signal.
Usually receive 1/1,000,000 the transmitted signal.
15
Artifacts - Shadowing
• Can’t see well beyond bones
(e.g. near ribs, spine).
16
1. The attenuation coefficient (decibels/cm) of ultrasound is largest in:
A. Liver
B. Lung
C. Fat
D. Blood
E. skull bone
2. The absorption (dB) of an ultrasound beam is:
A. Independent of frequency in most tissues.
B. Higher in soft tissue than bone.
C. Inversely proportional to the frequency.
D. Higher in materials having high viscosity.
E. Proportional to the frequency in most tissues.
3. The attenuation of ultrasound in soft tissue, in dB/cm/MHz, is
A. 0.1
B. 1.0
C. 10
D. 100
E. 1000
Match the attenuation of a 1 MHz ultrasound beam with the material
4. Air
5. Blood
Absorption (dB/cm)
a. 40
b. 12
c. 1
d. 0.02
e. 0.002
6. A 4 MHz ultrasound beam travels through 7.5 cm of soft tissue with an
attenuation of 1 dB/cm/MHz. The original intensity is reduced by _____ dB.
a. 3
b. 30
c. 100
d. 300
e. 1000
17
7. The intensity of a reflected 5MHz sound beam is 40 dB. The incident intensity is 20 mW/cm2. The echo intensity is about
_____ mW/cm2.
a. 20 x 104
b. 4 x 102
c. 4 x 10-2
d. 20 x 10-2
e. 20 x 10-4
8. if the ultrasound intensity is attenuated by 50% in a pulse echo experiment, the relative intensity of the pulse to the echo
is _______ dB.
a. 2
b. 3
c. 6
d. 10
e. 50
9. If an ultrasound beam is attenuated by 99%, the intensity of the original beam to the echo is
a. 1 dB
b. 2 dB
c. 20 dB
d. 99 dB
10. A 5 MHz ultrasound beam passes through 4 cm of soft tissue (absorption coefficient is 1 dB/cm/MHz). The original
intensity is reduced by a factor of
a. 20
b. 40
c. 100
d. 1000
e. 10,000
11. You are using a 5 MHz ultrasound transducer and you decrease the noise by 100 times. If the attenuation coefficient is
0.5 dB/cm/MHz, how much deeper can you penetrate?
a. 4 cm
b. 8 cm
c. 12 cm
18
1. B
Air is a large attenuator of ultrasound, particularly in alveolar structures.
2. E
The amount of absorption in ultrasound is measured in dB and is equal to the absorption coefficient for a particular tissue at 1 MHz times
the frequency.
3. B
4. B
5. D
6. B
For a 4 MHz transducer, the attenuation is 4 dB/cm x 75 cm = 30 dB.
7. E
The reflected intensity (I) is lower than the incident intensity (I0). The relative power in decibels (dB) is
dB = 10 log10 (I0/I)
40 = log10 (20/I)
(20/I) 4 = 10
I=20 x 10-4
8. B
The ratio of I1 to I2 is 2, Relative intensity of the pulse to the echo dB = 10 log (I1/I2) = 10 x log 2 = 10 x 0.3 = 3.
9. C
Here, the ratio of the incident intensity to the transmitted intensity (I1/I2) is 100/1. By definition, dB = 10log (I1/I2) = 10 log(100/1)=20
10. C
The attenuation in dB is given by 1dB/MHz/cm x 5 MHz x 4 cm = 20 dB.
11. A
They are assuming you have the same SNR. You can detect 100x smaller signal. 100 x smaller => -20 db
20 db = 5 MHz * d * 2 * .5
19
Reflection
20
Reflection & Transmission
Z1
Z2
Transducer
• At each interface, some sound is reflected and some is transmitted.
• This is called a specular reflector
• The relative amounts depend on the acoustic impedences Z1 and Z2.
• Z = ρv where ρ is the density
21
Reflection & Transmission
% reflected = 100 * (Z2-Z1)2/(Z2+Z1)2
% transmitted = 100 - %reflected = 100 * 4 * Z1Z2/(Z2+Z1)2
Material
Air
Fat
Water Muscle
Soft Tissue
Bone
Acoustic Impedence (MRayl)
0.0004
1.38
1.54
1.7
1.62
7.8
22
Reflection Examples
• What is the reflection at a fat-muscle interface?
reflection = 100*(1.7-1.38)2/(1.7+1.38)2 = 1.1%
• What is the reflection at a muscle-bone interface?
reflection = 100*(7.8-1.7)2/(7.8+1.7)2= 41%
23
1. The acoustic impedence of a type of tissue depends on:
A. Ultrasound frequency and tissue viscosity
B. Ultrasound wavelength and tissue density
C. Tissue elasticity and susceptibility
D. Ultrasound velocity and tissue density
2. Ultrasound moves with the highest velocity in
a. fat (Z=1.38)
b. blood (Z=1.61)
c. muscle (Z=1.7)
d. bone (Z=7.8)
3. The acoustical impedance of water to a 1 MHz ultrasound beam is 1.54
rayls. The acoustical impedance of water at 5 MHz is _____ rayls.
a. 0.31
b. 1.54
c. 7.7
d. 38.5
e. 1540
4. In ultrasound, the largest reflections will occur between:
a. liver and muscle
b. blood and water
c. kidney and muscle
d. blood and kidney
e. fat and liver
24
5. The reflection coefficient of an ultrasound beam traveling from soft tissue to bone is about 0.23. What will be the
reflection coefficient if the beam is traveling from bone to soft tissue? (Note acoustic impedance of bone is about three times
that for soft tissue.)
a. 0.07
b. 0.12
c. 0.23
d. 0.46
e. 0.69
6. Approximately what fraction of an ultrasound beam is reflected from an interface between two media with Z values of
1.65 and 1.55?
a. ½
b. 1/10
c. 1/100
d. 1/500
e. 1/1000
Match the reflected ultrasound intensity with the interface in the question:
7. Air-muscle interface
8. Bone-muscle interface
a. 0.999
b. 0.412
c. 0.050
d. 0.008
e. 0.0002
25
1. D
Acoustic impedance, by definition, is the product of the velocity and the tissue density. The unit of impedance is the Rayl.
2. D
The velocity of sound is inversely related to the compressibility of the conducting material. All body tissues have about the
same compressibility and velocity (1540 m/s) except bone, which has less compressibility than other tissues. The velocity is
about 4080 m/s.
3. B.
The acoustic impendance (Z) is equal to the product of the density of the material (ρ) and the velocity (v) of the sound in the
medium, Z=ρv. Both are independent of the frequency of the ultrasound beam, and therefore Z does not change when the
frequency is increased.
4. E
The largest reflections (echoes) will occur when there is the greatest difference in acoustic impedance (Z) since
R=(Z2-Z1)2/(Z2+Z1)2 .
5. C
The reflection coefficient R=(Z2-Z1)2/(Z2+Z1)2 between two tissues with acoustic impedances Z2 and Z1 is unaffected by an
interchange of Z2 and Z1.
6. E
Ir/Ii = (Z2-Z1)2/(Z2+Z1)2 = (1.65 - 1.55)2 / (1.65 + 1.55)2 = 1/1024.
7. A
Reflected intensity = (Z2-Z1)2/(Z2+Z1)2
8. B
26
Review of several concepts
1. In ultrasound, the intensity of a reflected echo depends on all of the following except
a. depth to interface
b. transducer diameter
c. tranducer frequency
d. difference in acoustic impedance at the interface.
2. In ultrasound imaging, it is important to insure good contact between the transducer and the patient’s skin. This is to
1. avoid reflections of the incoming wave from the patient’s skin.
2. get as close as possible to the area of interest.
3. remove the air between transducer and skin which will absorb incoming ultrasound waves.
4. increase the velocity of sound in the tissue.
A. 1,2,3
B. 1,3
C. 2,4
D. 4 only
E. all of the above
3. The acoustic impedance of fat and muscle are 1.38 and 1.7 Rayls, respectively. Calculate the intensity reflection coefficient
for the perpendicular incidence of a 2.5 MHz ultrasound pulse going from fat to muscle.
A. 0.001
B. 0.01
C. 0.1
D. 1.0
E. None of the above
4. An ultrasound wave is attenuated as it penetrates tissues. What is the approximate half value thickness for a 3 MHz
ultrasound beam in muscle?
A. 1 cm
B. 2 cm
C. 3 cm
D. 4 cm
E. None of the above
5. Opthalmology units use higher frequencies than abdominal units
A. True
B. False
27
Review of several concepts
6. The acoustic impedance of body tissue to ultrasound waves depends upon the ______.
A. frequency
B. velocity
C. tissue density
D. B and C
E. A and B
7. The amplitude of ultrasound echo depends upon the following factors_______.
A. frequency
B. type of tissue
C. depth of object
D. number of different tissue interfaces
E. all of the above
8. An ultrasound unit with a 5 MHz transducer that has a -60 dB dynamic range has a maximum imaging depth in muscle of
about ____cm.
A. 20
B. 12
C. 6
D. 3
E. Cannot be determined
9. In ultrasound imaging, which of the following is the most likely result of switching to a higher frequency transducers?
A. increased ultrasound attenuation
B. increased maximum depth of penetration
C. longer ultrasound wavelength in tissue
D. requires lower pulse repetition frequency
E. increased speed of sound in soft tissue
10. The velocity of a 2 MHz ultrasound beam is 1540 m/s in soft tissue. If the frequency is increased to 4 MHz, the ___ will
also increase.
A. velocity
B. wavelength
C. acoustic impendance
D. absorption
E. viscosity
28
Review of several concepts
11. In an ultrasound transducer, one quarter wavelength refers to the thickness of the:
A. backing material
B. electrical insulator
C. matching layer
D. radiofrequency shield
E. transmit crystal
29
Review of several concepts answers
1. B
Transducer diameter affects only lateral resolution.
2. B
Air/tissue interfaces reflect almost all of the incident beam. To remove the air trapped between the skin and transducer and
minimize skin reflection, gel is applied at the skin during imaging.
3. B
R=(Z2-Z1)2/(Z2+Z1)2 where Z1 and Z2 are the tissue impedances and R is the reflection coefficient.
4. A
The ultrasound attenuation in muscle is approximately 1 dB/cm/MHz. Therefore, the attenuation for a 3 MHz ultrasonic pulse
will be 3 dB/cm. The half value thickness is the distance required to reduce the intensity by a factor of 2. The is by definition
the 3 dB point, since (relative intensity) dB = 10 log(I1/I2), or in this case 10 log (2) = 3dB.
5. A
High frequency transducers (8-20 MHz) produce high resolution, shallow penetration beams suitable for ophthalmology.
6. D
The acoustical impedance is a product of the speed and density in the specified tissue.
7. E
Frequency , type of tissue and the depths affect the amount of attenuation of beam strength. The number of interfaces affect
the loss in power due to echoes at the interfaces.
30
Review of several concepts answers
8. C
An ultrasound beam in muscle is attenuated at a rate of 1 dB/cm/MHz. Hence a 5 MHz beam is attenuated 5 dB/cm in both
the transmit and receive directions. Thus, the ultrasound beam is attenuated 10 dB for each cm in depth. A -60 dB unit at 5
MHz could image 6 cm in depth.
9. A
Attenuation (dB/cm) is approximately proportional to frequency. Increased frequency results in a decreased depth of
penetration and a shorter wavelength (frequency is inversely proportional to wavelength). Axial resolution is improved with
higher frequency due to the shorter spatial pulse length. Higher ultrasound frequency has limited penetration depth, allowing
higher pulse repetition frequency. The speed of sound is constant in tissue. It is independent of the frequency.
10. D
Absorption is proportional to frequency. The velocity of sound is independent of frequency in the diagnostic ultrasound range
(1 to 5 MHz). Since the velocity is constant, the wavelength must decrease. Acoustic impedance and viscosity are
independent of frequency.
11. C
One quarter wavelength matching layer ensures the best energy transfer from the transducer material to soft tissue (and vice
versa).
31
Axial Resolution
32
Axial Resolution
Lateral Resolution
Resolvable
Resolvable
Not
Resolvable
Not
Resolvable
33
Transducers
Backing Block
Piezoelectric Crystal d = λ/2
Matching Layer d = λ/4
• A matching layer is used to prevent large reflections due to acoustic
impedance mismatches between tissue and piezoelectric material.
• Zm is between those of tissue and piezoelectric material.
• The matching layer is λ/4 thick.
34
Transducers
Backing Block
Piezoelectric Crystal d = λ/2
Matching Layers d = λ/4
• Traditionally, each transducer has a preferred frequency (resonant
frequency): d = λ/2
• Recent transducers can now allow wider bandwidths
(for harmonic imaging).
• The backing block dampens vibrations so transducer is ready to
receive. Typically 3 cycles.
35
SPL
• Spatial Pulse Length = nλ
where n = #cycles/pulse,
typically 3
1.5 MHz
time
λ = 1mm
SPL =
3 mm
36
Axial Resolution
Sound
packets
Objects you
want to
resolve
Resolvable
Not
Resolvable
Resolvable
How to get a smaller spl? Higher frequency
37
Resolution is determined by SPL
SPL
2x
x
To resolve x, the spatial pulse length must be < 2x.
nλ < 2x
Axial Resolution x = nλ/2
• Ex. n=3, f = 1.5 MHz, λ = 1 mm, resolution = 1.5 mm
• Ex. n=3, f = 5 MHz, resolution = 0.46 mm
38
Example
• What is the axial resolution for a 3 MHz transducer in which
the backing block limits the number of cycles to four?
Axial Resolution = nλ/2
= 4 * 1540m/s / 3MHz / 2
= 3000 m/s / 3 MHz
= 1 m / 1000
= 1 mm
velocity = distance/time
velocity = λf
39
Resolutions
• Axial resolution = 1/2 spatial pulse length = nλ/2
Important Tradeoffs
• High frequency -> shorter spatial pulse width
-> better axial resolution
But
• High frequency -> increased attenuation -> may not get depth
• So, use highest frequency that will give you the depth you need.
40
1. Modern ultrasound units have spatial resolution limits of about ___ in both axial and lateral dimensions.
a. 0.5 cm
b. 0.1 cm
c. 0.05 cm
d. 0.01 cm
e. 90.005 cm
2. If an ultrasound pulse is about three wavelengths long and the wavelength in soft tissue is 0.33 mm. What is the achievable
axial resolution in mm?
a. 0.5
b. 1.0
c. 3.3
d. 6.3
e. 9.9
3. In diagnostic ultrasound using a phased array, the axial resolution depends on the
a. number of scan lines used
b. imaging depth
c. length of the Fresnel zone
d. transducer diameter
e. pulse length
4. The choice of ultrasound frequency in a particular examination involves a trade-off between _____.
a. convenience in imaging vs. poorer penetration into tissue.
b. spatial resolution vs. transducer cost
c. spatial resolution vs. penetration into tissue
d. temporal resolution vs patient exposure
41
1. b
The spatial resolution of ultrasound units is similar to that of CT scanners. The spatial resolution is around 1 mm.
2. a
Axial resolution = ½ spatial pulse length. The pulse length = wavelength x number of cycles in a pulse = 0.33 mm x 3 = 1 mm.
Axial resolution = ½ x 1 mm = 0.5 mm. Echoes from surfaces further apart than that will not overlap, and can be resolved.
3. e
The axial or depth resolution is ½ the pulse length.
4. c
Axial resolution increases with increasing frequency, but depth penetration goes down.
42
Beam Formation
and Lateral Resolution
43
Beam Formation
Transducer
Multiple Elements
Single Element
5 Elements
44
5 Elements
with Delays
Beam Formation
Transducer
Multiple Elements
Single Element
5 Elements
45
5 Elements
with Delays
Beam Formation
Transducer
Multiple Elements
Single Element
5 Elements
46
5 Elements
with Delays
Beam Sensitivity
Transducer
Multiple Elements
Single Element
5 Elements
5 Elements
Inner Elements Delayed
Sidelobes
47
Thinnest Beam
Imaging Transducers
Linear
Curved
Linear
Square image
48
Fan Shaped Image
Beam Profiles
Remember how when we use many elements and phased them,
we got a better beam profile?
49
Phased Array Transducers
Focal zone
Programmable time delay
• Electronic beam focusing - appropriately delay each element.
• Focal depth set by the programmable time delays.
50
Focusing
51
Multiple Focal Depths
Programmable time delay
• Focus at multiple depths in successive images, combine them
(loss of temporal resolution).
52
Focal Zones
Near (Fresnel) Zone Far(Fraunhofer) Zone
Better Resolution
Worse Resolution
NFL= d2/4λ
Focal spot
Width
w∼λ
High Frequency:
Better lateral resolution
& longer NFL
Near Zone:wavy lines
Far Zone:smooth,
divergent lines
53
Longer NFL:
High frequency & larger
diameter transducer
Dynamic Receive Focusing:
Phased
Array
Proximal Echoes
Variable Time delays
Distal Echoes
54
Slice Thickness
Conventionally, one
element in the slice
direction, focuses at a
particular depth.
Multi elements provide
a better slice.
Not a full 2D array, so call it a 1.5 D array.
55
1. The resolution of a diagnostic ultrasound system is _____.
a. independent of the frequency of operation
b. usually better along the axis than perpendicular to the axis
c. independent of the length of the ultrasound pulse
d. independent of the slice thickness
2. The width of an ultrasound beam measured perpendicular to the image plane determines the
a. axial resolution
b. depth resolution
c. lateral resolution
d. slice thickness
3. When simultaneous multiple focal zones are used in ultrasound, the frame rate generally decreases. The reason for this is ____ .
a. to reduce the acoustical exposure to the patient.
b. multiple transmit pulses are used for each beam line.
c. to give the operator a longer time to view each image plane.
d. extra time is needed to set electronic time delays for transmission.
4. An ultrasound transducer which utilizes a series of elements that transmit sequential sound waves in order to form a rectangular
image is called:
a. annual array
b. phased array
c. linear array
d. sector scanner
e. variable focus scanner
5. The ultrasound transducer which utilizes a series of elements that are time sequenced in order to steer the beam is called a_______.
a. rotating sector
b. linear array
c. phased array
d. color doppler
e. A-mode
56
6. All of the following are types of ultrasound transducers except:
a. coded aperture
b. rotating sector
c. phased array
d. linear array
e. annular ring
7. Concerning ultrasound transducers
a. 2 MHz endorectal transducers are used for imaging the prostate
b. 5 MHz curvilinear arrays are used for large superficial organs, such as a large breast
c. 8 to 20 MHz transducers are used for abdominal imaging
d. low Q-factor transducers use a crystal without backing material
e. a small diameter transducer has a longer focal zone
8. The ultrasound operator can control lateral resolution by changing
a. pulse repetition frequency
b. TGC settings
c. transmit power setting
d. matrix size
e. transmit focal distance depth
9. Elevational (slice thickness) resolution in ultrasound imaging
a. is better with larger transducer height
b. improves with smaller transducer width
c. can be improved using 1.5D transducer focusing arrays.
d. can be improved using shorter pulse duration.
57
1. B
Axial resolution is dependent on the spatial pulse length (SPL), which is dependent on the frequency and Q factor. Lateral resolution
is dependent on the beam diameter, which changes considerably with depth. Axial resolution is better than lateral resolution.
2. D
3. B
4. C
An annular array is a circular arrangement of piezoelectric crystals which can be focused to provide optimal lateral resolution at
different depths in the body. A phased array activates the elements in a manner so that the ultraound beam can be steered in different
directions and produces a truncated triangular image. A linear array fires the ultrasound elements in a sequence and produces a
rectangular picture. A sector scanner has a single transducer which is on a moving head that produces an image that looks like a
triangular section. A variable focus scanner would be similar to an annular array.
5. C
A rotating sector utilizes a single transducer. The linear array fires beams sequentially and it does not steer the beam. Color doppler
measures frequency shifts due to blood flow in vessels. A-mode is a single beam measuring depth.
6. A
Coded aperture is a method of imaging with scintillation cameras without using standard collimators.
7. B
5 MHz transducers produce very good axial resolution when used for large organs such as the breast. Endorectal transducers use 5-7
MHz with limited range to image the prostate. 2.5-5MHz transducers with good depth penetration are used for abdominal imaging.
High Q-factor transducers have no backing material behind the crystal. The near zone (focal or useful zone) increases with transducer
size and frequency.
8. E
Transmit focal depth is selectable in array transducers. The lateral resolution is best in the focal zone.
9. C
Multiple array transducers with 5 to 7 rows, known as 1.5 dimensional transducer arrays, have the ability to steer and focus the beam
in the elevational dimention.
58
Imaging
59
How is Imaging Done?
• Pulse, listen
• Sound bounces back from reflectors all along the way
measured
sound
time
distance = v t
• Amplitude is given a greyscale value: one line of the image
• Repeat with a new line to build up the image
60
How is Imaging Done?
• Pulse, listen, move focal beam, repeat.
• Time to listen to a 20 cm depth is
t = d/v = 40cm / 1540 m/s = 260 µs
• Can repeat pulse how many times/s?
#pulses = 1s/260µs = 3846
• This allows a bunch of lines many times a second,
eg
225 lines at 17 fr/s
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Example
For a depth of 10 cm, what is the PRF?
PRF = 1/travel time = 1/(d/v) = v / (2 * depth)
= 1540 m/s / (2 * .1 m)
= 7. 7 kHz
velocity = distance/time
velocity = λf
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Imaging Tradeoffs
• Pulse, listen, repeat.
• Listen to a 20 cm depth for 260 µs.
• Repeat pulse 3846 times/second.
• Eg. 225 lines at 17 fr/s.
Low frame rate
How to increase frame rate
• reduce depth
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• reduce FOV
1. In ultrasound imaging, the pulse repetition frequency refers to the
a. number of waves per pulse
b. bandwidth frequency of the pulse
c. ratio of the pulse width to echo listening time
d. number of pulses produced per second
e. number of A-lines per image sector
2. In brightness mode ultrasound imaging:
a. axial resolution is generally equal to the pulse length
b. the pulse repetition rate determines the maximum penetration depth
c. axial resolution decreases as the frequency increases
d. tissue penetration increases as wavelength decreases
e. the pulse repetition rate is generally 1 MHz
3. Contrast resolution in ultrasound images depends on several interrelated factors including all of the following except
a. atomic number differences
b. acoustic impedance differences
c. attenuation differences
d. variation in density and size of scatterers within tissues or organs
e.
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1. D
2. B
The pulse repetition rate (PRR) refers to the number of separate pulses of sound that are transmitted per second. Its value is
generally 1000 pulses/sec (1 kHz). High PRR limits the penetration depth because the transducer receiving time decreases.
Echoes from distant parts may still be out, or in the way when the next pulse starts. The spatial pulse length is the number of
sonic waves in the pulse times the wavelength. Two objects will be resolved if the pulse length is less than one-half the
separation. The means that depth, or axial resolution is limited to half the pulse length. If the frequency increases, wavelength
decreases, and so does the pulse length, but the axial resolution increases. For a given transducer, as the wavelength
decreases, the frequency increases and tissue penetration will decrease.
3. A
Ultrasound contrast resolution depends on several interrelated factors. Acoustic impedance differences give rise to reflections
that delineate tissue boundaries and internal architectures. The density and size of scatterers within tissues or organs produce
specific “texture.” With proper signal processing, attenuation differences result in gray-scale differences among the tissues.
Atomic number differences do not produce acoustic contrast.
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Artifacts
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Attenuation
5 MHz
Huge loss of signal
Worse at high frequency
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3.5MHz
Time Gain Control
TGC - Amplifies the signal based on its depth - equalizes the
signal across the image.
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Artifactual Enhancement
• Some structures do not have much attenuation - cysts, bladder.
• Can have high signal beyond these structures.
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Artifacts - Reverberation
• Spurious images are
formed when echoes
bounce off the back
surface of materials.
Transducer
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Angle of Incidence
• When passing into a medium with a diffe
velocity, the angle changes according to
Snell’s law:
sinθ1/sinθ2 = v1/v2
θ1 θ1
Reflection
Tissue 1
(v1)
Tissue 2
(v2)
Refraction
θ2
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Refraction Examples
If a sound wave traveling through muscle hits a fat plane at an
angle of 30° from the perpendicular, the transmitted wave will
change its direction by how much?
sinθ2 = sinθ1*v2/v1
θ2 = sin-1(sin(75°)*1450/1540)
θ2 = 65°
Δθ = 10°
Refraction can be a problem when the beam hits an initial large
fat plane - it can get defocused.
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Air v. Bone
Material
Air
Soft Tissue
Bone
Impedence
0.0004
1.6
7.8
Attenuation
12
1
20
Reflection (ST -> Air) = 100%
Can get reverberation artifacts
looks like noise and stuff beyond air bubble
Reflection (ST -> Bone) = 43%
You do transmit, but highly attenuated
low signal beyond bone
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1. In ultrasound imaging, “Shadowing” refers to decreased echo amplitudes _______ a highly attenuating structure.
a. distal to
b. within
c. proximal to
d. lateral to
2. A shadowing artifact in ultrasound may be due to the reduction of reflected intensity behind all of the following except
a. a strong attenuator
b. a highly reflective interface
c. gas or air
d. water
Match the ultrasound artifact with its mechanism
a. occurs distal to objects of high attenuation, resulting in a hypointense signal
3. mismapping
b. occurs distal to objects of low attenuation, manifested by a hyperintense signal
4. shadowing
c. occurs between two strong reflectors, resulting in equally spaced signals of diminishing amplitude
5. side lobe energy emission
in the image
6. enhancement
d. variation of the speed of sound in the tissues causes displacement of returning echoes from distal
7. reverberation
anatomy
e. anatomy outside the main beam is mapped into the main beam
8. The primary use of time gain compensation is to increase the transducer intensity at different times during the pulse travel
True or False
9. Which of the following are not image artifacts in ultrasound images?
a. shadowing
b. reverberation
c. side lobes
d. multipath reflections
e. frequency modulation
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1. A
2. D
3. D
A shadowing artifact is caused by a lack of reflection from an area. This can be caused by an incident beam being highly
attenuated, or if the beam is strongly reflected from an overlying interface, such as between air or gas and tissue. Water has a
low absorption coefficient, and acts as a window, generally producing no shadowing.
4. A
5. E
6. B
7. C
8. True
TGC is necessary to compensate for increased signal attenuation with depth. TGC causes the signal gain to increase as the
echo-return time increases.
9. E
Shadowing is a low ultrasound signal caused by high attenuation, strong reflection or refraction of the beam. Reverberation
arises from multiple echoes from closely spaced surfaces which yields a “ring-down artifact.” Side lobes are off-shoots of the
sonic beam sich produce mispositioned echoes. Multipath reflections and refractions cause signals to be geometrically
misplaced. Frequency modulation is one of the means of producing radio broadcasts.
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Doppler
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Quality Factor - “Q” = f/BW
Long Pulses
High Q
Used for Doppler
Short Pulses
Low Q
Used for imaging
Low BW
High BW
Frequencies
Transmitted
f
f
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Doppler Shift
v = λ*f
Δf ∝ Reflector Speed
Transducer
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Doppler Shift
v = λ*f
Δf ∝ Reflector Speed
Transducer
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Doppler Shift
v = λ*f
Δf ∝ Reflector Speed
Transducer
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Doppler Shift
The frequency shift is given by
fshift = 2 * (v/c) * fo * cos(θ)
where
θ
v is the velocity of the reflector
c is speed of sound
fo is the transducer frequency
θ is the angle of the velocity wrt/ the sound wave.
Example: If a 5 MHz transducer is used, what is the shift
measured from blood moving at 20 cm/s at an angle of 60°?
f = 2 * (20 cm/s / 1540 m/s) * 5 MHz * cos (60°)
= 650 Hz
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1. Decreasing the ultrasound pulse length will generally
a. decrease the Q factor
b. reduce the lateral resolution
c. reduce axial resolution
d. decrease the acoustical impedance of the tissue
e. increase the speed of sound in tissue
2. Concerning ultrasound imaging, a transducer with a high Q factor would be desirable for Doppler applications.
True or False?
3. The largest Doppler ultrasound shifts are from scatterers in _______
a. a large artery
b. arterial branches
c. the aorta
d. capillaries
e. vessel walls
4. In Doppler ultrasound, the frequency change depends on:
a. the nominal transducer frequency
b. the speed of sound in the medium through which the beam passes
c. the angle between the beam and the moving structure
d. a and c
e. a, b, and c
5. An ultrasound beam passes from soft tissue to moving arterial blood. Which would give the largest Doppler shifts?
a. 5 MHz, 45° angle to artery
b. 10 MHz, 45° angle
c. 10 MHz, 90° angle (perpendicular)
d. 10 MHz, 0° angle (parallel)
e. 5 MHz, 0° angle (parallel)
6. Doppler frequency shifts in ultrasound are observed in the audible range. What is the Doppler shift for a 5 MHz transducer
used to reflect an ultrasound wave at a 45° angle wrt a vessel with an average speed of blood flow of 30 cm/sec?
a. 0.7 kHz
b. 1.4 kHz
c. 7.0 kHz
d. 14.0 kHz
e. none of the above
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1. a
Q factor = operating frequency/bandwidth. The Q factor increases with increased pulse length due to narrow bandwidth. A
decreased (short) pulse length has a much broader bandwidth and therefore decreased Q factor.
Lateral resolution is independent of pulse length. The beam width and number of scan lines are the major factors determining
the lateral resolution in the image.
A decreased pulse length will improve the axial resolution.
The acoustic impedance of tissue depends only on the density and the speed of sound. The speed of sound in tissue depends
on the tissue, and is independent of the pulse length.
2. True
The Q of a transducer is a measure of its frequency bandwidth. A high Q factor implies a narrow BW, hence almost all of the
ultrasound is emitted at or near the operating frequency. Doppler applications look for small changes in the frequency of
reflected beams. Therefore, a high Q is advantageous for Doppler work.
3. c
In Doppler ultrasound, the shift is proportional to the velocity of the red blood cells in the vessel. The highest velocity, about
100 cm/s is found in the aorta. In arterial branches, it is only about 4.5 cm/s.
4.e
The Doppler shift Δf = (2f0 v/c) cos Θ where f0 is the incident frequency, v is the velocity of the object, c is the speed of
sound in the medium, and Θ is the angle between the sound beam and the moving object.
5. d
The greatest Doppler shift is at the highest frequency with the beam parallel to the direction of blood flow.
6. b
Δf = (2f0 v/c) cos Θ
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Doppler
• Continuous Wave Doppler
• Pulsed Doppler • Color Doppler
• Power Doppler 84
Doppler - Continuous Wave (CW)
transmitter receiver
• 2 piezoelectric elements used
• sensitive volume is overlap of two beams
• lacks depth resolution and is sensitive to
weak signals - can get signal from
multiple vessels
• often limited to superficial structures and
• obstetrics - listening to fetal heart
• can be hooked to an audio speaker
comparison
sound signal
audio range
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Doppler
• Continuous Wave Doppler - no specific depth
velocity spectrum averaged over space
• Pulsed Doppler - specific depth
velocity spectrum
• Color Doppler
• Power Doppler
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Pulsed Doppler - “Duplex”
• Single transducer, sound is pulsed
1. Choose Depth
“Range Gate”
• based on transit time
2. Set the Vessel
Angle
vmax
vres
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Doppler - Pulsed
• Use a number of pulsed measurements. From each measurement,
get a phase.
φ
Ref
Sig 1
Sig 2
t
Sig 3
~µs
~ms
• Use all together to calculate the frequency shift waveform:
Signal(time) ---> FFT ---> Signal(frequency)
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Doppler - Vmax
• The maximum
frequency shift we can
detect is half the PRF
(the factor of 2 is needed
to prevent aliasing):
fshift = fPRF/2
2 * v / c * fo = fPRF / 2
vmax = fPRF / 2 / 2 * c / fo
substitute fPRF = c/z
vmax = c2/(8 * z * fo)
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Doppler - Pulsed
vmax = c2/(8 * z * fo)
The maximum velocity you can measure is determined by
the depth of the range gate and the transducer frequency.
Higher velocities will “alias”.
• Example: z = 10 cm, 3 MHz, what is vmax?
vmax = (1540 m/s)2/(8 * .1 m * 3MHz)
= .98 m/s
• Lower frequencies allow measurement of higher velocities.
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Doppler
• Continuous Wave Doppler - no specific depth
velocity spectrum averaged over space
• Pulsed Doppler - specific depth
velocity spectrum
• Color Doppler - more than one depth,
only one measurement (average)
• Power Doppler
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Color Doppler
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Doppler - Color Flow
• Uses multiple gates allows velocity
measurements in many
locations.
• But only measure a
mean in each location fewer measurements/
location.
• Overlay a color on a
greyscale image.
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Doppler
• Continuous Wave Doppler - no specific depth
velocity spectrum averaged over space
• Pulsed Doppler - specific depth
velocity spectrum
• Color Doppler - more than one depth,
only one measurement (average)
• Power Doppler - not directional, very sensitive
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Power Doppler
• Any measurable velocity is encoded with color. The color
depends on the amplitude of the doppler signal.
• More sensitive than color flow.
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Signal Magnitude
Velocity
Color Flow
Display is more
sensitive to slow flow.
Power
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Choice of Frequency
High frequency -- more signal scattered back from blood
But, attenuation.
So, use high frequency close to transducer, lower frequency
at depth.
May even use a lower frequency than for a B-mode image.
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1. When operating in the Doppler mode
a. one can employ a pulse echo or continuous wave technique
b. the frequency shifts are independent of the ultrasound frequency
c. meaningful information can only be obtained from moving interfaces
d. the principal application is measurement of midline shifts of the brain
e. only qualitative measurements can be obtained.
2. In Doppler ultrasound, select the one correct answer:
a. a low Q transducer is desirable
b. continuous Doppler uses a single transducer
c. the Doppler frequency shift is dependent on the velocity of sound
d. typical Doppler frequency shifts for moving blood are in the range of 2-3 MHz.
e. all of the above
3. B-mode scanning is used in placental localization
True or False
4. The best axial resolution is obtained with a ______ transducer.
a. 3.5 MHz pulsed
b. 5 MHz continuous
c. 5 MHz pulsed
d. 10 MHz continuous
e. 10 MHz pulsed
5. “Duplex” ultrasound refers to
a. simultaneous recordings of A lines from two transducers
b. doubling the pulse repetition frequency to enhance resolution
c. anatomical images and a selectable Doppler window gate simultaneously displayed
d. simultaneous measurement of ultrasound attenuation and acoustic impedance
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6. Color flow Doppler ultrasound displays the largest flow rates when the __________
a. frequency increases
b. wavelength decreases
c. power increases
d. angle to the vessel decreases
e. transducer is closer to the vessel
7. Color Doppler ultrasound measures blood flow based upon information from all of the following factors except
a. impedance
b. wavelength
c. ultrasound speed
d. angle of incidence
e. frequency shift
8. In pulsed Doppler ultrasound, the maximum blood velocity determined from the maximum Doppler shift is increased with a _____.
a. lower pulse repetition frequency (PRF)
b. lower operating frequency, f0
c. smaller Doppler angle
d. shorter operating wavelength
9. Which of the following best describes Power Doppler ultrasound?
a. It is less sensitive than color Doppler but provides better directionality of blood flow.
b. It requires a major hardware upgrade to implement on modern US equipment.
c. It delivers significantly more power deposition than standard Doppler techniques.
d. It cannot be used with existing transducers because directional information is lost.
e. It is chiefly implemented by software changes on existing US equipment.
10. In Doppler ultrasound:
a. Doppler imaging detects changes in the speed of sound from a moving sample.
b. Low Q-factor transducers are desirable.
c. Doppler systems tend to use lower frequencies than B-mode systems
d. Doppler shift decreases when the angle between transducer and direction of flow is decreased.
e. Continuous Doppler employs one transducer used as a transmitter and one receiver.
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1. a
Continuous and pulsed Doppler are both available. Meaningful information is displayed by moving and non-moving interfaces (e.g.,
normal vasculature vs clot).
2. C
Frequency shift in Doppler ultrasound increases with increasing incident beam frequency and the speed of the moving object (blood flow).
Δf, however, decreases with increasing velocity of sound in blood and the beam angle with the direction of the moving blood. High Q
transducers are desirable in Doppler images because they produce a narrow range of ultrasound frequencies. In continuous Doppler two
transducers are needed, one to continuously transmit, and the other is to continuously receive. Frequency shifts for moving blood will vary
depending on the factors mentioned earlier. Typical shifts are usually less than 1 kHz.
3. True
Brightness (B) mode displays a static image of a two dimensional tissue section. It is the common display mode in ultrasound imaging.
4. e
Axial (depth) resolution is the ability of the ultrasound beam to separate two objects lying in tandem along the axis of the beam. The pulse
length equals the wavelength of the beam ties the number of wavelengths used. The axial resolution is defined as ½ of the pulse length. The
highest frequency (10 MHz) has the shortest wavelength and a continuous beam has an infinite pulse length.
5. c
Duplex ultrasound refers to the simultaneous acquisition of dynamic B scan greyscale information and Doppler information within a
predetermined “gate.”
6. d
7. a
8. b
In pulsed Doppler, the maximum blood velocity Vmax that is accurately determined is increased with larger PRF, lower operating
frequency (longer operating wavelength), and larger Doppler angle.
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9. e
Power Doppler is a signal processing method that relies on the total strength of the Doppler signal (amplitude) and ignores directional
(phase) information. It does not require any major hardware changes and uses existing transducers. The power mode of signal acquisition is
dependent on the amplitude of all Doppler signals, regardless of the frequency shift. Power Doppler produces images that have more
sensitivity to motion (good for subtle and slow blood flow) and are not affected by Doppler angle (largely nondirectional). In power Doppler,
the transmit power levels are typically the same as those in a standard color flow procedure.
10. c
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Harmonic
102
Harmonic Imaging
• Transmit at a low frequency within transducer bandwidth
• Sound is reflected at frequency f (fundamental) and
also at 2f (harmonic), 3f, 4f, ...
• Receive at 2f to get only the harmonic
Transducer
f
f
3f
f
2f
2f
103
Harmonic Imaging of Contrast Agent
• micro-bubbles scatter sound at transmit frequency (fundamental)
and at 2nd and higher harmonics.
Fundamental imaging w/out contrast agents
Harmonic flow imaging with intravascular encapsulated bubbles
104
Tissue Harmonic Imaging
Cardiac scans:
• reduced reverberation noise between probe and rib bones
• Improved visualization of weak myocardial borders
105
Tissue Harmonic Imaging
Reverberations
Beam Aberration
Beam sidelobes
Abdominal Wall
fo
Conventional Imaging
reverb noise at fo
2fo
Harmonic Imaging
Receiver removes fo
& reverb noise
106
Cleaner Tissue
Image & Better
Cystic Clearing
Tissue Harmonic Imaging
Sound Propagation in tissue is nonlinear
⇒ Pulse becomes progressively distorted in shape
⇒ Backscatter occurs at harmonic frequencies
c+Δc
c ~1540 m/s
c-Δc
direction of propagation
More nonlinear effect at high intensity - less at low intensity where the
Artifacts are.
107
108
109
110
Tissue Harmonic Imaging
Cardiac scans:
• reduced reverberation noise between probe and rib bones
• Improved visualization of weak myocardial borders
111
1. In comparison with conventional ultrasound images, tissue harmonic images have all of the following characteristics
except:
a. improved lateral spatial resolution
b. faster image acquisition
c. reduced side lobe artifact
d. removal of reverberation artifacts caused by anatomy adjacent to the transducer
e. enhanced contrast agent imaging
2. Concerning ultrasound harmonic imaging, the following are true except
a. harmonic images are formed from integral multiples of the transducer fundamental frequency.
b. the transmitted frequencies and the returned harmonics have the same frequency but different intensity.
c. no harmonics are generated at near field.
d. harmonic pulses travel half the distance of the fundamental pulse.
e. harmonic imaging improves lateral resolution.
3. In ultrasound, harmonic imaging:
a. decreases clutter and improves lateral resolution near surface of patient.
b. decreases clutter and improves lateral resolution at depth.
c. decreases clutter and improves axial resolution near surface of patient.
d. decreases clutter and improves axial resolution at depth.
e. is only used with microbubble contrast agents.
112
1. b
2. b
3. b
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Safety
114
Bioeffects and Safety
• Heating
• At high power, cavitation is the creation and collapse of small
bubbles.
• Watts/kg that the body can dissipate by normal means.
• At low power, no apparent problems.
115
1. Possible adverse biological effects from ultrasound procedures include:
a. cavitation
b. genetic transmutation
c. ionization
d. production of radicals
e. none of the above
2. Acoustic exposure to the patient will be increased by increased:
a. receiver amplification
b. transmit power
c. frame averaging
d. TGC settings
e. threshold levels
116
1. a
2. b
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