-
The Webbook of Medical Imaging
By Jens E. Wilhjelm (ed.)
Markus Nowak Lonsdale
Mikael Jensen(Ver. 1 10/4/03) © 2001-2003 by J. E. Wilhjelm,
Markus Nowak Lonsdale and Mikael Jensen)
1 IntroductionMedical imaging is a collection of technologies,
all having the purpose of visualization of the interior
of the intact, living human body for the purpose of diagnosis.
The importance of medical imaging can perhaps best be understood
when remembering that human vision is by far the channel with the
largest bandwidth for communication between the human brain and the
outer world. Medical imaging serves to open the interior of the
body to the enormous analytical data handling performance of the
the eyes and the visual cortex.
Medical imaging is probably the largest area of diagnostic
technology available at the hospital. It deals with visuallization
of both structure and function of the internal of the body. Medical
imaging uses a wide range of different techniques as can be seen
from Table 1, which shows the main imaging modalities used
clinically at the hospital today.
The present book will try to explain the physical principle
behind each of these imaging modalities, together with a
description of how these are implemented. The book with also
provide some clinical images.
2 GlossaryTomography Image of a slice of the body[1]
Table 1: Overview of the physical principles of common imaging
modalities
Common name Physical basis
Ultrasound Sound
Planar X-ray X-ray
CT X-ray
SPECT Radioactive tracers
PET Radioactive tracers
MRI Nuclear magnetic resonance
1/2
-
3 References[1]Dorland’s Illustrated Medical Dictionary. 27th
edition. W. B. Saunders Co., Philadelphia, PA,
USA. 1988.
2/2
-
X-ray imaging:Fundamentals and planar imaging
Mikael Jensen1 and Jens E. Wilhjelm2
1Hevesy Laboratory, DTU Nutech, 2Biomedical Engineering, DTU
Elektro
Technical University of Denmark(Ver. 4 28/8/14) © 2004 - 2014 by
M. Jensen and J. E. Wilhjelm
1 IntroductionX-ray imaging is the most widespread and
well-known medical imaging technique. It dates back to
the discovery by Wilhelm Conrad Röntgen in 1895 of a new kind of
penetrating radiation coming from an evacuated glass bulb with
positive and negative electrodes. Today, this radiation is known as
short wavelength electromagnetic waves being called X-rays in the
English speaking countries, but “Roengten” rays in many other
countries. The X-rays are generated in a special vacuum tube: the
X-ray tube, which will be the subject of the first subsection. The
emanating X-rays can be used to cast shadows on photographic films
or radiation sensitive plates for direct evaluation (the technique
of pla-nar X-ray imaging) or the rays can be used to form a series
of electronically collected projections, which are later
reconstructed to yield a 2D map (thus, a tomographic image). This
is the so-called CAT or CT technique (see the chapter on CT
imaging).
1.1 Characterization of X-rayX-rays are electromagnetic
radiation (photons) with wavelengths, 10 pm < < 10 nm. They
travel
with the speed of light, c0 300 000 km/s and has a frequency of
= c0/ [Hz]. The energy of the individual photon is E = h [J], where
h = 6.62×10-34 Js is Planck’s constant. The energy of an X-ray is
typically measured in electron volts (eV). 1 eV is the energy
increase that an electron experiences, when accelerated over a
potential difference of 1 V. Thus, 1 eV = qeV = 1.602×10-19 J,
where the charge of an electron is qe (both qe and V is negative in
this context).
I
U+
Glass envelope
Anode
Cathode(heated filament)
Electrons
X-rays
Protective lead
Figure 1 X-ray tube showing cathode and anode with electrons
accelerated from cathode towards anode. The tube generates X-rays
in all directions, but due to the encapsulation most are lost and
only a fraction is used for imaging.
Problem 1 Calculate the frequency and energy for monochromatic
x-rays with = 1 nm.Answer: = c0/ 3×1017 Hz = 300 000 THz. E = h
1.99×10-16 J = 1.24 keV.
1/9HEVESYBME
-
1.2 X-ray generation: The X-ray tube
0
200
400
600
800
1000
1200
0 20 40 60 80 100 120 140 160Photon energy (keV)
Rel
ativ
e in
tens
ity a
t fix
ed e
lect
ron
curr
ent
V=50kVV=90kVV=140kV
Characteristic X-ray emission from Wolfram
1 mm Al filter cut-off
Figure 2 X-ray emission from Wolfram anaode X-ray tube. Observe
that for a given tube voltage, the higher the energy of the
photons, the less there are. And if the number of photons are to
increase, then the tube voltage should increase. Data from [1].
A typical X-ray tube is depicted in Figure 1. It consists of an
evacuated glass bulb with a heated fil-aments (Danish: glødetråd)
as the negative electrode and a heavy metal positive anode. Thermic
elec-trons emitted by the heated filament are accelerated across
the gab to the anode. If the voltage between cathode and anode is U
volts and the current in the tube being I amperes, each electron
will be hitting the anode with a kinetic energy of U eV. The power
deposited in the anode will be I times U, and the total energy
transferred to the anode in an exposure lasting t seconds will be
IUt.
The electrons will be slowed down in the anode material, mainly
releasing their energy as heat, but to a small degree (few percent)
the energy is transformed to either Bremsstrahlung or
characteristic X-rays. The Bremsstrahlung originates from the
sudden deacceleration and direction changes of the primary
electrons in the field of the anode atoms, the characteristics
X-rays originates from the knock-out and subsequent level filling
of inner electrons in the atoms of the anode material. The highest
pos-sible quantum energy of emanating X-rays (measured in eV) will
be equal to U. Typical energy spectra as a function of voltages are
shown in Figure 2. Please note that the spectra are all taken at
the same current, only the voltage has been varied. This
demonstrates that the total number of X-ray pho-tons are heavily
dependent on tube voltage. In addition to the information in Figure
2, a general rule of thumb says that 15 keV increase in voltage
corresponds to a doubling of the photon output. For practical
medical applications, the low energy part of the photons are
normally not used but removed by filtering either inside or just
outside the X-ray tube. Normal filter materials are either
aluminium or copper. The thicker the filter and the higher the
atomic number of the filter, the greater the cut-off of low energy
photons.
The description of the exposure characteristics of a given X-ray
tube will comprise the voltage (in units of kV), the current (in
units of mA), the time of exposure (in units of s) and the degree
of filtering (for example a plate of Al, one mm thick next to a
plate of Cu, 0.5 mm thick). As the total number of photons produced
for a given high voltage setting only depends on the product of
current and time this is often stated as a product in units of
mAs.
2/9HEVESYBME
-
1.3 Anode material, power dissipationHeavy elements are normally
preferred for anode materials as the high Z-number gives efficient
pro-
duction of the part of the X-ray that originates from
Bremsstrahlung. The characteristic X-ray lines, which add to the
total energy spectrum, normally appear in the range 50-70 keV,
which ofte is in the middle of the medical useful energy range.
The thermal load on the anode material both during the short
exposures and averaged over time when performing rapid, multiple
exposures heats the anode dramatically. For this reason, normally a
high melting point material is used. Anodes made out of Tungsten
(Danish: Wolfram), abbreviated W, are very common. The area of
thermal dissipation can be enlarged by rotating the anode during
exposure. An example of such a device is shown in Figure 3.
2 Typical X-ray system
Figure 3 Rotating anode X-ray tube. "RTM" anode designates a
Molybdenum anode mixed with 5% Rhenium to improve the thermal
stability. The metal anode is supported by graphite to improve the
total thermal capacity. Source: Siemens.
Figure 4 shows a typical X-ray system. The X-ray is generated by
the X-ray tube (Danish: Rønt-genrør). Low energy photons are
removed by the Al filter, since as they cannot penetrate the object
and contribution to the information on the film, they would only
add needless to the dose received by the object. X-ray radiation
outside the image region on the film is removed by the collimator
(Danish: primærblænde). Attenuation (what is measured on the film)
and Compton scattering take place at the object. Only photons
moving directly from the source to the film are allowed through the
grid at the bottom (Danish: sekundærblænde).
3/9HEVESYBME
-
Figure 5 provides an interactive illustration of a simple X-ray
system allowing for translation of a simple homogeneous phantom. As
the distance is changed, different edge effects can be observed. By
pressing the “Lines” button, the geometry of the edge effect is
visualized.
3 Geometrical considerations
Al-filter (removes low energy radiation)
Secondary radiation(Compton scattering)
Object
ScreensFilm
Protective shield of lead
Collimator
Grid (removes Compton scattering)
X-ray tube
Figure 4 Schematic illustration of a typical X-ray system.
Referring to Figure 5 we can define the distance from the X-ray
origin (the focus1) to the objects as FOD. The distance from the
focus to the film2 or any other medium of radiation detection can
be de-fined as FFD. Any object will to some degree attenuate the
X-ray, and variation in X-ray absorption across the objects will
create a corresponding variation in the radiation impinging on the
film. An un-avoidable and sometimes desirable geometrical
magnification of the image relative to the object can be deduced
from the triangle in Figure 5. The enlargement factor F, can be
defined as:3
F = size of film image / size of object = FFD / FOD (1)
If near to normal picture size and little variation in
enlargement is sought for organs having different depths in the
body, the geometrical magnification should be minimized by using a
large focus to film distance (FFD) and a small object to film
distance. A small object to film distance also improves im-age
contrast, as blurring by scattering increases with increasing
distance between object and film. This is because the origin of the
scatter is mainly inside the object: The longer the scattered
radiation is
1. The "focus" is in this context the source of the X-ray
photons, as it is the name for the electron spot on the anode of
the X-ray tube.
2. The term "film" is still common language, even though the
conventional x-ray film to a large degree has been replaced by
various other imaging plates.
3. Here “size” means any distance e.g. the length of a given
object.
4/9HEVESYBME
-
allowed to travel between the object and the film, the more this
radiation diverges from the true un-scattered photons.
Figure 5 Simple illustration of the geometry of the planar X-ray
system corresponding to Figure 4. The X-ray image of the green
homogeneous box is shown to the right. The focal point is the
origin of the X-rays. The film is identical to the detector. FOD =
focus to object distance. FFD = focus to film distance.
Similar geometrical considerations (i.e. similar triangles) can
demonstrate that extended size of the focus will generate blurring
on the film.
4 Origins of contrast in the X-ray imageX-rays are attenuated
according to the normal linear attenuation law:
I(x) = I0 exp(-x) = I0 exp(-/ x) (2)
where x is the distance transversed in the material and is the
so-called linear attenuation coefficient in units of m-1. I0 is the
intensity at the entrance to the material (x = 0) and I(x) is the
intensity at dis-tance x. In the latter part of (2), is the density
of the material. By giving the attenuation coefficient in units of
/ and the thickness in length times density (area weigth, e.g.
g/cm2) the attenuation co-efficient (now called mass attenuation
coefficient) becomes independent on the physical state of the
material. Mass attenuation coefficients for some common tissues are
given in Table 1.
The microscopic description of the attenuation comprises photo
electric effects and Compton scatter-ing, which are both described
in the chapter on nuclear medicine. For the understanding of the
X-ray technique, it suffices to say that the linear attenuation
coefficient for human tissues varies approxi-mately as the electron
density. Thus, it varies roughly proportional with the physical
density (kg/m3). Air has the lowest density, lung tissue has lower
density than fat, fat has lower density than muscle,
5/9HEVESYBME
-
again having much lower density than the bone mineral of the
skeleton. The X-ray attenuation varies accordingly. X-rays
transversing parts of the body having high absorbing material will
be much more attenuated, and the film or radiation capture device
will in this region not receive as much radiation. It should be
remembered that the X-ray image is a negative (bright areas
correspond to high attenua-tion) and that the image is a 2D
projection of the 3D distribution of attenuation.
0,1310,1860,424Bone =1,92 g/cm3
0,1370,1710,227Water =1,00 g/cm3
0,1360,1690,226Muscle =1,05 g/cm3
0,9995,5498,041Lead =11,35 g/cm3
0,1360,1690,212Adipose tissue =0,95 g/cm3
0,1220,1540,208Air =0,0013 g/cm3200 keV100 keV50 keVµ given in
cm2/g
Table 1: Mass attenuation coefficients for typical tissues in
/From [2].
Figure 6 Normal chest X-ray image. This image is recorded with a
tube voltage of 150 kV to minimize the contribution from bone.
Press “?” to try to identify tissues.
An example of a normal X-ray image of the chest (one of the most
common medical imaging proce-dures) is seen in Figure 6. Notes that
the most attenuating areas (ribs, vertebral column, heart) appear
white while the lungs with little attenuation appear black on a
conventional planar X-ray image.
Problem 2 Does the planar X-ray image have arbitrary units (or,
put in other words, are the pixel val-ues relative or
absolute)?
6/9HEVESYBME
-
The image information in the planar X-ray is mainly anatomical,
actual densitometric measurements on the film are only performed
for quality assurance programs and yields little medical
information. Today, all planar X-ray images are evaluated by a
human observer.
5 Film, intensifyer foils and screensOriginally, the radiation
was captured by a normal photographic film. In the film, the
energetic X-
ray photons are absorbed in the silver halide (NaB-NaI)
crystals, generating very small amounts of free silver. During film
processing, any grain with small amounts of free silver are
completely con-verted to metallic, nontransparent silver, while the
remaining unreduced silver halide is removed by the fixative. X-ray
films are of course made to the size necessary for the anatomical
situation in ques-tion and can be very large. To increase
sensitivity and thus lower radiation dose, the photo sensitive film
emulsions are often thicker and occasionally coated on both sides
of the film, in contrast to nor-mal photographic film. The silver
contents of the films makes X-ray films rather expensive. Any film
has a specific range of optimal sensitivity (exposure range from
complete transparency to completely blackened). Although modern
equipment are normally assisted by electronic exposure meters, the
correct choice of film, exposure time, exposure current and high
voltage is still left to the judgement of the X-ray technician.
To improve the sensitivity and thus lower radiation exposure to
the patient, the film is often brought in contact with a sheet of
intensifying screen. The screen contains special chemical compounds
of the rare earth elements, that emits visible blue-green light
when hit by X-rays or other ionizing radiation. This permits the
use photographic film with thinner emulsions and more normal
sensitivity to visible light. While increasing the sensitivity, the
use of intensifying screen on the other hand blurs the im-ages as
the registration of X-ray radiation is no longer a direct, but an
indicted process.
The patients or the object is not only the source of X-ray
absorption but also of X-ray scattering, mainly due to Compton
effect (please see the chapter on Nuclear Medicine). Any part of
the patient exposed to the primary X-ray beam will be a source of
secondary, scattered, X-rays. These X-rays will have lower energy
than the original ray, but as no energy discrimination is used in
the registration, also the secondary scattered radiation adds to
the blackening of the film. The scattered radiation car-ries no
direct geometrical information about the object and thus only
reduces the contrast by increas-ing the background gray level of
the film. Scattered radiation can to some degree be avoided by the
use of special collimators called raster. The raster can be a
series of thin, closely lying bars of lead, only allowing radiation
coming from the direction of the focus point to hit the film while
other direc-tions are excluded. A typical example of a complete
radiography cassette content is shown in Figure 7.
6 Digital radiography and direct captureDuring the last years
the conventional film based radiography has gradually been replaced
by newer
digital techniques. The end points is of course the acquisition
and storage of the X-ray image infor-mation as computer files. It
should be noted that X-ray images are normally of very high
resolution (more than four thousand by four thousand pixels) with
large dynamic range (12 to 16 bits). The cor-rect handling and
display of such image information without loss or compression is
still the subjects of specialized workstations. The digital X-ray
images are normally stored and displayed in so-called PACS systems
(Picture Archiving and Communications System). The image
information is either temporarily captured on phosphor plates for
subsequent transfer to digital storage by so-called phos-phor plate
readers, in function much related to the old X-ray film processors)
or by direct, position
7/9HEVESYBME
-
sensitive electronic X-ray detection devices, the so-called
direct capture systems. At present (2003) the geometrical
resolution of the various digital techniques is still somewhat
inferior to the best pos-sible film technique. However, the
benefits of rapid viewing, interactive image availability,
postpro-cessing and digital transmission often outweighs the
reduction in image quality. For special applications, like breast
cancer detection by mammography, the photographic film is still the
system of choice.
X-ray not passing through the raster
X-ray passing through the raster
Raster
ISFilm
Figure 7 X-ray cassette, containing double coated film,
intensifying screen (IS) and lead raster.
7 Analysis of image of phantomConsider the conventional planar
X-ray image of a typical phantom in the course shown in Figure
8.
This X-ray system has a high dynamic range, thus the image has
many levels. To include as many details as possible, this X-ray is
shown - very untraditional - in color. Normal X-ray images are
shown in shades of gray.
It is not trivial to analyze such an image, unless one is very
careful. First one should remember that what one see (the contrast
in the image) is total attenuation which again is dependent on what
kind of material is present and how much of this material is
present in the trajectory between focus and a given detector. And
since the colorbar is representing relative values, both of these
observations should be considered relative to other places in the
image.
The analysis goes as follows:
Since the image shows more than the box of the phantom itself,
the connectors can be identified. Those appear dark yellow and
since they obviously must attenuate more than the nearby air (which
appears nearly white) it can be concluded, that a high pixel value
corresponds to low attenuation (in contrast to normal X-ray). Be
aware of a potential pitfall: if the film is larger than the area
that is ex-posed (as controlled by the collimator at the x-ray
tube), then the image has an additional outer frame, which,
obviously, should not be used in the analysis.
Next consider the fiducial markers. Since the lid was in place
during recording of the image, we have two series of materials that
are penetrated by the X-rays: one through the markers and another
between the markers:
• agar, acrylic base plate
• acrylic lid, agar, acrylic base plate.
8/9HEVESYBME
-
Problem 3 Please make a cross-sectional drawing of this, and be
sure that the total distance penetrat-ed by the X-ray is the same
(for this particular problem, it is assumed that the X-rays hits
perpendic-ular to the surface).
Since air-free agar is mainly water and thus has very much the
same density as water and since acryl-ic has a higher density, the
fiducial markers must attenuate less, and thus appear brighter than
the sur-roundings. This is also the case in the image in Figure
8.
Horizontal (mm)
Ver
tical
(mm
)
40 60 80 100 120
20
40
60
80
100
120
140
160
300
400
500
600
700
800
900
1000
Figure 8 Left: Top photo of phantom 4 from 2009. Right: Planar
X-ray image of same phantom. The image is shown in color to enhance
contrast. The two images are sought aligned as well as
possible.
Next consider the tube. The tube contains air inside and thus
the wall is the most attenuating. But even if the tube contained
water inside, the wall would still be the most attenuating. Thus,
the tube periphery should be darker than the center when looking at
the projection image. The shape of the actual attenuation profile
through the tube is considered in Problem 13 in the exam of year
2008.
8 References[1] Johannes Jensen og Jens Munk: Lærebog i
Røntgenfysik, Odense 1973.2.udg. s.121.
[2] Data from
http://physics.nist.gov/PhysRefData/XrayMassCoef/cover.html
9/9HEVESYBME
-
CT scanning
By Mikael Jensen & Jens E. Wilhjelm
Risø National laboratory
DTU Elektro(Ver. 1.4 20/8/18) © 2002-2018 by M. Jensen and J. E.
Wilhjelm)
1 OverviewAs it can be imagined, planar X-ray imaging has an
inherent limitation in resolving overlying struc-
tures as everything seen in the images are the result of a
projection. It is, however, possible to resolve the 3D distribution
of X-ray attenuation from a set of projections.
This is actually what we do mentally when we access the 3D
structure of an object, for example the head of a person, by
walking around the object and looking at it from all different
angles. The CT scan is exactly such a reconstruction of the 3D
distribution based on a large set of X-ray projections ob-tained at
many angles covering a complete circle around the patient. CT is an
abbreviation of comput-ed tomography. In Anglo-American literature
one also occasionally finds the abbreviation CAT denoting computed
axial tomography. Tomography by itself means the rendering of
slices: naturally the 3D information cannot easily be displayed 3
dimensionally on a screen, instead it is most often displayed as a
series of axial slices.
The CT scanner was developed in the early 1970ies by Geoffrey
Hounsfield and his colleague Alan Cormack in England (actually
working for EMI on funds stemming from music record sales). For
this they were awarded the Nobel Prize in Medicine in 1979.
The basic three components of a CT scanner are still the same as
in planar X-ray imaging: An X-ray tube, an object (patient) and a
detection system. In the earliest scanners the output of the X-ray
tube was collimated to a narrow, pencil-like beam and detected by a
single detector. X-ray tube and detec-tor were translated in unison
(see Figure 1
Figure 1 Early CT scanner geometry. In (a), the X-ray tube and
detector is moved in unison across the object from right to left
in, say 100 discrete steps. This gives 100 intensity values
measured by the detector. Then the entire system (but not the
object) is rotated, say one degree, and the process is repeated.
This is done for at least 180 degrees. This yield 180 profiles each
with 100 values.
X-ray tube
Detector
Thin
X-ra
y be
am
X-ray tube
Detector
Thin
X-ra
y be
am
X-ray tube
Detector
Thin
X-ra
y be
am
X-ray tube
Detector
Thin
X-ra
y be
am
X-ray tube
Detector
Thin
X-ra
y be
am
X-ray tube
Detector
Thin
X-ra
y be
am
(a) (b) (c)
) across the object making a linear scan. After each scan,
1/8
-
typically lasting 10 seconds, the entire setup was rotated a few
degrees, the scan repeated, and so forth. From a set of such 180
scans the final image (a single slice) would be reconstructed by
overnight com-puting. This reconstruction - which derives an image
from a large set of projections - will be consid-ered in Subsection
2.3.
Modern scanners are now essentially of two types: the
rotate-rotate system and the rotate-fixed sys-tem. These are
illustrated in Figure 2
Rotating arcof detectors
Patient
Rotating X-ray tube
Figure 2 Geometry of gantry in CT scanner. Left: The third
generation is of type rotate-rotate, where both X-ray tube and
detectors rotate. Right: The fourth generation is of type
rotate-fixed, where only the X-ray tube rotate. The x-ray tube
emits a fan-shaped beam.
Static ringof detectors
Activedetectors
Patient
Rotating X-ray tube
. Both systems use narrow fan-shaped beams collimated to spread
across the full width of the patient. In the rotate-rotate system
as many as 700 detectors may be placed in an arc centered at the
focal spot of the X-ray tube. The tube is run continuously as both
it and the detectors revolve around the patient. The fast
electronics of the detectors take as many as thousand readings per
detector for a total of 700 000 readings in one second. In the
rotate-fixed system as many as 2000 fixed detectors form a circle
completely around the patient. The X-ray tube is rotating
con-centrically within the detector ring. The detectors are
normally focused at the center of the ring. Ac-quiring the detector
responses every one third of a degree produces more than 2 000 000
readings per second.
1.1 Hounsfield valueUsing mathematical algorithms (as will be
shown later in Subsection 2.3), the computer can calculate
the linear attenuation coefficient for each point (pixel) in the
object and assign an attenuation value to it. Normally, this
attenuation is not depicted as attenuation coefficients, instead
radiology uses a spe-cial unit, now called Hounsfield unit (HU).
The corresponding Hounsfield value is defined as fol-lows:1
HV = 1000 (μm-μw)/ μw (1)
where μm is the (average) linear attenuation coefficient within
the voxel it represents and μw is the linear attenuation
coefficient for water at the same spectrum of photon energies. The
Hounsfield unit is dimensionless in terms of The International
System of Units (Système international d'unités, SI).
1. Strictly speaking, the denominator should be µw - µair.
However, most scanners normalize the image with a “blank scan” in
air (i.e. no additional absorption in the field of view). The error
in omitting µair is small, since µair is about 800 times smaller
than µw.
2/8
-
From the above definition, one should think, that the Hounsfield
values are very well-defined. This is not so, as can be seen from
Figure 3, which represent data from two different teaching books.
For instance, the Hounsfield values for kidney span from 20 HU to
40 HU in the first textbook while it spans from 30 HU to 50 HU in
the second textbook. There can be a number of reasons for these
dif-ferences:
• Different spectra of emitted energy (e.g., the center
frequencies and the shape of the spectra dif-fers).
• Different definitions of what a given tissue type actually
represents.
Figure 3 Hounsfield values according to different text books:
(a) is from [2] while (b) is from [3]. As can be seen, the values
does not fully agree.
(a)
(b)
3/8
-
• Tissue types seldom consist of just one component (e.g.,
muscular tissue can contain various amount of lipid, but still be
described as “muscular tissue”).
1.2 Single slice versus multi-slice systemOriginally the CT
scanner only acquired one slice at a time, making extended axial
field of view a
time consuming process. Today the scanners, whether of the third
or fourth generation, acquire many slices (16 to 256) at a time
using an X-ray tube with an extended axial beam and multiple
stacked de-tector chains. Rotation time is now down to fractions of
a seconds making acquisition of of multi-slice representation of
the heart, almost motion arrested. If the patient is continuously
slid through the gan-try ring during the rotation, a so-called
spiral CT scan is acquired. Proper reconstruction can thus yield
large series of closely lying slices over extended parts of the
body, in principle from head to foot.
2 System details
2.1 CT scanner X-ray tubeProper reconstruction of the CT scans
is only possible if a very large number of photons are
available
for the detectors. If the acquired projections are not
statistically well-determined, the reading from a detector will be
noisy and the reconstruction algorithm will propagate this noise,
leading to unaccept-able high noise in the final image. Thus,
normally, the CT scan is done with a high output from the CT tube
corresponding to large kilovolts and milliampere settings. As the
scan are normally extended for many slices and many revolutions,
the final dose can be as high as 50 to 100 millisievert (see
defi-nition of Sievert in nuclear medicine chapter of this book).
As the number of CT scans has been in-creasing, so has the
contribution from CT scans to the total medical radiation dose.
Each CT scan gives many times the radiation dose of a single planar
X-ray. CT scans now constitutes a major frac-tion of the total
medical radiation exposure.
The high current and voltage and the extended exposure time,
deposits very large amounts of prima-ry electron beam energy in the
anode of the X-ray tube. Special tubes have been developed for
these X-ray scanners, with large, fast rotating anodes of high
melting point materials. Special problems are related to the
technology of bringing electricity of high voltage forward to the
X-ray tube, rotating at an orbital diameter of more than one meter
with the speed of more than two revolutions per second. Modern
multi-slice CT scanners operate at about 60 - 240 RPM. The rapid
revolution of X-ray tube and perhaps detector chain also puts a
large mechanical strain on the entire X-ray gantry, which must be
of extraordinary sturdy construction. Examples can be seen
here:
CT (open) rotating at full speed
GE 16 slice brightspeed scanner max speed spinning
2.2 Detector chain technologyToday, at least three types of
detectors are used. These detectors can be classified according to
the
type of material stopping the X-rays:
• Gas (Xenon)
• Scintillator (transforms the X-ray energy into visible light,
detected by a photo diode)
• Solid state semiconductor
The gas detectors are less efficient than the other two types of
detectors, but by using high pressure, and extended radial
dimensions efficiencies as high as 40 % can be achieved. These
“deep” detectors
4/8
https://www.youtube.com/watch?v=ra7sw0kNvTwhttps://www.youtube.com/watch?v=a1i8KFEtnok
-
has the important property of being most sensitive to radially
incoming X-rays thus providing inherent protection against too much
scattered radiation. With the other two detectors, which are more
like sur-face detectors, the scattered radiation cannot be
separated, and must be removed by the mathematical reconstruction
algorithm. This is possible, because the scattered radiation has
little spatial structure, and can thus be detected and subtracted
as a uniform blanket in the image matrix.
With many detectors in each chain and many slices the total data
sampling rate of a modern CT scan-ner is extremely high. At
present, it is exactly this data sampling rate which limits the
performance of
μ11 μ12
μ21 μ22
I0
I0
I0 I0
Ir1 = I0 exp(-μ11dx -μ12dx)
Ir2= I0 exp(-μ21dx -μ22dx)
Ic1 = I0 exp(-μ11dx -μ21dx)
Ic2 = I0 exp(-μ12dx -μ22dx)
Figure 4 For a medium assumed to consist of four different types
of materials, four measure-ments will allow enough information to
obtain four equations with four unknowns.
state-of-the-art CT scanner technology.
2.3 ReconstructionThe reconstruction - using a large set of
projection values to calculate an image that represents the
tissue that originally gave cause to the projection values - is
the heart of the CT system. This will now be explained with two
different approaches.
The reconstruction of the slices from a large number of
different projections forms an algebraic prob-lem. This can be seen
by considering a CT image with two by two pixels (actually,
voxels). If the ob-ject is irradiated with X-rays from two
perpendicular directions, the detectors will measure the four
values indicated in Figure 4. The four attenuation values of the CT
image can now be found by solving four equations of four unknowns.
The corresponding equations are:
ln(I0/Ir1) dx–1 = μ11 + μ12 (2)
ln(I0/Ir2) dx–1 = μ21 + μ22 (3)
ln(I0/Ic1) dx–1 = μ11 + μ21 (4)
ln(I0/Ic2) dx–1 = μ12 + μ22 (5)
Note that the basic physics does not require sampling of
projections for more than 180°, as the mea-surement is basically a
transmission measurement covering the entire depth forwards to
backwards of
5/8
-
the object. However, because of system stability, artifact
suppression and noise reduction, scans are normally acquired based
on 360° acquisitions. However beautiful the algebraic
reconstruction looks the practical application is difficult due to
the larger number of equations and unknowns. Reconstruct-ing a
single slice represented by a 512 by 512 matrix corresponds to the
diagonalization of such a ma-trix, which is no simple task. Some
algorithms, however, obtain this goal by iterative measures: first
making a guess of the distribution of attenuation values,
subtracting the corresponding projections from the actual
projections measured and then iteratively minimizing this error
difference.
7
35
176543212354321
7
35
176543212354321
For all projections, the measured values are added to all
contributing pixels
Figure 5 Back projection. Each attenuation value is put back
into the cells of the im-age that are located at the line of sight.
The same values are put into each cell.
2.4 Filtered backprojectionBecause it is computationally more
effective, the most used algorithm is the so-called filtered
back-
projection. Consider an image matrix whit pure zeros.
Backprojection by itself simply fills the atten-uation values of
individual projections into each cell of the matrix along the line
of sight. The values filled in, are added to those already in the
image matrix. This is sought illustrated in Figure 5. When the
backprojection is performed on a large number of projections, the
final image begins to emerge, as seen in Figure 6. However, the
image is blurred: a single point object with high attenuation (e.g.
a thin tube of water in air) will by this reconstruction be
depicted as a “1/r” distribution. By filtering the measured
projections before backprojection, this blurring can be reduced.
The filtering is actually a convolution of the individual
projection with a suitable spatial filter, amplifying high spatial
frequen-cies and damping low spatial frequencies. The final
reconstruction algorithm is often called LSFB, an abbreviation for
linear superposition of filtered backprojections. The exact choice
of filter function should be matched with the scanner
characteristics, field of view and object of interest. There is no
need to reconstruct with filters using higher spatial frequencies
than the inherent limits given by the finite detector size in the
detection chain. The reconstructed image represents the attenuation
coeffi-cients. These are re-calculated to Hounsfield units, and
this image is displayed as gray values on the screen. However, the
range of Hounsfield units (or attenuation) can be very large, and
if only soft tis-
6/8
-
sue is to be visualized,
Figure 6 Evolution of backprojection. The first five images are
derived from filtered projections, while the last is derived from
raw projections.
only a small window of values are displayed, as illustrated in
Figure 7.
Houndsfield units0 500 1000
Gre
ysca
leva
lue
Window width
Windowcenterline
LL
UL
Figure 7 Windowing. Only HU between -300 and 600 are visualized
in the gray scale bar. LL = lower level. UL = upper level (drawing
not fully to scale).
Because of the large dynamic range of the CT scanner, it is
often better from the beginning of the reconstruc-tion to limit the
interest area of the image values to a suitable range. For this
reason reconstruction is often formed in “brain window”, “lung
window” or “bone window”.
3 Example of “clinical” CT imageFinally, a comparison between an
anatomical photograph and a CT image from exactly the same
plane is included in Figure 8. The data is from the Visible
Human Project[1] in which a cadaver was scanned from toe to top
with CT and later thinly sliced. From the CT image, it is very
clear which types of tissue, that is best distinguished in the
(raw) CT image. No windowing have been applied to
7/8
-
the data, however. Applying windowing to a region of soft tissue
would improve the visualization of this type of tissue.
4 Points to pay attention toIf you write a short resume of this
chapter, then please carefully consider, if the order of
presentation
should be the identical to what you see here.
5 AcknowledgmentsStudent Jonas Henriksen is gratefully
acknowledged for the help with the tables for Hounsfield val-
ues.
Figure 8 An anatomical photograph and corresponding CT image at
a hori-zontal scan plane of the head. Data from: [1]
6 References[1] The visible human project:
http://www.nlm.nih.gov/research/visible/visible_human.html
[2] Willi A. Kalender, "Computed Tomography", 2005, 2nd edition,
Publicis Corporate Publishing, Erlangen.
[3] Erich Krestel, "Imaging Systems for Medical Diagnostics",
1990, Siemens Aktiengesellschaft, Berlin and Munich.
8/8
-
Medical diagnostic ultrasound- physical principles and
imaging
By Jens E. Wilhjelm, Andreas Illum, Martin Kristensson and Ole
Trier Andersen
Biomedical Engineering, DTU ElektroTechnical University of
Denmark
(Ver. 3.1, 5 December 2016) © 2001-2013 by J. E. Wilhjelm
Preface
This document attempts to introduce the physical principles of
medical diagnostic ultrasound to a broad audience ranging from
non-engineering students to graduate level students in engineering
and science. This is sought achieved by providing chapters with
different levels of difficulty:
Chapters with no asterisk can be read by most.
* These chapters are directed towards bachelor students in
engineering.
** These chapters are directed towards graduate students in
engineering.
The document can be studied at a given degree of detail without
loss of continuation.
To help understanding, a number of Flash animations and quizzes
are included. In order for these to work, the computer used for
viewing this document must have one of the newest Flash players
installed (www.adobe.com). The version of the current Flash player
is written in the box to the right. If no version number appear at
all, you must update Flash.
If viewing this text in a browser, please use one that supports
Adobe pdf-files with embedded Flash such as e.g., Internet explorer
or Mozilla Firefox. Also note that internet access might be
required for some animations to work. Please also note that if you
start an animation and then move onto another page, the animation
might still run in the background slowing the computer.
This document contains a number of quizzes that will pop up in
individual windows when activated by the reader. If the window
appears difficult to read, do this: right click on the quiz icon
and enter the window size (written in parenthesis after the quiz)
in the window that appear and set “Play back style” to “Play
content in floating window”. If it is necessary to move the quiz
window, drag the win-dow at the black frame.
This chapter does not consider blood flow imaging with
ultrasound, which is treated excellently else-where[5].
1 IntroductionMedical diagnostic ultrasound is an imaging
modality that makes images showing a slice of the body,
so-called tomographic images (tomo = Gr. tome, to cut and
graphic = Gr. graphein, to write). It is a diagnostic modality,
meaning that it gathers information about the biological medium
without modi-fication of any kind1.
1/21B M EDTU Elektro
-
Ultrasound is sound with a frequency above the audible range
which ranges from 20 Hz to 20 kHz. Sound is mechanical energy that
needs a medium to propagate. Thus, in contrast to electromagnetic
waves, it cannot travel in vacuum.
The frequencies normally applied in clinical imaging lies
between 1 MHz and 20 MHz. The sound is generated by a transducer
that first acts as a loudspeaker sending out an acoustic pulse
along a nar-row beam in a given direction. The transducer
subsequently acts as a microphone in order to record the acoustic
echoes generated by the tissue along the path of the emitted pulse.
These echoes thus car-ry information about the acoustic properties
of the tissue along the path. The emission of acoustic en-ergy and
the recording of the echoes normally take place at the same
transducer, in contrast to CT imaging, where the emitter (the X-ray
tube) and recorder (the detectors) are located on the opposite side
of the patient.
This document attempts to give simple insight in to basic
ultrasound, simple wave equations, some simple wave types and
generation and reception of ultrasound. This is followed by a
description of ultrasound’s interaction with the medium, which
gives rise to the echo information that is used to make images. The
different kinds of imaging modalities is next presented, finalized
with a description of more advanced techniques. The chapter is
concluded with a list of symbols, terms and references.
2 Basics of ultrasoundUltrasound (as well as sound) needs a
medium, in which it can propagate by means of local defor-
mation of the medium. One can think of the medium as being made
of small spheres (e.g. atoms or molecules), that are connected with
springs. When mechanical energy is transmitted through such a
medium, the spheres will oscillate around their resting position.
Thus, the propagation of sound is due to a continuous interchange
between kinetic energy and potential energy, related to the density
and the elastic properties of the medium, respectively.
The two simplest waves that can exist in solids are longitudinal
waves in which the particle move-ments occur in the same direction
as the propagation (or energy flow), and transversal (or
shearwaves) in which the movements occur in a plane perpendicular
to the propagation direction. In water and soft tissue the waves
are mainly longitudinal. The frequency, f, of the particle
oscillation is related to the wavelength, λ, and the propagation
velocity c:
λf c= (1)
The sound speed in soft tissue at 37°C is around 1540 m/s, thus
at a frequency of 7.5 MHz, the wave-length is 0.2 mm.
2.1 The 1D wave equation*Describing the wave propagation in 3D
space in a lossy inhomogeneous medium ((Danish: et inho-
mogent medium med tab) such as living tissue is very
complicated. However, the description in 1D for a homogenous
lossless medium is relatively simple as will be shown.
An acoustic wave is normally characterized by its pressure.
Thus, in order to obtain a quantitative relation between the
particle velocity in the medium, u, and the acoustic pressure, p, a
simple situation
1. To obtain acoustical contact between the transducer and the
skin, a small pressure must be applied from the transducer to the
skin. In addition to that, ultrasound scanning causes a very small
heating of tissue (less than 1°C) and some studies have
demonstrated cellular effects under special circumstances.
2/21B M EDTU Elektro
-
with 1D propagation in a lossless media will be considered, as
shown in Figure
u+ΔuuA
x x+Δx
p + Δp p
Figure 1 1D situation showing a liquid element inside a sound
wave.
1. This figure shows a volume element of length Δx and with
cross-sectional area A. The volume is thus V = AΔx. The den-sity of
the medium - a liquid, for instance, - is ρ and the mass of the
element will then be ρAΔx.
The pressure p is a function of both x and t. Consider the
variation in space first: There will be a pres-sure difference, Δp,
from the front surface at x to the back surface at x+Δx, thus the
volume element will be subject to a force –AΔp. By applying
Newton’s second law (F = ma):
AΔp– ρAΔxdudt------= (2)
or after performing the limit (Δ → d)
dpdx------ ρ– dudt
------= (3)
Next consider the variations over a time interval Δt. A
difference in velocity, Δu, between the front surface (at x) and
the back surface (at x+Δx) of the elemental volume will result in a
change in that volume which is:
ΔV A u Δu+( )Δt uΔt–( ) AΔuΔt= = (4)
which in turn is connected with a change in pressure, Δp,
according to
ΔV κ AΔx( )– Δp= (5)
where κ is the compressibility of the material (e.g. a liquid)
in units of Pa-1. Performing the same limit as above, gives the
second equation:
dudx------ κ– dpdt
------= (6)
Equations (3) and (6) are the simplest form of the wave
equations describing the relation between pressure and particle
velocity in a lossless isotropic medium.
3/21B M EDTU Elektro
-
3 Types of ultrasound waves
Figure 2 Dynamic visualization of plane wave (a) and spherical
wave (b) pressure fields. The pressure fields are monochromatic,
i.e., contains only one frequency. Pure black indicates zero
pressure, red indi-cates positive and blue negative pressure
values. The wavelength can be read directly from the plots. When
including the propagation velocity, c = 1500 m/s, the frequency of
the wave can also be found.
(a) (b)
The equations above describe the relation between pressure and
displacement of the elements of the medium. Two simple waves
fulfilling the above will now be considered. Both are theoretical,
since they need an infinitely large medium.
Since optical rays can be visualized directly, and since they
behave in a manner somewhat similar to acoustic waves, they can
help in understanding reflection, scattering and other phenomena
taking place with acoustic waves. Therefore, there will often be
made references to optics.
There are two types of waves that are relevant. They can both be
visualized in 2D with a square acryl-ic water tank placed on an
overhead projector:
• The plane wave which can be observed by shortly lifting one
side of the container.
• The spherical wave, which can be visualized by letting a drop
of water fall into the surface of the water.
When the plane wave is created at one side of the water tank,
one will also be able to observe the reflection from the other side
of the tank. The wave is reflected exactly as a light beam from a
mirror or a billiard ball bouncing off the barrier of the
table.
The spherical wave, that on the other hand, originates from a
point source and propagates in all di-rections; it creates a
complex pattern when reflected from the four sides of the tank.
3.1 The plane wave*The plane wave is propagating in one
direction in space; in a plane perpendicular to this direction,
the pressure (and all other acoustic parameters) is constant. As
a plane extends over the entire space, it is not physically
realizable (but within a given space, an approximation to a plane
wave can be ob-tained locally, such as in the shadow of a planar
transducer (see later)).
4/21B M EDTU Elektro
-
If the plane wave is further restricted to be monochromatic,
that is, it oscillate at a single frequency, f0, then the wave
equation in 1D is:
p(x,t) = P0 exp(–j (2πf0t – 2πx/λ)) (7)
where P0 is the pressure magnitude (units in pascal, Pa) x is
the distance along the propagation direc-tion and λ = c/f0. (7) is
a complex sinusoid that depends on space and time. The equation
will be the same in 3D, provided that the coordinate system is
oriented with the x-axis in the propagation direc-tion. The plane
wave travelling in the x-direction is sought illustrated by the
pressure animation in Fig-ure 2a.
A plane wave thus propagates in one direction, just like a laser
beam, however, it is merely the oppo-site of a beam.
Quiz 1 (Open in floating window of size 800 x 1100)
3.2 The spherical wave* The other type is a spherical wave. It
originates from a point (source) and all acoustic parameters
are constant at spheres centred on this point. Thus, the
equation is the same as in (7), except that the x is substituted
with r in a polar coordinate system:
p(r,t) = P0 exp(–j (2πf0t – 2πr/λ)) (8)
where r is the distance from the centre of the coordinate system
(i.e., the source) to any point in 3D space. The spherical wave are
sought illustrated by the pressure animation in Figure 2b.
Problem 1 With the animations in Figure 2, measure the
wavelength and calculate the centre frequen-cy of the waves.
Problem 2 There are a few aspects of Figure 2, that were too
difficult to visualize correctly, when using Flash as the
programming tool. Which?
3.3 Diffraction**An important concept in wave theory is
diffraction. Ironically, the term diffraction can best be de-
scribed by what it is not: “Any propagating scalar field which
experiences a deviation from a rectilin-ear propagation path, when
such deviation is not due to reflection or refraction (see later),
is generally said to undergo diffraction effects. This description
includes the bending of waves around objects in their path. This
bending is brought about by the redistribution of energy within the
wave front as it passes by an opaque body.”[3] Examples where
diffraction effects are significant are: Propagation of waves
through an aperture in a baffle (i.e. a hole in a plate) and
radiation from sources of finite size.[3] With the above
definition, the only non-diffracted wave is the plane wave.
4 The generation of ultrasoundThe ultrasonic transducer is the
one responsible for generating ultrasound and recording the
echoes
generated by the medium. Since the transducer should make
mechanical vibrations in the megahertz range, a material that can
vibrate that fast is needed. Piezoelectric materials are ideal for
this.
5/21B M EDTU Elektro
-
Figure 3 Example of modern ultrasound transducer of type 8820e
(BK Medical, Denmark) with frequency range 2 - 6 MHz. From
www.bkmed.com.
The typical transducer consist of a disk-shaped piezoelectric
element that is made vibrating by ap-plying an electrical impulse
via an electrode on each side of the disc. Likewise, the echo
returning to the disk makes it vibrate, creating a small electrical
potential across the same two electrodes that can be amplified and
recorded. In modern clinical scanners, the transducer consists of
hundreds of small piezoelectric elements arranged as a 1D array
packed into a small enclosure. The shape of this line can be either
linear or convex. An example of the latter can be seen in Figure 3.
The use of arrays with hundreds of elements, makes it possible to
electronically focus and steer the beam, as will be consid-ered
later in Chapter 7.
Figure 4 Left: Piezo electric crystal at different states of
compression. Right: Single element transducer consisting of
piezoelectric crystal with electrodes. This “sandwich” is placed
between a backing material and the matching layer towards the
medium.
CrystalBacking
g(t)
Acoustic axis
2a
Housing
λ/4 matching layer and wear plate
Shadow region
However, first the single-element transducer will be
considered.
4.1 PiezoelectricityThe acoustic field is generated by using the
piezo electric effect present in certain ceramic materials.
Electrodes (e.g. thin layers of silver) are placed on both sides
of a disk of such a material. One side of the disk is fixed to a
damping so-called backing material, the other side can move freely.
If a voltage is applied to the two electrodes, the result will be a
physical deformation of the crystal surface, which will make the
surroundings in front of the crystal vibrate and thus generate a
sound field. If the mate-rial is compressed or expanded, as will be
the case when an acoustic wave impinges on the surface, the
displacement of charge inside the material will cause a voltage
change on the electrodes, as illus-trated in Figure 4 (left). This
is used for emission and reception of acoustic energy,
respectively.
6/21B M EDTU Elektro
-
4.2 The acoustic field from a disk transducer*Since the
ultrasound transducer - or the piezoelectric crystal - has a size
comparable to or larger than
the wavelength, the field generated becomes very complex. Rather
than providing equations for de-scribing the field, it will now be
attempted visualised.
It is assumed that the piezoelectric, disk-shaped crystal is
fixed at the back, as illustrated in Figure 4 (right) and can move
freely at the front. Specifically, movement of the surface of the
transducer can be described by a velocity vector oriented
perpendicular to the surface. In short, the electrical signal
applied to the transducer is converted by the electro-mechanical
transfer function of the transducer to a velocity function
describing the movement of the transducer surface. Note the backing
material lo-cated behind the crystal; this is used to dampen the
free oscillation of the crystal (in the time period just after a
voltage is applied), thereby creating a short vibration, when an
impulse is applied to the crystal. The radius of the crystal is
denoted a. The thickness of the crystal is selected according to
the frequency of operation, so that it is λpiezo/2, where λpiezo is
the wavelength of sound in the crystal ma-terial.
In order to assess the pressure field generated by the
transducer, the surface of the crystal will be divided up into many
small surface elements, each contributing to the entire pressure
field. If the sur-face elements are much smaller than the
wavelength, they can be considered point sources. In the present
case, the point source will generate a semi-spherical wave in the
space in front of the trans-ducer. The waves are identical, the
only difference is the location of the point source. At a given
field point in front of the transducer, the total pressure will
then be the pressure due to the individual point sources. This is
an application of Huygens1 principle. Of course, these individual
pressure contribu-tions will interfere positively and negatively
dependent on the location of the field point. This inter-ference
will result in the final beam, which can be rather complex.
Rather than doing this calculation analytically, a graphical
illustration is provided in Figure 5 which shows point sources
along a diameter of the transducer disk (the remaining point
sources on the disk surfaces are ignored for simplicity). For each
point source, a bow shows the location of the equal-phase-fronts
(or equal-time-lag) of the spherical pressure wave generated from
that source at given instances in time. The equal-phase-fronts are
not the same as the pressure field; the latter can be cre-ated by
adding the pressure fields of each individual source time-shifted
according to the equal-time-lag. Hence, the moving bows in Figure 5
reveal how complicated the field is at a given point.
The “wave” fronts generated by the flat piston transducer in
Figure 5 (left) tend towards a (locally) plane wave inside the
shadow of the transducer. The pressure field is thus broad, and
unsuitable for imaging purposes, as will become clear, when the
imaging technique is considered later. In order to focus the
ultrasound field and obtain a situation where the acoustic energy
travels along a narrow path, a focused transducer is used, as
illustrated in Figure 5 (right). In this situation, the individual
spherical waves from the transducer are performing constructive
interference at the focal point, whereas at all other points, the
interference is more or less destructive. In order to make this
work efficiently, the wavelength must be much smaller than the
distance to the focal point. However, a typical depth of the focal
point for a 7.5 MHz transducer - 20 mm - will correspond to
100λ.
Notice here, that the key to understand this is the fact that it
takes a different amount of time to travel to a given field point
from two different source locations. The interference that is
caused by this is quite unique for ultrasound.
1. Christian Huygens, physicist from the Netherlands,
1629-95.
7/21B M EDTU Elektro
-
Example: The interference phenomena can be explored in everyday
life: if one positions oneself with one ear pointing into a
loudspeaker and turns up the treble, then the sound picture will
change if you move in front of the loudspeaker, especially when
moving perpendicular to the loudspeaker’s acoustic axis. What
happens is that the ear is moved to different points in space,
which exhibits different amounts of constructive and destructive
interference. This phenomenon is less distinct at low frequen-cies
(bass), because the wavelength gets larger. This is also the reason
that a stereo sound system can do with one subwoofer for the very
low frequency band, but needs two loudspeakers for the remaining
higher frequencies.
Figure 5 Left: Example of moving circles showing “wave fronts”
of equal phase (or equal travel time) at as a function of time from
selected point sources (=red dots). For simplicity, only point
sources located on a diameter are shown, making this drawing two
dimensional. Right: The same for a focused transducer. c = 1500
m/s. The radius of curvature of the disk surface can be deducted
from this Figure. What is it?
As noted above, dimensions give most insight, when they are
measured in wavelength. Consider the planar transducer in Figure 5
(left): The near field from this type of transducer is defined[4]
as the re-gion between the transducer and up to a range of a2/λ.
The far field region corresponds to field points at ranges much
larger than a2/λ. In Figure 5 (left), a is specified, but λ is not.
If the transducer fre-quency is f0 = 0.5 MHz, then a2/λ = 33 mm,
which is in the middle of the plot. If f0 = 7.5 MHz, then a2/λ =
0.5 m! The explanation is as follows: The far field is defined as
the region, where there is only moderate to little destructive
interference. If this should be possible, then from a given field
point in this region, the distance to any point on the transducer
surface should vary much less than a wave-length: Consider a given
field point not on the acoustic axis. Next, draw two lines to the
two opposite edges of the transducer. Now the difference in length
of these two lines - measured in wavelength - must be much less
than one, in order to have little destructive interference at this
field point. Thus, the higher the frequency, the lower the
wavelength, and the farther away one must move from the trans-ducer
surface in order to get differences between the length of the two
lines much less than one wave-length.
An ultrasound field from a physical transducer will always show
a complicated behaviour as can be sensed from Figure 5. Each point
source is assumed to emit exactly the same pressure wave (an
ex-ample of the temporal shape is given in Figure 8). Thus, the
circles in the animation in Figure 5 indi-cate spatial and temporal
locations of each of the individual waveforms. The contribution of
all these waveforms would have to be added in order to construct
the total pressure field in front of the trans-ducer (however, the
circles in Figure 5 only represent point sources on a single
diameter across the transducer; many more point sources would be
needed to represent the total field from a disk trans-ducer).
8/21B M EDTU Elektro
-
Problem 3 Huygens’ principle. How would you find - or calculate
- how many point sources are need-ed on the transducer surface in
Figure 5 in order to represent the pressure field in front of the
trans-ducer with a given accuracy?
Problem 4 Write a short summary of this chapter.
5 Ultrasound’s interaction with the mediumThe interaction
between the medium and the ultrasound emitted into the medium can
be described
by the following phenomena:
The echoes that travel back to the transducer and thus give
information about the medium is due to two phenomena: reflection
and scattering. Reflection can be thought of as when a billiard
ball bounc-es off the barrier of the table, where the angle of
reflection is identical to the angle of incidence. Scat-tering
(Danish: spredning) can be thought of, when one shines strong light
on the tip of a needle: light is scattered in all directions. In
acoustics, reflection and scattering is taking place when the
emitted pulse is travelling through the interface between two media
of different acoustic properties, as when hitting the interface of
an object with different acoustic properties.
Specifically, reflection is taking place when the interface is
large relative to the wavelength (e.g. be-tween blood and intima in
a large vessel). Scattering is taking place when the interface is
small relative to the wavelength (e.g. red blood cell).
The abstraction of a billiard ball is not complete, however: In
medical ultrasound, when reflection is taking place, typically only
a (small) part of the wave is reflected. The remaining part is
transmittedthrough the interface. This transmitted wave will nearly
always be refracted, thus typically propagat-ing in another
direction. The only exception is when the wave impinges
perpendicular on a large pla-nar interface: The reflected part of
the wave is reflected back in exactly the same direction as it came
from (like with a billiard ball) and the refracted wave propagates
in the same way as the incident wave.
Reflection and scattering can happen at the same time, for
instance, if the larger planar interface is rough. The more smooth,
the more it resembles pure reflection (if it is completely smooth,
specular reflection takes place). The rougher, the more it
resembles scattering.
When the emitted pulse travels through the medium, some of the
acoustic (mechanical) energy is converted to heat by a process
called Absorption. Of course, also the echoes undergo
absorption.
Finally, the loss in intensity of the forward propagating
acoustic pulse due to reflection, refraction, scattering and
absorption is under one named attenuation.
5.1 Reflection and transmission*When a plane wave impinges on a
plane, infinitely large, interface between two media of
different
acoustic properties, reflection and refraction occurs meaning
that part of the wave is reflected and part of the wave is
refracted. The wave thus continues its propagation, but in a new
direction.
To describe this quantitatively, the specific acoustic
impedance, z, is introduced. In a homogeneous medium it is defined
as the ratio of pressure to particle velocity in a progressing
plane wave, and can be shown to be the product of the physical
density, ρ, and acoustic propagation velocity c of the me-dium.
Thus, if medium 1 is specified in terms of its physical density,
ρ1, and acoustic propagation ve-locity c1, the specific acoustic
impedance for this medium is z1 = ρ1c1. The units become kg/(m2s)
which is also denoted rayl. Likewise for medium 2: z2 = ρ2c2. The
interaction of ultrasound with this
9/21B M EDTU Elektro
-
interface can be illustrated by use of Figure 6, where an
incident plane wave is reflected and transmit-ted at the interface
between medium 1 and medium 2. The (pressure) reflection
coefficient between the two media is:[2]
Rz2 θtcos( ) z1 θicos( )⁄–⁄z2 θtcos( )⁄ z1 θicos(
)⁄+------------------------------------------------------------=
(9)
where the angle of incidence, θi, and transmission, θt, are
related to the propagation velocities asθisinθtsin
------------c1c2-----= . (10)
Equation (10) is a statement of Snell’s law,[2] which also
states that: θr = θi. The pressure transmission coefficient is T =
1 + R.
It should be noted here, that Snell’s law applies to optics,
where the light can be considered to travel in rays. For a planar
wave in acoustics, which only have one direction, the above
formulation of Snell’s law applies as well. However, when the
acoustic wave travels like a beam, Snell’s law is only
approximately valid. The validity is related to the properties of
the beam, namely to which degree the wave field inside the beam can
be considered locally plane (which again is related to the
thickness of the beam, measured in wavelengths).
Figure 6 Graphical illustration of Snell’s law describing the
direction of an incident plane wave (pi), reflected plane wave (pr)
and transmitted (refracted) plane wave (pt) from a large smooth
interface. The three arrows indi-cate the propagation direction of
the plane waves; the three parallel lines symbolizes that the wave
is planar. The pressure amplitudes of the reflected and transmitted
waves are not depicted, but their relative amplitude can be
calculated from R and T. θr = θi.
Strictly speaking, if the field incident on an interface is not
fully planar, and the interaction is to be modelled quantitatively,
then the field should be decomposed into a number of plane waves,
just like a temporal pulse can be decomposed into a number of
infinite tone signals. The plane waves can then be reflected one by
one, using (9) and (10).
In the human body, approximate reflection can be observed at the
interface between blood and the intima of large vessel walls or at
the interface between urine and the bladder wall.
Quiz 2 (Open in floating window of size 800x700)
Quiz 3 (Open in floating window of size 800x400)
10/21B M EDTU Elektro
-
5.2 Critical angle**Depending on the speed of sound of the two
media, some special cases occur.[2]
If c1 ≥ c2, the angle of transmission, θt, is real and θt <
θi, so that the transmitted wave is bent towards the normal to the
interface. This can be studied with the interactive Figure 6.
If c1 < c2, the so-called critical angle can be defined
as
θcsinc1c2-----= . (11)
If θi < θc, the situation is the same as above, except that
θt < θi, i.e., the transmitted wave is bent away from the normal
to the interface. This can be studied with the interactive Figure
6.
If θi > θc, the transmitted wave appear to have a very
peculiar form. In short, the incident wave is to-tally
reflected.[2] The interested reader can find more details in larger
textbooks[2].
5.3 Scattering*While reflection takes place at interfaces of
infinite size, scattering takes place at small objects with
dimensions much smaller than the wavelength. Just as before, the
specific acoustic impedance of the small object must be different
from the surrounding medium. The scattered wave will be more or
less spherical, and thus propagate in all directions, including the
direction towards the transducer. The lat-ter is denoted
backscattering.
The scattering from particles much less than a wavelength is
normally referred to as Rayleigh scat-tering. The intensity of the
scattered wave increases with frequency to the power of four.
Biologically, scattering can be observed in most tissue and
especially blood, where the red blood cells are the predominant
cells. They have a diameter of about 7 μm, much smaller than the
wave-length of clinical ultrasound.
5.4 Absorption*Absorption is the conversion of acoustic energy
into heat. The mechanisms of absorption are not ful-
ly understood, but relate, among other things, to the friction
loss in the springs, mentioned in Subsec-tion 2. More details on
this can be found in the literature.[2]
Pure absorption can be observed by sending ultrasound through a
viscous liquid such as oil.
5.5 Attenuation*The loss of intensity (or energy) of the forward
propagating wave due to reflection, refraction, scat-
tering and absorption is denoted attenuation. The intensity is a
measure of the power through a given cross-section; thus the units
are W/m2. It can be calculated as the product between particle
velocity and pressure: I = pu = p2/z, where z is the specific
acoustic impedance of the medium. If I(0) is the intensity of the
pressure wave at some reference point in space and I(x) is the
intensity at a point xfurther along the propagation direction then
the attenuation of the acoustic pressure wave can be writ-ten
as:
I(x) = I(0)e–αx (12)
where α (in units of m-1) is the attenuation coefficient. α
depends on the tissue type (and for some tissue types like muscle,
also on the orientation of the tissue fibres) and is approximately
proportional with frequency.
11/21B M EDTU Elektro
-
As a rule of thumb, the attenuation in biological media is 1
dB/cm/MHz. As an example, consider ultrasound at 7.5 MHz. When a
wave at this frequency has travelled 5 cm in tissue, the
attenuation will (on average) be 1 dB/cm/MHz x 5 cm x 7.5 MHz =
37.5 dB. For bone, the attenuation is about 30 dB/MHz/cm. If these
two attenuation figures are converted to intensity half-length (the
distance corresponding to a loss of 50 %) at 2 MHz, it would
correspond to 15 mm in soft tissue and 0.5 mm in bone.
Absorption
Scattering
Atte
nuat
ion
time
Transducer
Reflection, 90°
Refraction
Diffuse scattering
Voltage
Z1 = ρ1c1
Z3 =ρ3c3
Reflection, ≠90°
Z2 =ρ2c2
Figure 7 Sketch of the interaction of ultrasound with tissue.
The left drawing shows the medium with the transducer on top. The
ultrasound beam is shown superimposed onto the medium. The right
part of the drawing shows the corresponding received echo
signal.
Problem 5 Consider a scanning situation, with two interfaces.
One located at a depth of 1 cm. There is water between this and the
transducer. The other is located at a depth of 2 cm and there is
oil from 1 cm to 2 cm. From 2 cm there is water again. The
attenuation of water is 0 dB, while it is 1.5 dB/cm/MHz for oil.
The transducer frequency is 5 MHz. What is the pressure magnitude
at the receiving transducer of the second, relative to the first?
(Hint: put the information into a drawing.)
5.6 An example of ultrasound’s interaction with biological
tissueWhen an ultrasound wave travels in a biological medium all
the above mechanisms will take place.
Reflection and scattering will not take place as two perfectly
distinct phenomena, as they were de-scribed above. The reason is
that the body does not contain completely smooth interfaces of
infinite size. And even though the body contain infinitesimally
small point objects, the scattered wave from these will be
infinitesimally small in amplitude and thereby not measurable!
The scattered wave moving towards the transducer as well as the
reflected wave moving towards the transducer will be denoted the
echo in this document.
So the echo is due to a mixture of reflection and scattering
from objects of dimension:
• somewhat larger than the wavelength (example: blood media
interface at large blood vessels)
12/21B M EDTU Elektro
-
• comparable to the wavelength
• down to maybe a 20th of a wavelength (example: red blood
cells).
The effects in Subsection 5.1 - 5.5 are illustrated in Figure
7.
The absorption continuously takes place along the acoustic beam,
as media 1 and media 2 (indicated by their specific acoustic
impedances) are considered lossy.
Consider the different components of the medium: Scattering from
a single inhomogeneity is illus-trated at the top of the medium.
Below is a more realistic situation where the echoes from many
scat-terers create an interference signal. If a second identical
scattering structure is located below the first, then the
interference signal will be roughly identical to the interference
signal from the first. The over-all amplitude, however, will be a
little lower, due to the absorption and the loss due to the first
group of scatterers. Notice that the interference signal varies
quite a bit in amplitude.
The emitted signal next encounters a thin planar structure,
resulting in a well-defined strong echo.
Next, an angled interface is encountered, giving oblique
incidence and thus refraction, according to (10) and Figure 6. The
change in specific acoustic impedance is the same as above, but due
to the non-perpendicular incidence, less energy is reflected back.
The transmitted wave undergoes refraction, and thus scatterers
located below this interface will be imaged geometrically
incorrect.
Problem 6 The example in Figure 7 is not totally correct. What
is wrong?
6 Imaging
Figure 8 Left: The basic principle behind pulse-echo imaging. An
acoustic pulse is emitted from the transducer, scattered by the
point reflector and received after a time interval which is equal
to the round trip travel time. The emitted pulse is also present in
the received signal due to limitations of the electronics
controlling the transducer. Right: the signal processing creating
the envelope of the received signal followed by calculation of the
logarithm yielding the scan line.
Imaging is based on the pulse-echo principle: A short ultrasound
pulse is emitted from the transduc-er. The pulse travels along a
beam pointing in a given direction. The echoes generated by the
pulse are recorded by the transducer. This electrical signal is
always referred to as the received signal. The later an echo is
received, the deeper is the location of the structure giving rise
to the echo. The larger the amplitude of the echo received, the
larger is the average specific acoustic impedance difference
between the structure and the tissue just above. An image is then
created by repeating this process with the beam scanning the
tissue.
13/21B M EDTU Elektro
-
All this will now be considered in more detail by considering
how Amplitude mode, Motion mode and Brightness mode work.
6.1 A-modeThe basic concept behind medical diagnostic ultrasound
is shown in Figure 8, which also shows the
simplest mode of operation, A-mode. In the situation in Figure 8
(left) a single point scatterer is lo-cated in front of the
transducer at depth d. A short pulse is emitted from the
transducer, and at time 2d/c, the echo from the point target is
received by the same transducer. Thus, the deeper the point
scat-terer is positioned, the later the echo from this point
scatterer arrives. If many point scatterers (and reflectors) are
located in front of the transducer, the total echo can be found by
simple superposition of each individual echo, as this is a linear
system, when the pressure amplitude is sufficiently low.
The scan line - shown in Figure 8 lower right - is created by
calculating the envelope (Danish: ind-hyllingskurve) of the
received signal followed by calculation of the logarithm, in order
to compress the range of image values for a better adoption to the
human eye. So, the scan line can be called a gray scale line. The
M-mode and B-mode images are made from scan lines.
6.2 Calculation of the scan line*The received signal, gr(t), is
Hilbert transformed to grH(t) in order to create the corresponding
ana-
lytical signal g̃ r(t) = gr(t) + jgrH(t). Twenty times the
logarithm of the envelope of this signal, 20log| g̃ r(t)|, is then
the envelope in dB, which can be displayed as a gray scale line, as
shown in Figure 8 (right). Such a gray scale bar is called a scan
line, which is also the word used for the imaginary line in tissue,
along which gr(t) is recorded. Note, that because the envelope
process is not fully linear, the scanner does not constitute a
fully linear system.
Unfortunately, clinical ultrasound scanners do not feature
images in dB. More image improvements takes place in the scanner
(typically proprietary software) and the gray scale is thus - at
best - a pseudo dB-scale, in this document denoted “dB”.
Quiz 4 (Open in floating window of size 800x700)
6.3 M-modeIf the sequence of pulse emission and reception is
repeated infinitely, and the scan lines are placed
next to each other (with new ones to the right), motion mode, or
M-mode, is obtained. The vertical axis will be depth in meters
downwards, while the horizontal axis will be time in seconds
pointing to the right. This mode can be useful when imaging heart
valves, because the movement of the valves
14/21B M EDTU Elektro
-
will make distinct patterns in the “image”.
Figure 9 Screen dump of clinical ultrasound scanner used to
image the carotid artery in the neck. Upper: the B-mode image.
Lower: the M-mode image recorded along the vertical line in the
B-mode image. Notice in the lower image, the change in location of
the vessel walls due to the heart beat.
An example is shown in Figure 9.
6.4 B-modeBrightness or B-mode is obtained by physically moving
the scan line to a number of adjacent loca-
tions. The principle is shown in Figure 10. In this figure, the
transducer is moved in steps mechani-cally across the medium to be
imaged. Typically 100 to 300 steps are used, with a spacing between
0.25λ and 5λ. At each step, a short pulse is emitted followed by a
period of passive registration of the echo. In order to prevent
mixing the echoes from different scan lines, the registration
period has to be long enough to allow all echoes from a given
emitted pulse to be received. This will now be consid-ered in
detail.
Assume that the average attenuation of ultrasound in human soft
tissue is α in units dB/MHz/cm. If the smallest echo that can be
detected - on average - has a level of γ in dB, relative to the
echo from tissue directly under the transducer, then the maximal
depth from where an echo can be expected is γ = α f0 2Dmax or
Dmaxϒ
2αf0-----------= (13)
Example: According to a rule of thumb, the average attenuation
of ultrasound in human soft tissue is 1 dB/MHz/cm. Assume that γ =
80 dB. At f0 = 7.5 MHz (13) gives Dmax = 5.3 cm.
The time between two emissions will then be Tr = 2Dmax/c, which
is the time it take the emitted pulse to travel to Dmax and back
again. If there are Nl scan lines per image, then the frame-rate
(number of images per second produced by the scanner) will be
fr = (Tr Nl)–1. (14)
15/21B M EDTU Elektro
-
Emission
&
reception
Control
unit
Scan conversion
TransducerM
ediu
m
Ult
raso
und i
mag
e
Measurement situation Ultrasound system
Figure 10 The principle of a simple B-mode ultrasound system. At
this particular point in time, half of the image has been
recorded.
Example: For Nl = 200, fr = 70 Hz a good deal more than needed
to obtain “real-time” images (some 20 frames per second). However,
an fr of 70 Hz might not be an adequate temporal resolution, when
studying heart valves. If the total image width is 40 mm, then the
distance between adjacent scan lines is 40 mm / 200 = 0.2 mm.
Please note that this number is not directly reflecting the spatial
resolution size of the scanner, which is considered in Chapter
8.
Problem 7 If the frame rate is fr = 20 Hz (a typical number for
clinical use), how long time will be available for recording half
an image as shown in Figure 10?
16/21B M EDTU Elektro
-
In order to better appreciate the dynamics of the recording
situation, Figure 11 shows the recording situation in extreme slow
motion. It will be wise to consider this animation in detail. To
help with this, a number of problems and quizzes are provided
below:
Figure 11 Schematic live illustration of the recording of a
B-mode image. Left: The ultrasound transducer scanning a piece of
animal tissue in oil. The photograph is made by later slicing the
tissue and photographing the slice where the scanning took place.
The red dot represents the emitted pulse, which decreases in
amplitude the more tissue it pene-trates. The green dots represents
the echoes. Right: The screen of the scanner. The scan line is
updated from left to right. Not all in this “drawing” is to
scale.
Problem 8 Use a ruler (Danish: lineal) to check, if the green
dots in Figure 11 are located correctly, when the red dot is at the
location shown?
Problem 9 How much slower is the scanning performed in Figure
11, compared to normal clinical use?
Quiz 5 (Open in floating window of size 800x700)
Quiz 6 (Open in floating window of size 800x1000)
Examples of clinical B-mode images can be seen in the chapter on
clinical imaging in this Webbook.
7 Array transducersThe recording of a B-mode ultrasound image by
mechanical movement of the transducer is now an
old technique. Today most ultrasound systems apply array
transducers, which consist of up to several hundreds of crystals,
arranged along a straight or curved line. The elements of the
transducer array, or a subset of elements, are connected to a
multichannel transmitter/receiver, operating with up to sev-
17/21B M EDTU Elektro
-
eral hundred independent channels. The shape, direction and
location of the ultrasound beam can then be controlled
electronically (in the newest scanners completely by software)
thereby completely elim-inating mechanical components of the
transducer. In the most flexible systems, the amplitude, wave-form
and delay of the pulses can be controlled individually and
precisely.
Two different types of transducer systems exist: Phase array
systems, where all elements are in use all the time. The beam is
then steered in different directions to cover the image plane. In
the linear array systems, a subset of elements is used for each
scan line. From this subset a beam is created, and then translated
by letting the subset of elements “scan” over the entire array. The
latter can be ob-served (schematically) in Figure 11: The blue dots
show all the crystals. The light blue dots show the active
crystals, which are used for emitting a focused beam and receiving
the echoes along the same beam.
8 Resolution size and point spread functionThe resolution size
of an imaging system can be assessed in many different ways. One
way is to re-
cord an image of a small point target. The resulting image is
called the point spread function (psf), i.e.an image which shows
how much the image of a point target is “spread out”, due to the
limitations of the imaging system. The point target should
preferably be much smaller than the true size of the psf. Another
related way is to image two point targets with different
separations, and see how close they can be positioned and still be
distinguishable.
Figure 12 The principle of spatial compound imaging for Nθ = 3.
Three single-angle images are recorded from three different angles
and then averaged to form the compound image. Inside the triangular
region, the image is fully compounded, outside, less
compounded.
Transducer Transducer
z
Transducer0
Transducer
+ + =
compoun-ded region
Partly
zmax
max
scan lines
xD0
Fully comp-
regionpounded
The –3 dB width of the psf in the vertical and horizontal image
direction will then be a quantitative measure for the resolution
size. The two directions correspond to the depth and lateral
direction in the recording situation, respectively.
The resolution in the depth direction (axial resolution) can be
appreciated from the echo signal in Figure 8. This echo signal was
created by emitting a pulse with the smallest possible number of
peri-ods. The resolution size is equal to the length of the echo
pulse from a point target, which in the pres-ent noted is assumed
identical the emitted sound pulse. Thus, if the axis resolution
size should be improved (decreased) the only possible way is to
increase the centre frequency of the transducer. But increasing f0
will increase attenuation as well, as discussed in Subsection 5.5.
The consequence is that centre frequency and resolution size is
always traded off.
18/21B M EDTU Elektro
-
The resolution size is treated in more detail in the chapter on
image quality in this webbook.
Figure 13 Left: Conventional image of a porcine artery. Right:
Spatial compound image of the same por-cine artery (average image
of single-angle images from the angles: -21°, -14°, -7°, 0°, 7°,
14°, 21°).
9 Spatial compounding*The array technique described in
Subsection 7 can be used to implement so-called spatial
compound-
ing. In this technique, several images are recorded from
different angles and then combined, to yield an image with some
desirable properties, relative to the conventional B-mode image.
The technique is illustrated in Figure 12. Because a single
compound image consists of Nθ single-angle images, the frame-rate
will be reduced by a factor of Nθ compared to B-mode imaging.
An example of a conventional B-mode image and the corresponding
compound image is shown in Figure 13. If compared to the B-mode
image, a number of (desirable) features become apparent:
The B-mode image has a quite “mottled” appearance, in the sense
that the image consists of dots - roughly the size of the psf - on
a black background. This is the result of the before mentioned
con-structive and destructive interference from closely spaced
scatterers and reflectors, as illustrated in Figure 7. The
phenomenon is commonly referred to as speckle noise. Speckle noise
is a random phe-nomenon, and a given combination of constructive
and destructive interference from a cloud of close-ly spaced
scatterers is closely related to beam size, shape, orientation and
direction. Thus, the interference pattern will change for the same
tissue region when imaged from a different direction. If the change
in view-angle is large enough, this interference patterns will be
uncorrelated; so averaging of several uncorrelated single-angle
images, will yield a reduction in speckle noise.
Because the ultrasonic echoes from interfaces vary in strength
with the angle of incidence, the more scan angles used, the larger
the probability that an ultrasound beam is perpendicular or nearly
perp