the webbook of medical imaging - biomedical...

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

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3 References[1]Dorland’s Illustrated Medical Dictionary. 27th edition. W. B. Saunders Co., Philadelphia, PA,

USA. 1988.

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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.

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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.

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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).

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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.

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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,

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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)?

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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

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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.

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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

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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,

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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.

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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)

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• 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

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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

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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-

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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.

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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.


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 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)).


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.


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.


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.


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).


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


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)



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.


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)


• 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.


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”.


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


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)


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?


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?



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-


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.


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

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