Section VII 29 Remote Sensing

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Remote Sensing 29. Remote sensing

Content

29.1 Production and use of X-rays

29.2 Production and use of ultrasound

29.3 Use of magnetic resonance as an imaging technique

Learning outcomes

Candidates should be able to: (a) explain in simple terms the need for remote sensing (non-invasive

techniques of diagnosis) in medicine

(b) explain the principles of the production of X-rays by electron

bombardment of a metal target (c) describe the main features of a modern X-ray tube, including

control of the intensity and hardness of the X-ray beam

(d) show an understanding of the use of X-rays in imaging internal

body structures, including a simple analysis of the causes of

sharpness and contrast in X-ray imaging 1


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(e) show an understanding of the purpose of computed tomographyor CT scanning

(f) show an understanding of the principles of CT scanning (g) show an understanding of how the image of an 8-voxel cube can

be developed using CT scanning

(h) explain the principles of the generation and detection of ultrasonicwaves using piezo-electric transducers

(i) explain the main principles behind the use of ultrasound to obtaindiagnostic information about internal structures (j) show an understanding of the meaning of acoustic impedance and

its importance to the intensity reflection coefficient at a boundary

(k) recall and solve problems by using the equation I = I0exfor the

attenuation of X-rays and of ultrasound in matter

(l) explain the main principles behind the use of magnetic resonanceto obtain diagnostic information about internal structures

(m) show an understanding of the function of the non-uniformmagnetic field, superimposed on the large constant magnetic field,

in diagnosis using magnetic resonance.

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Remote Sensing

Remote sensing is the investigation of an object using equipment that hasno direct contact with the object being investigated

e.g. an orbiting satellite may be designed so that it can detect small changesin mean sea level. These small changes can then be interpreted to determinethe nature of the rocks under the sea-bed. This investigation enablesinformation to be gathered without actually drilling into the sea-bed

Medical diagnosis for over 100 years used 2 risky techniques

Observe the patient externally for fever, pulse-rate, breathing, vomiting,

skin condition etc. This was part science and part art Carry out investigative invasive surgery which involved a high risk

causing many patients to die either from trauma of surgery or infection

Although there is some risk versus benefit with any procedure, nowdiagnostic imaging techniques have been developed that enable externally

placed equipment to obtain detailed information about internal bodystructures without surgery i.e. non-invasive, from under the skin i.e. a formofremote sensing

Some of the techniques make use of X-rays

Ultrasound

Magnetic resonance imaging(MRI ) 3


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THE DUCK IS DEAD - diagnosis

A woman brought a very limp duck into a veterinary surgeon. As shelaid her pet on the table, the vet pulled out his stethoscope and listenedto the bird's chest.

After a moment or two, the vet shook his head sadly and said, "I'msorry, your duck, Cuddles, has passed away."

The distressed woman wailed, "Are you sure?"

"Yes, I am sure. The duck is dead," replied the vet.

"How can you be so sure?" she protested.. "I mean you haven't doneany testing on him or anything. He might just be in a coma orsomething."

The vet rolled his eyes, turned around and left the room.?????????????

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cont

He returned a few minutes later with a black Labrador Retriever.As the duck's owner looked on in amazement, the dog stood on hishind legs, put his front paws on the examination table and sniffed theduck from top to bottom. He then looked up at the vet with sad eyesand shook his head.

The vet patted the dog on the head and took it out of the room.

!!!!!!!!!!!!!!!!!

A few minutes later he returned with a cat.

The cat jumped on the table and also delicately sniffed the bird fromhead to foot. The cat sat back on its haunches, shook its head, meowedsoftly and strolled out of the room.

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Verdict

The vet looked at the woman and said, "I'm sorry, but as I said, this ismost definitely, 100% certifiably, a dead duck."

The vet turned to his computer terminal, hit a few keys and produced

a bill, which he handed to the woman.

The duck's owner, still in shock, took the bill.

"$250?" she cried, "$250 just to tell me my duck is dead?"

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Do you know why?

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Verdict justified

The vet shrugged, "I'm sorry. If you had just taken my word for it, thebill would have been $20, .

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but with the LabReport and the CatScan, it's now $250."

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X-rays

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X-ray machine

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Dental X-ray

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Mobile X-ray machine

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Properties of X-rays

Produce fluorescence in materials e.g. zinc sulphide Produce latent image (developed to give visible image)

Penetrate substances opaque to light

Ionize and excite atoms and molecules

Have biological effects in living organism

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The production of X-rays

Whenever a charged particle is accelerated, electromagnetic radiation is emitted

The frequency of the radiation or emitted photon is proportional to the magnitude

of the acceleration Whenever high speed electrons are stopped in a metal target, X-ray photons are

emitted

X-rays production process: A metal filament(cathode) in an evacuated tube made of material that is opaque to X-rays,

is heated using a low voltage supply causing electrons to be emitted through the thermioniceffect(the opaque material also reduces background radiation)

The electrons are then accelerated through a potential difference of between 20 90 kV sothat they have high energy and high speed, but this acceleration is insufficient to cause X-ray radiation to be emitted

These high energy and high speed electrons are then accelerated towards the target(anode),constructed of metal with a high melting point and high atomic number

When these electrons are bombarded and strike or are stopped by the target metal, largedecelerations are involved, causing the electrons to lose kinetic energy very rapidly giving

rise to the emission of X-ray photons known as Bremmstrahlung radiation or brakingradiation(slowing down) This X-ray beam is made to pass out of the tube through a window that is transparent to X-

rays

Not all of the energy of the electrons is emitted as X-rays as the majority istransferred as thermal energy in the target rotating metal anode and cooling isnecessary

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Simplified design of an X-ray tube

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X-ray spectrum A typical X-ray spectrum of the variation with wavelength of the intensity

has 2 distinct components:

A continuous distribution of wavelengths with a sharp cut-off at theshortest wavelength,0

Sharp peaks may be observed corresponding to the emission linespectraof the target material and therefore a characteristic of the target

Continuous X-rays Characteristic X-rays

0

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Continuous spectrum

The continuous distribution comes about because the electrons when

incident on the metal target, will not all have the same decelerations

but will instead have a wide range of values.

Since the wavelength of the emitted spectrum is dependent on the

deceleration, there will be a distribution of wavelengths

The cut-off wavelength corresponds to an electron that is stopped in

one collision in the target so that all of its kinetic energy is given upas one X-ray photon

KE energy is lost in the form of X-ray photons

Energy of photon depends on how much KE losthence a continuous

range

Max energy of photon occurs when all KE of electrons is converted toX-rays.

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Characteristic/discrete X-rays

Superimposed on the continuous spectrum Produced when an incident electron knocks electrons out of the K-

shell(lowest shell) of the target atom

An electron from the L or M shell may

move into the vacancy in the K-shell, emitting characteristic X-rays

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X-ray spectra

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Discrete X-rays

M shell

K shell

L shell

K

K L

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Difference between continuous and characteristic

spectrum

Continuous Characteristic

(i) A continuous range of wavelength A discrete wavelength

(ii) Produced by loss of KE of incident Produced by electron

electron transition from higher shellto inner shell

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Quality of beam

Quality of beamdescribes how penetrating the beam is For monochromatic radiation, the quality is completely described by

the wavelength

High quality refers to a very penetrating beam. High quality beam is

also known as hard X-raywhile low quality is called soft X-ray.

Quality is specified/measured by HVT (half-value thickness)

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Control of the X-ray beam

In order that the optimum X-ray image may be obtained, there are 2 factors

that need to be controlled Hardness is the penetration of the X-ray beam, which determines the

fraction of the intensity of the incident beam that can penetrate the part of

the body being X-rayed. In general the shorter the wavelength of the X-rays,

the greater their penetration

Intensitythis is the wave power per unit area, and this affects the degreeof blackening of the image

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Hardness/penetration

The kinetic energy Ek of an electron is equal to the energy gained by the electronwhen it is accelerated from the cathode to the anode i.e.

Ek = eV where e is the charge of an electron and V the accelerating potentialdifference

Using E = hc/ at the cut-off wavelength, eV = hc/0hence 0 = hc/(eV)

i.e. the accelerating potential V thus determines the cut-off wavelength

The larger the potential difference, the shorter the wavelength

Therefore the hardness(penetration) of the X-ray beam is controlled byvariation of the accelerating potential difference between the cathode and theanode

A continuous distribution of wavelengths implies that there will be X-ray

photons of long wavelengths that would not penetrate the person beinginvestigatedand hence would not contribute towards the X-ray image

Such X-rays would add to the radiation dose received by the person without

serving any purpose, so the X-ray beam emerging from the tube frequentlypasses through aluminium filters that absorb these long-wavelength photons

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Intensity

The intensity of the beamdepends on the number of photons emittedper unit time and hence the number of electrons hitting the metaltarget per unit time

Since the electrons are produced by thermionic emission, increasingthe heater or filament current in the cathode will increase the rate of

production of electrons and hence increase the intensity of the X-raybeam

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Example

The accelerating potential difference between the cathode and the anodeof an X-ray tube is 30 kV. Given that the Planck constant is 6.6 x 10-34 J

s, the charge on the electron is 1.6 x 10-19 C and the speed of light in free

space is 3.0 x 108 m s-1, calculate the minimum wavelength of photons in

the X-ray beam.

Solution

For the minimum wavelength,

Energy gained by electron = energy of photon

eV = hc/0

Therefore 0 = 4.1 x 10-11 m

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The X-ray image

The image is based on the penetration of body parts by X-rays.

The transmitted X-rays produce a latent image on a photographic film. The

latent image is developed to give a visible image of the internal organ

scanned.

The image is not really an image in the sense of the real image produced by a

lens

X-ray is an ionising radiation. It loses energy mainly due to

(a) absorption and (b) scattering

The reduction in energy is an exponential decay to the distance traveled.

When an X-ray beam is incident on the body part of the patient, it can

penetrate soft tissues(skin, fat, muscleetc) with little loss of intensity

A photographic film after development will show a dark area corresponding

to these soft tissues Bone however, causes a greaterattenuation(reduces the intensity by a greater

extent) than soft tissues and therefore the photographic film will be lighter incolour in areas corresponding to the positions of bones

The quality of the shadow imageproduced depends on its sharpness andcontrast

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X-ray images

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X-ray images

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X-ray images

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A nebula: Image from Chandrasekhar observatory

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Quality of imaging

Quality of an X-ray imageis described by sharpnessand contrast Sharpness refers to a clear boundary between different tissues or the

ease with which the edges of structures can be determined

A shadow image where the bones and other organs are clearly

outlined is said to be a sharpimage

But although an image may be sharp, it may still not be clearly visiblebecause there is little difference in the degree of blackening between

e.g. bone and surrounding tissue

An image having a wide range of degrees of blackening is said tohave good contrast

Contrast refers to different intensity (brightness) in the image ofvarious parts of the internal organ.

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Sharpness

Image is sharp if the boundary is clearly visible

A sharp image requires a parallel X-ray beamwhich can be achievedby

(1) reducing the area of the target anode in the X-ray tube

(2) limiting the size of the aperture through which the X-ray beampasses

(3) reducing scattering of the emergent beam

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Sharpness(1) reducing the area of the target anode

in the X-ray tube

Secondary/partial shadows/penumbra can cause images to be blur

The full shadow produces an area that is white on the film

Where there is no shadow, the image will be black

In the region of partial shadow or greyness, the image gradually changes from

white to black

If the image is to be sharp, this area of greyness must be reduced as muchas possible

The area of the target anode should be kept to a minimum

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Sharpness(2) limiting the size of the aperture

through which the X-ray beam passes

A reduction in the grey area at the edge of the image can also beachieved by limiting the size of the aperture through which the X-ray

beam passes

This is achieved by using overlapping metal sheet plates, throughwhich the X-ray beam passes after leaving the tube

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Sharpness - (3) reducing scattering of the emergent

beam As a result of interactions between photons

and any substance through which the beampasses(even air), some photons will be

scattered resulting in loss of sharpness

These scattered or stray photons may be

absorbed in a metal lead grid placed in

front of the photographic film Scattering can also be reduced by reducing

the distance between patient and film

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Contrast

Good contrast is achieved when neighbouring

body organs and tissues absorb the X-rayphotons to very different extents e.g. bone andmuscle

Not the case e.g. if stomach or blood vessels

are being investigated

In such a case to improve contrast, especiallyfor soft tissues, a contrasting mediumis used

The patient is asked to swallow a solution of

barium sulphate(barium meal taken orally)

which is a good absorber of X-ray photons,

causing the outline of the stomach to show upclearly

Blood vessels can be made to show up visibly

by injecting a radio-opaque dye into the

bloodstream

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Contrast

Contrast also depends on other factors such as increasing exposure

time and the use of intensifying cassettes or backing the film with

fluorescent materials

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The attenuation of X-rays

When a parallel beam of X-ray photons passes through a medium, absorption

processes occur that reduce the intensity of the beam

The intensity is reduced(attenuation) by the same fraction each time thebeam passes through equal thicknesses of the medium no matter what thestarting point is chosen. This thickness of medium is called the half-value-thickness(HVT) and denoted by the symbol x

The decrease in transmitted intensity is an exponential decrease

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Linear attenuation coefficient

Consider a parallel beam having an incident intensity I0 and themedium(absorber) having thickness x and the transmitted energy is I

The transmitted energy is given by the expression

I = I0 e-x or I = I0exp(-x)

where is a constant depending on the medium and on the energy of theX-ray photons, known as the linear attenuation coefficient or the linearabsorption coefficientof the medium. The unit is mm-1 or cm-1

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Linear attenuation coefficient and HVT

For a thickness x (the half-value-thickness) of the medium, the

intensity I will be equal to I0hence I0 = I0 e

-x or I0exp(-x)

which gives x= ln 2 where x is HVT

In practice, x does not have a precise value as it is constant onlywhen the beam has photons of one energy only.

Approximate values of linear absorption coefficient for somesubstances are:

Substance /cm-1

copper 7

water 0.3

fat 0.9

bone 3

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Example

The linear absorption coefficient of copper is 0.693 mm-1.

Calculate:

(a) the thickness of copper required to reduce the incident intensity by

50%

(b) the fraction of the incident intensity of a parallel beam that is

transmitted through a copper plate of thickness 1.2 cm

Solution

(a) using I = I0 e-x or I = I0exp(-x)

I/I0 = 0.50 = exp(-0.693x)

ln 0.50 = - 0.693x therefore x= 1.0 mm

(b) I/I0 = exp(-0.693 x 12)

I/I0 = 2.4 x 10-4

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C d h (C i )


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Computed tomography(CT scanning)

The image produced on a photographic plate or film is a shadow or

flat image, and there is little or no indication of depth i.e. theposition within the body is not apparent. This is a 2-dimensional view

Soft tissues lying behind structures that are very dense also cannot be

detected

Tomography is a technique whereby a 3-dimensional image is

obtained or constructed by slicingor sectioning the body using aCT scanner through different angles using computer technology andtechniques

Data from each individual X-ray image and angle of viewing is fedinto a powerful computer enabling a 3-D image of the entire object to

be reconstructed, which can then be viewed from any angle

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B i i i l f CT CAT


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Basic principles of CT or CAT scan

Can be illustrated using a simple cube with the aim of producing an image

of a slice or section through the body from measurements made about its

axis The section or cube is divided into a series of smaller units called voxels

which absorbs the X-ray beam to different extents due to its structure

The intensity transmitted through each voxel alone is given a number

referred to as a pixel, and these various pixels are built up from

measurements of the X-ray intensity along different directions through theslice or section

Operated by using a moving X-ray emitter and detector and a powerful

computer to store data, reconstruction of these pixels in their correct

positions is done to display the 3D image of the internal organ being

scanned

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R i f h li i


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Re-construction of the slice or section

Suppose a simple cube showing a four-voxel section

The image of each voxel would have a particular intensity, known asa pixel. The pixels are built up from measurements of X-ray intensitymade along a series of different directions around the section of the

body

The number on each voxel is the pixel intensity that is to be

reproduced Pixel(picture element) is actually a two dimensional unit based on the

matrix size and the field of view.

When the CT slice thickness is also factored in, the unit is known as aVoxel, which is a three dimensional unit.

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Detector readings(illustrations - clockwise)


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g ( )

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Fi l tt


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Final pattern

In order to obtain the original pattern of pixels, two operations must

be performed.- 1. The background intensity must be removed. The background

intensity is the total of each set of detector readings. In this case, 14 is

deducted from each pixel.

- 2. After deduction of the background, the result must be divided by

three to allow for the duplication of the views of the section since 4

sets of readings were taken

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CT cont

In practice, the image of each section is built up from many small

pixels, each viewed from many different angles.

The greater the number of voxels, the better the definition, similar toa digital camera

In order to build up an image of the whole body, the procedure would

be repeated for further sections through the body. All the data for all the sections can be stored in the computer memory

to create a three-dimensional image. Views of the body from different

angles may constructed

The collection of the data and its construction into a display on a

screen requires a powerful computer and complicated softwareprogramming and programs. In fact, the reconstruction of each pixel

intensity value requires more than one million computations.

The computer allows for the contrast and brightness of the image to be

varied so that an optimum image can be obtained50

Ad t /di d t


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Advantages/disadvantages

CT scan is expensivebut is indispensable in todays medical practice

Although historically the images generated were in the axial or

transverse plane, orthogonal to the long axis of the body

Modern scanners allow this volume of data to be reformatted in

various planes or even as volumetric (3D) representations of

structures. The image of the organ can be viewed in any direction. You can look

at the organ from the front, top, back, and side. Hence a 3 D image ofthe organ

Higher resolution compared to Ultrasound

Excellent for bone imaging

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E i


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Exercise

1. Compare the image produced during an X-ray investigation and that

produced in CT scanning

2. The principles of CT scanning have been understood for some time.

However scanners could not be developed until large powerful computers

were made available. By reference to the image produced in a CT scan,

suggest why such a computer is necessary

3. A simple object consists of 4 voxels out of 8 voxels (i.e. cube can be thought

of as 2 slices). The object is scanned from 4 different directions, each at 45to the next. The detector measurements for each individual voxel are summed

and the result is as shown here. The total of the readings of the detectors in

any one position is 22. Determine the pattern of the pixels in the voxels

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Ultrasound

Incident wave

Reflected wave

Transmitted wave

Boundary between media

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Recappiezo-electric effect Piezo-electric devices contain a crystal which can expand and compress

when external pressure is varied e.g. quartz

The crystals structure is such that the centre of positive charges coincideswith the centre of negative charges when not stressed

When expanded, both centres will not coincide.

When compressed, the centres will be in the opposite direction ascompared to under expansion.

The separation results in a voltage across the crystal surface and this effectis known as the piezo-electric effect

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Piezo-electric transducer A piezo-electric device is a sensor that detects differences in pressure

(sound wave)

Variation in pressure will result in an ac voltage The magnitude of the voltage generated depends on the magnitude of the

pressure on the crystal and the polarity depends on whether the crystal iscompressed or expanded i.e. whether the pressure is greater than or lessthan the ambient pressure

A transducer is any device that converts energy from one form to another

The piezoelectric transducer converts mechanical energy (vibration) intoelectrical energy in the form of ac voltages

It also can convert electrical voltages back to vibration.

Hence it acts as a receiver as well as an emitter.

To detect the voltages, opposite faces of the crystal are coated with a metal

(silver) and electrical connections are made to these metal films and sincethe voltages are very small they are amplified

The crystal and its amplifier may be used as a simple microphone forconverting sound signals into electrical signals

Ultrasound waves may be generated using a piezo-electric crystal such asquartz, as it can convert electrical voltages to vibrations

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Ultrasound


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Ultrasound

When a potential difference is applied between the electrodes of the crystal,

an electric field is set up in the crystal which causes forces to act on the ions

Quartz has a tetrahedral silicate structure with the oxygen ion negativelycharged and the silicon ion positively charged, and as these ions are notheld rigidly in position, they will be displaced slightly when an electric fieldis applied

The positive ions will be attracted to the negative electrode, and the negative

ions will be attracted to the positive electrode and depending on the direction

of the electric field, the crystal will become slightly thinner or thicker

An ac voltage applied across the electrodes will cause the crystal to vibrate

with a frequency equal to that of the applied voltage with a small amplitude

If the frequency of the applied voltage is equal to the natural frequency of

vibration of the crystal, resonance will occur and the amplitude of vibration

will be a maximum The dimensions of the crystal can be such that the oscillations are in the

ultrasound region(> 20 kHz) and this will give rise to ultrasound waves inany medium surrounding the crystal

In the medical field the ultrasound frequency is in the megahertz region

Human range of hearing is from 20 Hz to 20,000 Hz56

The reflection and absorption of ultrasound


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The reflection and absorption of ultrasound

Ultrasound is typical of many types of waves in that , when it is incident on a

boundary between 2 media, some of the wave power is reflected and some

transmitted For a wave of incident intensity I , reflected intensity IR and transmitted

intensity IT, by conservation of energy,

I = IR + IT Although for a beam of constant intensity, the sum of the reflected and

transmitted intensities is constant, their relative magnitudes depends not onlyon the angle of incidenceof the beam on the boundary but also on the mediathemselves

Hence the relative magnitudes of IRand IT are quantified by reference to thespecific acoustic impedance, Z of each media

The specific acoustic impedance Z is defined as the product of the density

of the medium and the speed c of the wave in the medium i.e. Z =c For a wave incident normally on a boundary between 2 media having specific

acoustic impedances ofZ1 and Z2, the ratio of the reflected intensity IR to theincident intensity I known as the intensity reflection coefficient symbol

i.e. intensity reflection coefficient = IR/I = (Z2-Z1)2/(Z2+Z1)

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Material Density

r/kg m3Speed of sound

c/m s1

Acoustic impedance

Z/kg m2

s1

air 1.3 330

blood 1060 1570

muscle 1075 1590

bone 1600 4000

The specific acoustic impedance Z is defined as the product of the density

of the medium and the speed c of the wave in the medium i.e.

Z =c

Typical values of specific acoustic impedance and


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Typical values of specific acoustic impedance and

speed of ultrasound

Medium Speed/m s-1 specific acoustic impedance/kg m-2 s-1

Air 330 430

Water 1500 1.5 x 106

Blood 1600 1.6 x 106

Fat 1500 1.4 x 106

Muscle 1600 1.7 x 106

Soft tissue 1600 1.6 x 106

Bone 4100 5.67.8 x 106

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Example


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Example

Calculate the intensity reflection coefficient for a parallel beam of ultrasoundincident normally on the boundary between:

(1) air and soft tissue, specific acoustic impedances of 430 kg m-2 s-1and 1.6 x 106 kg m-2 s-1 respectively

(2) muscle and bone, specific acoustic impedance of 1.7 x 106 kg m-2 s-1

and 6.5 x 106 kg m-2 s-1 respectively

Solution

Using = IR/I = (Z2-Z1)2/(Z2+Z1)

2

(1) = 0.999 almost 1!

(2) = 0.34

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Linear absorption coefficient


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Linear absorption coefficient

The intensity reflection coefficient for a boundary between air and soft tissue

from the last example is almost unity which means that when ultrasound is

incident on the body, very little ultrasound is transmitted into the body In order to overcome this, it is important that there is no air between the

transducer and the soft tissue(skin) and this is achieved by using a waterbased jelly whose specific acoustic coefficient is approximately 1.5 x 106kgm-2 s-1

Once the ultrasound is within the medium, the intensity of the wave is

reduced by absorption of energy as it passes through the medium

This causes heating and appropriate frequencies of ultrasound are actually

used in physiotherapy to assist in sprains and similar injuries

For a parallel beam, the absorption is approximately exponential and for abeam of ultrasound that is incident normally on a medium of thickness x, the

transmitted intensity I is related to the incident intensity I0by the expressionI = I0 e

-kx or I = I0exp(-kx)

where k is a constant depending on the medium known as the linearabsorption coefficient. The unit is cm-1

The coefficient k depends not only on the medium but also the frequency of

the ultrasound61

Typical values of linear absorption coefficient k of


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Typical values of linear absorption coefficient, k ofultrasound

Medium Linear absorption coefficient/cm-1

Water 0.0002

Bone 0.13

Muscle 0.23

Air 1.2

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Example


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Example

A parallel beam of ultrasound is incident on the surface of a muscle and

passes through a thickness of 3.5 cm of the muscle. It is then reflected at

the surface of a bone and returns through the muscle to its surface.Calculate the fraction of the incident intensity that arrives back at the

surface of the muscle given that the linear absorption coefficient for

muscle is 0.23 cm-1 and the fraction reflected at bone-muscle interface is

0.34(from the last example)

Solution

The beam passes through a total thickness of 7.0 cm of muscle

For the attenuation in the muscle,

using I = I0exp(-kx) = I0exp(-0.23 x 7.0) = 0.20I0Given that the fraction reflected at the bone-muscle interface i.e. is 0.34,

therefore the fraction received back at surface = 0.34 x 0.20 = 0.068 = 1/15

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Obtaining diagnostic information using ultrasound


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Obtaining diagnostic information using ultrasound

The ultrasound transducer is placed on the skin with the water based

jelly excluding any air between the transducer and the skin

Short pulses of ultrasound are transmitted into the body where theyare partly reflected and partly transmitted at the boundaries between

media in the body

The reflected pulses or echoes return to the transducer where they are

detected and converted into voltage pulses which are amplified and

processed by electronic circuits to be displayed on a screen oroscilloscope

The time between the transmitted and reflected pulses givesinformation as to the distance of the boundary from the transducer

The intensity of the reflected beam gives information as to the

nature of the boundary

2 techniques are in common use for the display of an ultrasound scan:

A-scan

B-scan

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A-scan


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A-scan

A short pulse of ultrasound is transmitted into the body through the coupling

medium

At each boundary some of the energy of the pulse is transmitted and some isreflected

The transducer(generator/detector) detects the reflected pulses as it now acts

as a receiver

The signal is amplified and displayed on a c.r.o.

Reflected pulses received at the transducer from deeper in the body tend tohave lower intensity than those reflected from boundaries near the skin

This is due not only to the energy being absorbed by the various media but

also on the return of the reflected pulse to the transducer, some of the energy

of the pulse will again be reflected at intervening boundaries

To allow for this, echoes received later at the transducer are amplified more

than those received earlier

A vertical line is observed on the c.r.o. corresponding to the detection of each

reflected pulse

The time-base of the c.r.o. is calibrated so that knowing the speed of the

ultrasound wave in each medium, the distance between boundaries can be

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B-scan


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B-scan

This consists of a series of A-scans all taken from different angles so

that a 2-D image can be formed

The ultrasound probe consisting of a generator/detector, for a B-scandoes not consist of a single crystal, but rather an array of smaller

crystals each one at a different angle to its neighbours

The separate signals received from each of the crystals in the probe is

processed and a pattern of spots is built up to create a 2-D image for

immediate viewing, photographing or to be stored in computermemory

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Advantages of ultrasound scanning


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Advantages of ultrasound scanning

Health risk to patient and operator is very much less compared to use

of X-rays

The equipment is much more portable and easy to use

Using higher frequency ultrasound enables greater resolution to be

obtained i.e. greater details to be seen

Modern techniques allow for the detection of very low intensity

reflected pulses, hence boundaries between tissues where there is little

change in acoustic impedance can be detected

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MRI

Magnetic resonance imaging

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Proton spin


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Proton spin Many atomic nuclei have a property

known as spin which causes the

nuclei to behave as if they were

small magnets

Such nuclei have an odd number ofprotons and/or neutrons e.g.hydrogen, carbon, phosphorus

The proton is not stationary but

spins about an axis and the spinning

acts like a tiny current loop which

generates a magnetic field along the

spin axis

The spinning proton behaves like a

spinning top

When perfectly balanced, it spins

about a vertical axis

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Hydrogen atoms


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The magnetic fields of hydrogen atoms are random if there is no external

magnetic field around. Their individual fields tend to cancel out

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Precession


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ecess o

When a proton is slightly displaced, it wobbles or precesses about thevertical axis.

When a external magnetic field Bo

is applied to these spinning protonnuclei, they tend to line up along the field but this alignment is not perfectand the nuclei rotate about the direction of the magnetic field due to theirspin

The alignment is either parallel or anti-parallel.

Parallel alignment: lower energy states, E1

Anti-parallel alignment: higher energy state, E2 This motion can be modelled as the motion of a top spinning about the

direction of a gravitational field and this rotation is known as precession

The spinning about the direction of the magnetic field i.e. the precessabout the external field has a frequency of precession known as theLarmor frequency which depends on the nature of the nucleus and the

strength of the magnetic field which is the natural frequency of theprecessing proton.

The frequency is directly proportional to the external magnetic fieldstrength, Bo and is given by f = 42.57 Bo for the hydrogen atom

For large magnetic fields like 1 T or 2 T, f is of the order of 50 MHz,which is in the radio frequency range. 71

Precession


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Precession

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Nuclear magnetic resonance(NMR) and magnetic


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g ( ) g

resonance imaging(MRI)

If an external stimulus pulse of electromagnetic radiation of radio frequency

of the same frequency as the Larmor frequency is incident on the precessingnuclei, the nuclei will resonate in phase with each other, absorbing energyand flipping to the higher energy state, E2.

This frequency is in the radio-frequency(RF) band with a wavelength shorter

than about 10 cm

A short time after the incident pulse has ended, the nuclei will return to theirequilibrium state, emitting RF radiation

This short time between the end of the RF pulse and the re-emitting of the

radiation is known as the relaxation time

This entire process of precessing and resonating in phase and emitting RF

radiation is known as nuclear magnetic resonance

In practice, there are 2 relaxation processes and it is the time between these 2

that forms the basis ofmagnetic resonance imaging(MRI)

Since hydrogen is abundant in body tissues and fluids, hydrogen is the atom

used in MRI and MRI monitors the concentration of hydrogen nuclei in thebody

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RF stimulus and MRI signal


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RF stimulus and MRI signal A small resultant longitudinal magnetic field MLis developed by the

protons, parallel to Bo.

This cannot be measured since it parallel to the large external magneticfield Bo

Before receiving the RF pulse, there in no resultant transversemagnetic field MTas the protons precess out of phase with each other.(cancel out)

However, a RF pulse will force them to precess in phase with each

other. This produces a resultant transverse magnetic field MT which precessesabout the main spinning axis at the same Larmor frequency as the

protons. This a signal that can be measured.

According to EM induction law, a changing magnetic field induces ane.m.f in the conductor.

The resultant MT precessing at 90o

to the external magnetic field caninduce a small e.m.f. (microvolts) in a detector coil positioned outsidethe patient.

This forms the MRI signal, the alternating voltage pulse having thesame frequency (the Larmor frequency) as the RF pulse detected.

Hence, all hydrogen atoms in body act as emitters of RF pulses74


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Principal of the MRI scanner


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The patient is positioned in the scanner between the poles of a large magnet thatproduces a very large uniform magnetic field in excess of 1 Tesla

This magnetic field causes all the hydrogen nuclei within the person to precess withthe same Larmor frequency

In order that the hydrogen nuclei in only one small part of the body may bedetected, a non-uniform magnetic field is also applied across the patient

This non-uniform field is accurately calibrated and results in a different magnitudeof magnetic field strength at each point in the body of the patient

Since the Larmor frequency is dependent on the strength of the magnetic field,the Larmor frequency will also be different in each part of the patient

The particular value of the magnetic field strength together with the radio-frequency that is emitted, enables the hydrogen nuclei in the part underinvestigation to be located

Radio-frequency pulses are produced in coils near the patient which pass into the

patient The emitted pulses produced as a result of de-excitation of the hydrogen nuclei are

picked up by the coils

These signals are processed to construct an image of the number density ofhydrogen atoms in the patient i.e. concentration of hydrogen nuclei

As the non-uniform magnetic field is changed, atoms in different parts of the

patients body are detected and displayed as a 3-D image

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Schematic of a MRI scanner


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Schematic of a MRI scanner

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MRI scanner


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MRI scanner

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Slice selection


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Slice selection

Larmor frequency is proportional to Bo.

If a slightly different Bo is applied to different sections of the body,

each section will have its own individual Larmor frequency f.

Hence, if a pulse of specific RF frequency is sent to the body, only

those protons having exactly the same precession frequency f will

resonate. Protons in neighbouring slices will be relatively unaffected

The variation in the value of external field applied is achieved by using

a gradient field coil.

Btotal = Bo + B

(fixed) (spatially varying)

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MRI Image


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g

Rf signals from the body are used to build up an image.

Their origins are located using three perpendicular gradient fields. The signal strength dependson:

(i) Proton density (hydrogen concentration)

(ii) tissue type (structure and surroundings)

(iii) pulse sequence

(iv) relaxation

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Resolution


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Resolution

Resolution can be improved using

(a) large and uniform main magnetic field B

(b) Large gradient fields

(c) Small receiver coils close to the body

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Advantages/disadvantages of MRI


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Advantages/disadvantages of MRI

Advantages

Safe since no ionising radiations are used

Any selected plane and orientation can be imaged

Excellent soft tissue contrast, no contrast agent needed

Versatile since image varies with many parameters

Body function and chemistry investigated besides structure

Disadvantages High capital and running costs

Image selection and interpretration is complex

Examination can be claustrophobic, noisy and long

Hazards with implants, (pacemaker)

Practical problems associated with large superconducting magnets. Takes time

For children it may not be possible as the patient has to be very still

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Section VII 29 Remote Sensing

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Transcript of Section VII 29 Remote Sensing