Ultrasound Flashcards
How is a sound wave different to an EM wave?
Unlike electromagnetic waves, sound waves cannot be transmitted through a vacuum; they require a medium in the form of a gas (such as air), a liquid (such as water or blood) or a solid (such as soft tissue, bone or metal).
An electromagnetic wave is transverse but a sound wave is longitudinal i.e. the pressure variation occurs in the direction of travel. Like all other sinusoidal waves, a sound wave is characterised by frequency, wavelength and speed (or velocity).
What is the pulse-echo principle?
The pulse-echo principle states that the distance of a reflecting object can be established by timing the go-and-return of a short pulse if the speed of the pulse is known. The returning pulse is known as the echo.
What is the rough speed of sound in tissue?
around 1540 m/s
What is ultrasound?
The answer is that it is simply sound at a frequency (pitch) which is too high for us to hear. In all other respects it is the same as normal sound.
Why are echoes smaller for deeper tissues?
some of the signal is lost (attenuated) on the way and hence there is a general tendency for echoes to get smaller as the depth increases.
What is TGC?
control which allows the operator to adjust the amount of gain (amplification) which is applied to echoes from different depths. This is called time gain compensation, and is an important control for the operator to identify and master. It normally needs to be adjusted during a scan and certainly between patients.
What is A-mode scanning?
This way of presenting information about the targets along a single direction is called the A-scan. it does not resemble a 2D cross-section image but may be regarded as a 1D image line
What is b-mode scanning?
The assembly of many A-scan lines forming a composite image in this way is called a B-scan. In a modern B-scan, the dots do not have the same brightness. Dot brightness depends on echo amplitude and is displayed on a greyscale with white representing very large amplitude echoes and black being used when no echoes are detected (Fig 2). Furthermore, each B-scan is acquired so rapidly that the image is obtained in ‘real time’.
Concerning the use of ultrasound in a medical setting:
A. Time gain compensation is used to correct for variations in the speed of sound between tissues
B. Medical ultrasound imaging typically uses frequencies in the range 3-150 kHz
C. Each single transmitted pulse may result in the creation of a large number of echoes
D. Soft tissues typically have sound speeds of the order of 1500 m/s
E. The angle at which the beam meets an organ boundary may have a large impact on the echo generated
F
F
T
T
T
Time gain compensation is not used to correct for variations in the speed of sound between tissues, but for the increase in echo attenuation with tissue depth.
Medical ultrasound imaging typically uses frequencies from around 1 MHz to around 10 MHz.
Which of the answers below is/are a correct description of TGC?
A. It is fitted to virtually all ultrasound scanners
B. It alters the way ultrasound is generated
C. It needs adjustment between patients and between views
D. It can create artefacts if incorrectly applied
E. It influences only the B-scan
Answers A, C and D are correct.
Time gain compensation does not alter the way ultrasound is generated.
Time gain compensation influences the A-scan as well as the B-scan.
Although some scanners are now fitted with automatic TGC, it is always present in some guise. Its effect is confined to processing signals from received echoes. Since the TGC is an attempt to correct for tissue attenuation, it follows that it needs to be altered between patients, but also within a scan on an individual patient when moving from one organ or section to another with different acoustic properties.
What is the piezoelectric effect?
If an electric voltage is applied to specific materials, they deform, and if a mechanical pressure is applied to them, they develop a voltage on their surface.
What materials are used for ultrasound transducers?
Some naturally-occurring materials (such as quartz crystals) have piezoelectric properties and these were originally used as ultrasound transducers. However, modern imaging transducers are made of synthetic crystalline ceramic materials such as lead zirconate titanate (PZT).
What is the curie temperature?
Immediately after manufacture, this material is not piezoelectric. However, heating to raise its temperature beyond a certain value (the Curie temperature) while applying an external voltage causes electric dipoles within the crystal to align and give the material piezoelectric properties. If the voltage is maintained as the material cools below the Curie temperature the dipole alignment and hence the piezoelectricity is preserved. The material is then cut into a suitable size and shape for use as an ultrasound transducer. Note that further heating above the Curie temperature destroys the piezoelectric properties.
What does the resonant frequency depend on?
the thickness of the transducer and so the same voltage spike will create different damped responses in different transducers.
At the resonant frequency, the wavelength of an ultrasound wave in the transducer is twice the transducer thickness. Therefore, for a given transducer material, resonant frequency and transducer thickness are inversely proportional to each other; a thick crystal gives a low resonant frequency and vice versa
What is the spatial resolution of ultrasound?
The spatial resolution of an imaging system may be thought of as the extent to which the system blurs the image of an object. Alternatively, it can be regarded as the ability of the system to distinguish two separate small objects lying close together. As the distance between the objects is reduced, there will come a point at which the system merges the two and they blur into a single image
What is axial resolution?
Axial resolution refers to a situation in which two objects lie along a single scan line i.e. along the axis of the ultrasound beam; it is sometimes called the longitudinal resolution.
when the object separation is half the length of the ultrasound pulse i.e. half the spatial pulse length the objects will not be resolvable. This distance is a measure of the axial resolution.
How does axial resolution change with frequency?
speed = frequency x wavelength; if ultrasound frequency increases, wavelength must decrease to compensate since speed is more or less constant (assumed to be 1540 m/s in soft tissue). Therefore, higher frequencies are associated with shorter wavelengths and so 2.5 cycles at a higher frequency will occupy a shorter distance (spatial pulse length).
What is lateral resolution?
resolving objects in the lateral (aka azimuthal plane)
The task of resolving these objects is a different one. Note firstly that it requires more than one beam. In fact we need a minimum of three:
The first beam must hit one object but not the other
The second beam must go through the gap between them
The third must hit the second object only
The beam width is a measure of the lateral or azimuthal resolution.
What is slice thickness resolution?
out-of-plane resolution or elevational resolution. Slice thickness is a measure of elevational resolution and it depends on the dimension of the transducer in this direction
What crtieria does the matching layer of the probe have to meet?
In front of the transducer is a matching layer. It serves as an interface between the PZT transducer element and the patient’s skin. Specifically, it reduces the effect of differences in a property called acoustic impedance, whose value is much greater for PZT than for soft tissue. For it to be effective, the matching layer has a thickness equal to a quarter of the ultrasound wavelength in the material of the layer and acoustic impedance equal to the geometric mean of the values for PZT and tissue. In the absence of a matching layer, much of the ultrasound would be reflected back towards the transducer at its interface with the skin.
What lies behind the PZT crystal in the transducer?
Behind the transducer there is a backing block and an acoustic absorber. The backing block is made of a material that is highly attenuating and has acoustic impedance similar to that of PZT, such as epoxy resin loaded with tungsten. This means that ultrasound energy emitted by the back face of the transducer is transmitted into the block and attenuated by it. This dampens the vibration of the transducer producing pulses of short spatial length and so improving axial resolution. The acoustic absorber attenuates stray ultrasound energy.
What is SPL determined by?
In imaging with pulsed ultrasound, a short spatial pulse length (SPL) improves axial resolution. The SPL is determined by the degree of transducer damping.
What is the Q factor of an ultrasound pulse?
The range of frequencies in the spectrum is known as the bandwidth of the pulse and the Q factor is defined as fo/bandwidth.
What properties does a lightly damped US beam have?
A lightly damped transducer is associated with long SPL, narrow bandwidth and high Q
What properties does a heavily damped US beam have?
a heavily damped transducer is associated with short SPL, broad bandwidth and low Q
What sort of imaging is a high Q pulse good for?
higher Q is more appropriate for Doppler flow measurement.
What sort of imaging is a low Q pulse good for?
A low Q transducer is best suited for B-mode imaging
What happens to beams from narrow transducers?
beams from smaller sources begin to diverge at shorter distances from the source surface.
What is the fresnel zone?
The region of the beam close to the transducer where it maintains its original width (the width of the transducer element) is called the near field or Fresnel zone.
What is the Fraunhofer zone?
Further from the transducer is a region in which the beam diverges; this is known as the far field or Fraunhofer zone.
What does the length of the near field depend on?
diameter of the source, wavelength
What is Huygens principle?
the surface of a transducer can be regarded as an infinite number of very small sources each of which emits spherical sound waves (wavelets).
What is the principle of superposition?
when two or more waves coincide, the net amplitude at a point is the sum of the individual wave amplitudes at that point
When does constructive interference happen?
Constructive interference occurs when the waves are in phase (0° phase difference) and destructive interference when they are out of phase (180° phase difference).
How does a change in source diameter effect near field?
a modest decrease in transducer diameter (from 2 cm ro about 1.5 cm) has reduced the near field by a factor of almost two (from 19 cm to 10 cm). The decrease in diameter also means better lateral resolution in the near field.
How does changing requency effect legth of near field?
For a fixed transducer radius, we can reduce the length of the near field by choosing a lower frequency transducer
For a fixed frequency, how can we reduce the length of the near field?
For a fixed frequency, we can reduce the length of the near field by choosing a transducer with reduced diameter - this also reduces overall beam width and so improves lateral resolution
For a fixed near field length, how can we reduce overall beam width and improve lateral resolution?
For a fixed near field length, we can reduce overall beam width and improve lateral resolution by choosing a smaller diameter transducer but the transducer also has to be of higher frequency - thus good lateral resolution is usually associated with higher ultrasound frequency
Where is lateral resolution of a beam best?
ultrasound beam has a degree of natural focusing. It is at its narrowest towards the end of the near field and so this is the point at which lateral resolution is at its best.
What is beamforming?
using electronic lenses. the sophistication of the various focusing techniques is now such that the term ‘beamforming’ is often used instead of beam focusing
At what frame rate can the eye not really distinguish?
Typically, if images are presented at a rate of more than about 20 frames per second, the eye and brain together interpret this as a continuously moving image.
What can limit the frame rate possible?
hen a pulse is sent out from one element, it is necessary to wait for all the echoes to come back from the region in front of that element before sending out another pulse from a nearby element. Failure to do this runs the risk of ambiguity.
If a pulse is sent out from a nearby element before the normal waiting time has elapsed, it is possible for returning echoes to be picked up by the wrong element and therefore misinterpreted. To avoid this, it is normal to wait between pulse firings. This has the effect of slowing down the imaging process and limiting the frame rate.
What is a ‘synthetic aperture’?
firing of elements in groups to avoid issues of very short near field and wide divergence and hence very poor lateral resolution due to <1 mm wide elements.
Why are electronic dealys used in the transducer?
simultaneous firing (excitation) of transducer elements generates an ultrasound beam whose axis corresponds to the position of the central element. This beam has a useful near field length but its overall width limits lateral resolution.
electronic delaying allows ‘electronic focusing of the beam. This symmetrical arrangement of delays produces wavefronts that are concave in the direction of travel and converging on the focal point.
It is at the focus that the beam is at its narrowest and lateral resolution at its best
How are electronic delats used in signal reception?
echoes that start from T in phase are out of phase when they reach the individual transducer elements and the signals they produce have small amplitude when summed by the scanner electronics.
However, the signals on the return journey can be brought into phase by applying exactly the same electronic delays as were applied to the excitation pulses on the outward journey (because the distances to and from T to individual transducer elements are the same in both cases). Addition of the individual delayed signals now gives a strong total signal
What is dynamic focusing?
the receiving electronic focus to change to a greater depth and repeat this process to steadily move out the focal point so that it is always in the right place for the echoes arriving at the instant. In fact, electronically it is possible to make a continuous smooth sweep of the delays so that the focal length of the receiving electronic lens is always correct
What is multiple zone focusing?
The transmitted pulse cannot be heavily focused if it is required to generate echoes over a wide range of tissue depths as this would cause image degradation through poor lateral resolution.
We have the option of selecting a preferred depth or range of tissue depths and accepting only echoes that arrive from those depths. we can cover the entire tissue depth of interest by transmitting not one but several ultrasound pulses per image line. For successive pulses we can focus at increasing depths in transmission (Fig 1). Focusing is also applied in reception but, by rejecting echoes from other depths (i.e. those outside the selected focal zone), we can gradually build up a composite image over the whole depth range.
What are the limitations of multiple zone focusing?
It produces an extended focal zone (Fig 2) and improves image quality but, unlike dynamic (swept) focusing in reception, it carries a performance penalty because it increases the time needed to acquire one image frame (and so reduces the frame rate). Thus the operator has a choice between good lateral spatial resolution and good temporal resolution.
What is dynamic aperture control?
To maintain the same beam width at the focus for increasing focal length, the synthetic aperture must be increased. Thus smaller groups of transducer elements are used for short focal lengths whereas larger groups are used at greater depths.
What are the typical number of elements in a linear array?
Typically there may be 256-512 elements that are electrically and acoustically isolated from each other. The elements are about 0.5 mm in width and about 5 mm in height.
How is a phased array different?
The construction of a phased array is similar to that of a linear or curvilinear array except that there are fewer elements and all of them are excited nearly (but not exactly) simultaneously to produce a single ultrasound beam. This beam is steered in such a way that it is swept through an arc to produce a sector-shaped FOV with an apex at the transducer
How is beam steering achieved in pahased arrays?
Beam steering is achieved with electronic delays that steadily increase (or decrease) from one end of the array to the other. The greater the difference in the delays, the greater the deviation of the beam from 90° to the transducer face. The beam is swept by rapidly changing the value of the delay difference. By making the differences unequal, the beam may be focused within the scan plane.
What does lateral resolution depend on?
A. Pulse length
B. Frequency
C. Scanner focus setting
D. Beam width
Answers B, C and D are correct.
The primary determinant of lateral resolution is ultrasound beam width. Focusing reduces beam width and so improves lateral resolution. Furthermore, in the focal region:
Beam width = focal length (distance to focus) x wavelength/transducer aperture
Therefore, short ultrasound wavelength (high frequency) and large aperture improve lateral resolution.
What does axial resolution depend on?
A. Pulse length
B. Frequency
C. Scanner focus setting
D. Beam width
Answers A and B are correct.
Axial resolution is equal to half the spatial pulse length.
Pulse length = number of waves in pulse x wavelength.
Therefore a small number of waves (achieved by heavy transducer damping and low Q factor) and a short ultrasound wavelength (high frequency) improve axial resolution.
What is acoustic impedance?
property of individual materials and it is often given the symbol Z. The Z of a material is the product of its density (ρ) and the speed of sound (c) in that material.
The international system (SI) unit of acoustic impedance is kgm-2s-1, which is given the special name of rayl.
What happens to sound waves at a boundary between materials with different Z at 90 degree incidence?
some of the wave will be reflected and some transmitted. the strength of the reflection is related to the change in Z value across the boundary. In other words, if the difference between the Z values on either side of the boundary is large, then that interface will be a strong reflector
What is the reflection coefficient?
The reflection coefficient (R) = fraction of incident intensity reflected at the boundary, where intensity is the energy per unit time per unit area in the ultrasound beam
What is the optimal strength of a reflection for the purposes of imaging?
The ability of the system to detect and display the existence of an interface is obviously enhanced if that interface creates a strong echo. However, any energy reflected at that interface is not available to interrogate tissue distal to it. Beyond a strong interface there will be a region of shadow in which little or no information can be obtained.
What is specular reflection?
In the example of normal incidence, the interface is large and smooth, i.e. its dimensions are much greater than the size and wavelength of the ultrasound beam. Such an interface is called a specular reflector.
What artefact can specular reflection cause?
Strong specular reflection is the origin of the artefact known as shadowing
During ultrasound imaging, a coupling gel is used between the ultrasound probe and the patient’s skin.
Regarding coupling oils and gels that are are normally used when scanning with ultrasound:
A. Coupling gels are only used to ensure a low friction scanning surface
B. The Z values of coupling gels will be similar to that of water
C. The presence of a small amount of air between transducer and skin is not important
D. Using more gel will normally enhance an image which is already acceptable
E. During the course of a scan, it may be necessary to provide extra gel
F. Some patients are allergic to scanning gels
G. Gels may cause damage to transducer surfaces
H. Gels should be sterilised for normal use
F
T
F
F
T
T
T
F
Most liquids have Z values similar to water. When a dry transducer is placed on the skin, air is trapped in the small space but because of its very low Z value, even a tiny amount of such air will cause serious image degradation with ultrasound energy being reflected back into the transducer rather than passing into the patient.
A key part of the couplant’s function is to displace this air, although it also serves as a lubricant. Some patients are allergic to some gels and some gels will attack specific transducer materials. If there is an open wound or a biopsy is planned, then couplant sterility is critical - otherwise it is not. However, cleaning of the transducer between patients is important to minimise cross-infection risk.
What is Snell’s law?
The transmitted beam is refracted at the boundary, i.e. it does not travel in the same direction as the incident beam, and t is called the angle of refraction. As with the refraction of light, Snell’s law applies to refraction of sound where the angle of refraction is dependent on the change in speed of sound across the boundary.
What characteristics does rayleigh scattering have?
Scattering has the following characteristics:
The scatterer is smaller than a wavelength
Most of the beam does not interact with it at all
The part of the beam which does interact is re-transmitted in almost all directions
The strength of the scatterer increases quickly as its radius increases (provided that it is still much less than the wavelength of the ultrasound)
The strength of the scatterer also increases as the ultrasound frequency increases (as the fourth power of the frequency)
As for reflection, the greater the acoustic impedance mismatch between the material of the scatterer and the surrounding medium, the greater will be the strength of the scatterer
Which of the following are real scatterers?
A. Gallstone
B. Red blood cell
C. Small air bubble
D. Liver metastasis
B and C are true.
Typical ultrasound wavelengths are between 0.1 and 0.5 mm. It follows that a true scatterer needs to be much smaller than this. The typical size of a red blood cell is 6-8 μm (i.e. 0.006-0.008 mm). Small gas bubbles will also be scatterers of ultrasound, provided they are as small as this.
Gallstones, metastases etc. which are detectable by ultrasound, would normally be at least 1 mm or so in size.
What phenomena do real tissue interfaces exhibit?
Echoes have an angular dependence due to reflection
The interfaces are detectable over a range of angles due to scattering
Interfaces between two soft tissues create weak but detectable echoes because of the small differences in acoustic impedance
Gas inclusions of any size in an interface create strong echoes
Gas inclusions and calcified regions in an interface create shadowing artefacts in the region distal to the interface
WHat is absorption in US due to?
direct conversion of ultrasound energy into heat.
All ultrasound exposures contribute some heating to the tissues through which the wave travels
What pattern does US attenuation follow in tissue?
Ultrasound attenuation follows an exponential pattern in tissue and other materials. In addition, higher frequencies are attenuated to a greater extent than lower ones.
What is the HVL in ultrasound?
the thickness of tissue required to reduce the intensity of the wave by half.
In soft tissue:
Frequency (MHz) -> Half value thickness (cm)
3 2.0
5 1.2
7 0.8
10 0.6
How many HVLs can scanners still detect at?
most scanners can cope with signal losses caused by at least eight half value layers.
What is the fundamental compromise in ultrasound?
If the frequency is too high, we will not penetrate to the most distal structures of interest
If the frequency is reduced, there will be a loss in overall image quality
What is the clock in a US transducer?
The general purpose of the clock is to synchronise the operation of the whole system. In particular, the clock determines the pulse repetition frequency (PRF).
WHat is the minimum pulse repitition time determined by?
minimum pulse repetition time period is determined by the maximum depth of imaged tissue and sets an upper limit on the PRF.
Why is it desirable to use high values of PRF?
The number of ultrasound scan lines which can be used to create the image is limited by the need for real-time display. The higher the PRF, the greater the number of lines which can be used and therefore, the better the image quality will be.
What is the minimum frame rate desired?
25
Why is interpolationused in US?
additional lines are introduced using values which are intermediate between adjacent real lines because A standard domestic TV monitor has 625 lines and so a 206 line image would look coarse and there would be a sort of ‘venetian blind’ effect.
What is the transmitter in an US probe?
This component responds to the instruction from the clock to begin a new line by creating a large voltage which is applied to the transducer.
These high-amplitude fast transmitter pulses are applied directly to the piezo-electric crystals in the transducer. The result is that the ultrasound pulse is generated and transmitted into the body.
What is an ADC in a US probe?
analogue to digital converter converts echoes received from the body into digital signals before any further processing takes place.
Why is a higher digitisation rate better?
More detail is preserved using a higher digitisation rate
If we think of the wave as containing a range of frequencies, then the Nyquist-Shannon Sampling Theorem tells us that we need a sample rate which is at least twice the highest frequency present in the signal.
For a pulse of, say 5 megahertz (MHz) ultrasound, this does not mean that we need to sample at 10 MHz because the pulse contains a range of high frequency components. In practice, it is often necessary to sample at about 10 times the basic frequency, e.g. 50 MHz sampling for a 5 MHz pulse.
How many grey levels should be used in US?
we do not want to lose any diagnostic information. The number of available grey levels is related to the contrast resolution of the system. In other words, if two echoes have similar but different amplitudes, we need them to have different grey levels.
Using this criterion, we shall see later that we need a very large number, possibly tens of thousands.
On the other hand, we are representing this information as grey levels on a viewing monitor of some kind. How many levels can the human eye and brain combination discern as being different? The answer to this is probably in the range 16 to 32.
WHat does the signal processor do in US?
In a modern B-mode ultrasound scanner, the signal processor is normally no more than sophisticated and fast software.
The processor has two main functions:
Apply time gain compensation
Compression
What is the upper and lower limit of US Dynamic range determined by?
Upper - saturation. n other words, a signal of a certain amplitude will cause the spot to be as white as it can be. Under these circumstances, an echo arriving with an amplitude greater than that will be assigned to the same level (peak white).
Lower - Monitor black level and Noise. The electronics may set a threshold at which all incoming echoes are set to black (lowest grey level). Alternatively, noise will set a fundamental limit. If an incoming signal is below the noise level it will not be detected.
What is the typical max dynamic range of an ultrasound system?
100 000 = 10^5 = 100 dB.
In order to express such a large value of DR, we often use a logarithmic scale called the decibel (dB) scale.
How does the max theoretical DR compare to DR of the monitor?
If we compare the DR of 100 dB for the incoming echoes with the capability of typical displays, we can see a problem.
Most viewing monitors have a DR of roughly 30-35 dB. Even if this were not the case, the human eye and brain in combination have a similar DR.
Why is image compression needed in US?
the DR of the echoes is well in excess of the DR of the viewing system. Therefore, some form of compression is required; this is done using relative echo amplitudes expressed in dB and greyscale transfer functions (curves). You might like to consider the role of ‘windowing’ in CT for example.
WHat is the problem with a linear transfer function in uS?
Most of the grey levels are assigned to small echoes, with the big echoes sharing a relatively small number of levels.
What is the image store in US?
We can consider the image store as a large matrix, of perhaps 500 x 500 cells, each of which forms a pixel (picture element) of the image.
Each echo signal arrives with a grey level assigned to it and this is the z or brightness value, which will be stored in the appropriate cell
How does the scanner calculate the x and y values?
the y value is simply the depth in tissue from which the echo originated. We know from the pulse-echo principle that this is calculated by measuring the time taken for the echo to be received and assuming a value for the speed of sound.
To find the x value we need to know which part of the transducer array was fired to send out the particular pulse that created the echo. We assume that sound travels in a straight line and hence all the returning echoes from a specific transmitted pulse are assumed to come from a vertical column of tissue underneath that transducer position. Therefore, for a given pulse, all the echoes will have the same x value.
The important key points for pre- and post-processing are:?
Pre-processing is part of the preset function of the scanner which allows for optimisation from the outset
Anything added or removed in pre-processing will not be retrievable later. If the TGC is wrongly set, the only way of recovering from this is to re-scan the section with the correct setting
Post-processing is non-destructive. This potentially allows for considerable experimentation
Some processes can be implemented either as pre or post-processing. There are pros and cons but it is important to know which is which. A good example is the ‘zoom’ facility which is available on many scanners
Why is optimising in pre-processing better than post?
One important aspect of the pre-/post-processing features is the way in which the displayed depth scale can be changed.
In pre-processing, this is normally done by altering the depth or scale in the image acquisition setting and then re-scanning the section(s) of interest. This will optimise the image quality because other image processing functions such as focus and TGC will follow the selection. In particular, the digital image memory will alter the relationship between pixel size and real dimensions.
For example, if the image is stored as a 512 x 512 grid and the depth selected is 120 mm, then each pixel will correspond to a depth of 120/512 = 0.23 mm in real tissue. If the depth is changed to 40 mm, for example, then each pixel will represent just 0.08 mm. The sharpness of the display will be improved.
If an image has its scale altered in post processing, this change in pixel size is not possible. It is often possible to ‘zoom’ in on a frozen image but this will not, in general, add new information. In the above example, the zooming of the 120 mm image by a factor of two will not alter the actual tissue size per pixel. In fact, if zooming is taken too far, the image will become rather ‘blocky’ as its true digital nature is revealed and this will make it more difficult to view or interpret.
In a B-mode ultrasound scanner, signal compression is applied to reduce:
A. Tissue depth
B. Dynamic range
C. Transducer thickness
D. Scattered ultrasound
E. Computer memory
The correct answer is B.
Signal compression is used to match the dynamic range of the signals to that of the image viewing system.
How does blood flow in a vessel?
Blood doesn’t move at a uniform velocity in a vessel. It is slow near the vessel wall and faster in the centre.
What does the doppler US signal measure?
The Doppler ultrasound signal is a measure of relative motion of moving scatterers (red blood cells) in the direction of the ultrasound beam. If ultrasound is scattered from stationary tissue, the frequency back towards the transducer is the same as the transmitted frequency. If the scatterers are moving, they receive a different frequency and re-transmit a different frequency by virtue of their motion
What is the equation for the doppler shift?
(2 x V x Ft x cosθ)/c
where:
V is the velocity of moving blood in the vessel
θ is the angle between the axis of the blood vessel and the ultrasound beam
c is the speed of sound in blood.
and transmitted frequency (Ft)
WHat increases doppler frequency?
If the speed of the moving blood is high
If the ultrasound beam is more aligned to the flow direction (i.e. θ is small)
If the transmit frequency of the ultrasound probe is increased
What are higher frequency probes more sensitive to in doppler?
Higher frequency probes are more sensitive to low flow velocities.
However, tissue attenuation is higher at high frequencies and so probe and frequency choice is usually dictated by the depth of blood vessels under investigation.
What are the pros and cons of continuous wave doppler?
Continuous wave Doppler devices are inexpensive, straightforward to use and sensitive to flow.
Their major disadvantage is that they insonate all vessels in the beam path (until the beam is attenuated). Because arteries and veins usually lie in close proximity, the CW Doppler audio and sonogram outputs often combine arterial and venous signals. This means that CW Doppler cannot be used to locate the flow information at a precise depth. As a result, for example, the technique cannot be used for colour flow imaging.
What is a sonogram?
display of the changing distribution of Doppler frequencies with time produced by analysing the distribution of frequencies in successive small time intervals (each of a few milliseconds (ms) duration) and displaying the results graphically
What are waveform indices?
non–dimensional numbers that provide a guide to differences in shape and which have proved useful in studying proximal and distal changes in vascular resistance within a sonogram.
The resistance (or resistive) index (RI) and pulsatility index (PI) are based on measurements of the outline of the sonogram
What are the units for blood flow and blood velocity?
velocity of blood; the units are usually cm s-1
Blood flow, on the other hand, is expressed in cm3 s-1.
Regarding Doppler ultrasound:
A. It is always best to use a high frequency transducer for Doppler ultrasound
B. Doppler frequencies typically lie in the range of 100 Hz and 8 kHz
C. In an artery, the Doppler frequencies at a beam/flow angle of 60° will be higher than at 30°
D. The Doppler spectrum shows velocities in vessels
A. False. Although high transmit frequencies give higher Doppler frequencies, lower frequencies may penetrate better and give an improved signal amplitude.
B. True. Doppler frequencies are in the audible range.
C. False. Doppler frequencies are higher the more aligned the ultrasound beam is to the direction of blood flow, i.e. the smaller the angle between the beam and the flow.
D. False. This is not necessarily the case. Doppler frequencies are related to blood velocities, but the sonogram can calculate velocities (using the Doppler equation) only if the beam/flow angle is determined. This is usually only possible in duplex ultrasound scanners and requires the operator to make the correct angle determination.
What is the difference between spectral and pulsed wave doppler?
Doppler ultrasound can be used to measure movement of blood along an ultrasound beam. The Doppler frequencies that result can be heard as an audio signal or analysed and displayed as a sonogram. This latter technique is known as spectral Doppler.
In an ultrasound scanner, Doppler ultrasound can be combined with brightness-mode (B-mode) imaging to display moving objects (usually blood cells) and to show the location of that movement.
In order to do this, the scanner uses pulsed wave (PW) Doppler ultrasound so that the exact position of the movement can be determined.
How is scatterer velocity deduced in pulsed wave doppler?
The change in phase as a fraction of 2π (one complete cycle) is equal to the interval between pulses (1/PRF) as a fraction of the ‘Doppler period’ (Fig 1). From this relationship, the Doppler frequency (Doppler shift) can be calculated as the inverse of the Doppler period; it is much lower than the ultrasound frequency.
The same quantity is equal to the distance moved by the scatterer in the direction of the ultrasound beam as a fraction of the ultrasound wavelength. This allows calculation of the component of the scatterer velocity in the direction of the beam (Vcosϴ) and hence the derivation of the equation for the Doppler shift, bearing in mind that the effect occurs twice (for the incident wave and the reflected wave).
Although calculated slightly differently, the Doppler equation for pulsed wave Doppler is the same as that for continuous wave Doppler.
How does PRF effect aliasing in doppler?
The correct use of PRF is essential in both spectral Doppler and CFI to control aliasing. In CFI, the PRF also affects colour sensitivity, the ultrasound scanner’s ability to assign a colour range to blood velocities.
The PRF determines the velocity scale, which is often just called the scale. This is the maximum velocity (in either direction) that can be measured without aliasing; the range of available colours is assigned to the range of measurable velocities from the most negative value in blue/green (usually away from the ultrasound beam) to the most positive value in red/yellow (usually towards the beam).
If the PRF/scale is too high, flows may not be detected. If the PRF is too low, aliasing may occur.
Why is frame rate reduced in CFI?
Colour flow imaging is the superimposition of a map of Doppler shifts onto a B-mode image.
Separate transmit pulses and processing are used for B-mode and colour flow. The need for two sets of pulses increases the time required to produce an image and frame rate is reduced as a consequence.
Colour flow imaging requires several pulses along each line to detect movement along the line and display it; correlation methods are used for velocity detection. This also reduces the frame rate.
How are the scan parameters different in CFI compared to B-mode?
The constraints of time mean that a lower line density across the image is used for CFI. Colour flow imaging also uses longer, narrowband pulses compared with B-mode. This results in axial and lateral resolution that is usually poorer than that for B-mode imaging.
how does a change in frequency effect CFI?
Transmit frequency: lower frequencies have better penetration but have decreased spatial resolution and may be less sensitive to low flows. Low transmitted ultrasound frequencies result in lower Doppler frequencies and so they are less prone to aliasing
What is persistence in CFI?
Persistence: averages successive colour frames to improve image quality
How can parallel vessles still be imaged with doppler?
An unsteered beam shows a confused colour pattern because the flow and beam directions are close to 90°.
By steering the linear array uniform flow is revealed
What effect does increasing the size of the colour box have?
Increasing the colour flow box size requires more time for the pulse-echoes to measure the Doppler shifts over a larger area. Frame rate may decrease as a consequence.
For large colour areas the line density may be reduced and the colour flow image may appear coarse
WHat is the ensemble length in CFI?
Colour flow images require several pulses for each line of the image to detect movement between pulses. The number of pulses for each line is known as the ensemble length. The quality of the colour information generally improves with increased ensemble length.
A. Reducing the PRF/scale increases sensitivity to low velocities
B. Increasing the colour box size reduces frame rate
C. Red always shows arterial flow
D. The colour image is dependent on beam alignment with flow
E. Decreasing transmit frequency reduces the risk of aliasing
F. Spatial resolution of the colour flow image is the same as B-mode
A. True. Reducing PRF allows more time between pulses for motion detection of low flows.
B. True. Increasing the colour box size requires more time for additional lines of colour. If no other parameters change then the frame rate decreases.
C. False. The colour hue is dependent on the direction towards or away from the transducer. Colours may be reversed at the discretion of the operator.
D. True. Doppler frequencies are higher and the colour signal shows higher velocities as the flow direction is more aligned to the beam.
E. True. By decreasing transmit frequencies, Doppler frequencies are reduced and are less prone to aliasing for a given PRF.
F. False. The narrower bandwidth/longer pulse length and poorer line density of CFI results in poorer axial and lateral resolution when compared with B-mode ultrasound.
What 5 steps should be taken to optimise CFI?
Select the appropriate applications/set-up key. This optimises parameters for specific examinations
Set power to within study limits. Adjust colour gain. Ensure focus is at the region of interest and adjust gain to optimise colour signal
Use probe positioning/beam steering to obtain a satisfactory beam/vessel angle
Adjust the PRF/scale to suit the flow conditions. Low PRFs are more sensitive to low flows/velocities but may produce aliasing. High PRFs reduce aliasing but are less sensitive to low velocities
Set the colour flow region to appropriate size. A smaller colour flow ‘box’ may lead to a better frame rate and better colour resolution/sensitivity
effective spectral Doppler interpretation is dependent on two major factors (golden rules):?
Obtaining a good sonogram: a good sonogram is obtained if the ultrasound beam is aligned to the flow direction. At angles approaching 90° the Doppler frequencies are low and the sonogram is confusing. Try to use an angle of 60° or less by moving the transducer around and checking the angle between the vessel direction and the beam as shown on the screen
Making an angle correction to measure velocities: To measure velocities you must ensure you make an angle correction to tell the scanner the direction of flow.
A. The quality of the sonogram depends on the alignment between ultrasound beam and flow direction
B. You must make an angle correction to measure velocities
C. You must make an angle correction to measure resistance or pulsatility index
D. Velocities are accurate as long as the beam/flow angle is kept to within 60°
E. To reduce aliasing, use a higher PRF scale, a lower transmit frequency or a higher beam/flow angle
A. True. Doppler frequencies are higher and the sonogram shows higher velocities/frequencies as the flow direction is more aligned to the beam.
B. True. The scanner does not know the direction of flow, the operator must make an angle correction to convert the Doppler frequencies obtained to true velocities.
C. False. Resistance index and pulsatility index are non-dimensional indices of flow waveform shape and do not rely on measurement of absolute velocities.
D. False. There is no cut-off to ensure accurate velocities. The accuracy depends on the accuracy of the beam/flow angle that is set. There is a gradual increase in the consequence of errors which becomes more severe at higher angles. 60° is a pragmatic cut-off balancing what is attainable against possible errors.
E. True. All these reduce the effect of aliasing. By increasing scale, the PRF is increased for better sampling of high Doppler frequencies. Reducing transmit frequency (possible in some probes) or increasing beam/flow angle reduces the Doppler frequency.
What false assumptions can lead to artefacts in ultrasound?
The speed of sound is constant
Sound always travels in a straight line along the axis of the beam
The beam has negligible thickness
The rate of attenuation in tissue is constant
Why does acoustic shadowing occur?
Acoustic shadowing occurs when the proportion of sound energy transmitted beyond a specific target is too small to produce detectable echo signals and thus contributes no further useful information to the image. This occurs when the target attenuates a high proportion of the ultrasound beam energy through reflection or absorption.
Why is shadowing a significant limiting factor in a number of ultrasound examinations?
Shadowing from overlying bone or bowel gas will often mean that a flexible scanning technique is required to scan around such obstacles.
The patient may be fasted to reduce bowel gas and organs such as the liver or a full urinary bladder are used as acoustic windows. However, it is not always possible to achieve adequate images due to these limitations.
Why may small calcified structures not cause a shadow?
For very small calcified structures (Fig 1), even a very large change in impedance may not produce convincing shadowing.
Since the area of calcification is small compared to the effective width of the ultrasound beam, not all of the sound energy will be reflected. Sound energy will pass to each side of the calcification and continue into the deeper tissues.
As the scanner assumes all echoes return from structures along the central axis of the beam, echo information obtained from tissue adjacent to the area of calcification will register in the same on-axis location. If the calcification is sufficiently small, it will not be seen within the final image due to volume averaging.
What is critical angle shadow?
Interfaces that are at a steep angle relative to the direction of the ultrasound beam, such as the lateral margins of a focal lesion, can therefore be difficult to demonstrate.
In addition, where there is a change in the speed of sound across the boundary, as the obliquity of the interface increases, Snell’s law predicts that a critical angle is reached at which the transmitted beam is directed along the interface. Beyond this angle, almost total beam reflection occurs. As no sound energy is transmitted beyond the interface, this results in an area of shadowing where no distal information is obtained.
Critical angle shadowing is characteristic at the lateral margins of a well-defined structure such as the lower pole of the kidney
What is increased through transmission?
As sound travels through solid tissue, it is attenuated as a result of absorption and scattering. Time gain compensation (TGC) is used to compensate for this attenuation by gradually increasing the amplification of echoes with increasing depth. Attenuation within a fluid-filled structure, such as a simple serous cyst, is minimal. Since the scanner is calibrated to assume uniform attenuation throughout the body, there will be an area of ‘over-compensation’ distal to such a lesion.
Echoes arising from tissues within this region are thus over-amplified
WHat is image compounding?
Compound imaging uses electronic beam steering to fire pulses from between three and nine separate transmit angles. Real-time imaging is then achieved by combining the data from multiple lines of sight to generate successive frames.
By scanning from a number of different angles, artefacts such as speckle, clutter, noise and shadowing are reduced and real structures are reinforced.
What are the disadvantages of image compunding?
The key drawback of this technique is that the system is unable to distinguish between artefacts that reduce image quality and ‘helpful’ artefacts that may alert the operator to the presence of pathology or help to characterise a lesion. Compound imaging at a number of different angles effectively removes or reduces the appearance of all shadowing.
To demonstrate shadowing behind small structures such as renal stones, compound imaging should be switched off
What is the mirror image artefact?
Specular reflection occurs at large, smooth interfaces. If the change of impedance is high (for example, at an air-soft tissue boundary) the interface can act as an acoustic mirror.
Mirror imaging works as follows:
A pulse of sound travels through the liver, bounces off the highly reflective lung interface and is directed back into the liver at an equal and opposite angle
Echoes from structures within the liver are subsequently reflected back to the diaphragm and on to the transducer
The equipment assumes a single, straight line journey, and will therefore place these echoes above the diaphragm. A mirror image of the liver is thus seen above the diaphragm
What is the slice thickness artefact?
Focusing of the beam in this ‘short axis’ plane (Fig 1) is achieved by use of a mechanical lens attached to the face of the transducer and is normally limited to one specific depth. The operator has no control of this and simply needs to be aware of resultant image appearances. The scattered echoes from targets lying at the same depth within this slice will be averaged, stored and displayed as a single pixel value within the image.
volume averaging results in the appearance of low-level echoes
When do grating lobe artefacts occur?
Grating lobe artefacts occur when a cuff of low-level intensity exists in the intensity profile of the ultrasound beam.
Why can the use of harmonic imaging reduce grating lobe artefact appearances significantly?
The use of tissue harmonic imaging (THI) reduces clutter within an image. The harmonic signal is generated by the high intensity central beam This improves lateral resolution and grating lobes are effectively eliminated.
When do reverberation artefacts occur?
Reverberation occurs where the sound pulse bounces back and forth between two highly reflective interfaces.
The equal time interval between returning reverberant signals results in a series of equally spaced echoes distal to the deepest reflecting interface
What are comet tail artefacts?
seen where reverberation occurs within a small, but highly reflective object. This short path reverberation produces a series of closely spaced echoes giving a characteristic banded appearance.
What are ring-down artefacts?
a small gas bubble resonates within the ultrasound field. This generates a continuous emission of ultrasound. The result is a streak of high-intensity echoes seen posterior to the location of the bubble. This can be distinguished from the comet-tail artefact by the absence of a banded appearance.
Why do refraction artefacts occur?
Refraction occurs when the ultrasound beam crosses a boundary between tissues that have different speeds of sound propagation; the term describes the phenomenon whereby the sound beam ‘bends’ from its original path.
The appearances generated within the image can be confusing. A classic example of this is the ‘disappearing twin’ artefact.
Echoes generated from the refracted beam as it continues its journey beyond the boundary will therefore be misplaced in the image. The equipment will assume that these echoes originated along the original beam pathway - echoes are therefore picked up (and displayed in the images) from ‘off-axis’ structures.
