Ultrasound

Describe the main method ultrasound images are created and how the depth and brightness/echogencity of an object is determined.

Pulse-echo mode is the main method ultrasound images are created:

• Piezoelectric effect = Burst of RF signal / voltage applied to the transmitter’s ceramic crystals – causes crystals to expand or contract depending on the polarity of the voltage. The crystal then resonates, converting the electricity to an ultrasound beam (spatial pulse length = number of pulses x wavelength of pulses). The frequency of the ultrasound waves is predetermined by the crystals and crystal thickness in the transducer
• Ultrasound waves are produced in pulses because the same crystals are used to generate and receive sound waves (can’t do both at the same time).
  • transducers generally emit ultrasound only 1% of the time, the rest of the time is spent receiving the returning echoes
• In the time between pulses, the ultrasound beam enters the patient and is reflected back to the transducer (echoes) which cause the crystals to deform again and produce an electrical signal that is converted to an image on the monitor
  • The time it takes for the echo to return (sound propagation time) is used to calculate the DEPTH of the object (longer time for echo means deeper object as the ultrasound travels a longer distance) (velocity m/sec) = frequency (cycles/sec) x wavelength (m)
  • The strength of the echo highlights the BRIGHTNESS/ECHOGENICITY of the object (stronger pulses = brighter = higher echogenicity = higher intensity of reflection)
• The ultrasound machine repeats this for the adjacent transducer elements and so on until all elements are done and hence the whole image is formed.

Describe with examples the two main ways an ultrasound wave interacts with tissue?

Reflection and attenuation are two main ways the ultrasound wave interacts with tissue

1. Reflection – ultrasound waves are bounced back to the transducer for image generation. The proportion of ultrasound beam reflected is determined by the DIFFERENCE in acoustic impedance between adjacent structure. Acoustic impedance = tissues’ density x velocity of the sound waves passing through it – hence denser the tissue, the greater the acoustic impedance. Large differences in density and sound velocity between air, bone and soft tissue create a large difference in acoustic impedance, causing almost all of the sound waves to be reflected at soft tissue-bone and soft tissue-air interfaces. On the other hand, because there is little difference in acoustic impedance between soft tissue structures, relatively few echoes are reflected the transducer from these areas.
2. Attenuation refers to the gradual weakening of the ultrasound beam as it passes through tissue. Attenuation can be caused by scattering, absorption or refraction of the sound waves and is compensated by specific controls
   • Scattering – refers to the redirection of ultrasound waves as they interact with small, rough, or uneven structures. This tissue interaction occurs in the parenchyma of organs, where there is little difference in acoustic impedance and is responsible for producing the texture of the organ seen on the monitor. Scattering increases with higher-frequency transducers, thus providing better detail or resolution.
   • Absorption – occurs when the energy of the ultrasound beam is converted to heat. This occurs at the molecular level as the beam passes through the tissues.
   • Refraction – occurs when the ultrasound beam hits a structure at an oblique angle. The change in tissue density produces a change in velocity, and this change in velocity causes the beam to bend, or refract. This type of tissue interaction can also cause artefacts that need to be recognized by the sonographer

Describe five assumptions the algorithm uses to form a real-time ultrasound image about how sound travels through and interacts with tissue and for two explain situations where the assumptions are not

1. Ultrasound travels at a constant speed of sound in the body 1540m/sec (this is why artefacts happen through fluid as ultrasound travels faster in fluid and in bone 4080m/sec)
2. Ultrasound beam travel in a straight line

(however, refraction occurs at an interface between tissues of differing velocities) 

3. All echoes detected by the transducer originate from the central axis of the beam
4. The time taken for an echo from a given interface to the transducer is directly related to its distance from the transducer
5. The rate of attenuation of the beam is constant with depth and throughout the field of view

Describe five artefacts that occur because one or more of the assumptions about how sound travels and interacts with tissue is not true.

Velocity artefacts occur because some tissues have velocities different from the assumed 1540m/sec which results in incorrect placement of echoes in the display: inaccurate depth calculation effects + refraction effects

• Low-velocity lesion within the liver compared to the surrounding liver, causes echoes to take longer to return to the transducer, therefore the structures deeper to the lesion are displayed erroneously deeper in the image
• Refraction effects such as “lens effect” caused by well-developed rectus abdominus muscles, results in two images of the same object displayed side by side in the transverse plane (eg. aorta, early gestation sac) causes displacement of the diaphragmatic echoes, deep the lesion

Mirror image – structures displayed twice, one being a mirror of the other. This occurs when there is a strong specular reflector that acts as a mirror.

• Occurs diaphragm/lung interface (show mirroring of the liver texture, above the diaphragm)

Multipath – placement of echoes in the display assumes that echoes return directly to the transducer after reflection, but this is not always the case as echoes can be further reflected and refracted. The component of the echo that is reflected can then reflect a second interface before returning which results in additional path length and time. This delayed echo is therefore displayed deeper than the first, along the same line of sight

Reverberation artefact – repeated reflection of sound between two interfaces with marked differences in acoustic impedance (lots of reflection) – evenly spaced lines such as deep to bowel gas, fat and muscle layers, comet-tail artefact

Posterior acoustic enhancement – increased brightness deep to a lesion with low attenuation, so all the returning echoes deep to it have a stronger signal

Explain the principles of and reason for Time-gain compensation

Time Gain Compensation (TGC) increases signal gain as the echo return time increase – ie. it compensates for increased attenuation with image depth

• Ultrasound echoes are more attenuated (weaker) from similar tissue further away from the probe as they travel through greater tissue compared to echoes from more proximal areas (ie tissue from 1cm travel 2cm versus 2cm away travel 4cm )
• Uncorrected echo data would show distant echoes much weaker then superficial echoes
• Makes similar tissues have the same brightness regardless of depth on the resultant ultrasound image
• Automatic + manual TGC controls

Explain the principles of Dynamic range compression in ultrasound. 

Dynamic range compression

• Dynamic range refers to the range of echo strengths used to create the image
• It may be as much as 60dB – ie the strongest echo has an intensity 1000000 times greater than the weakest echo
• The display and human eye can only deal with a dynamic range of 30dB (ratio of 1000:1)
• Dynamic range compression compresses the dynamic range of the echo signal (60dB) to fit what can be displayed (~30dB) / video monitors (20dB)
• Generally stronger echoes are compressed more than weaker echoes
  • Variations in the strength of strong echoes are usually not significant – often edges of interfaces
  • Variations in the strength of weaker echoes from soft tissue are more significant – to help find lesions
• Compression uses logarithmic amplification – to increase the smallest echo amplitudes and decrease the largest amplitudes

Compare the spatial resolution between axial and lateral spatial resolution.

• Axial resolution has 4x better than lateral resolution

Define and discuss the factors affecting the lateral and axial spatial resolution in real-time ultrasound imaging. 

Lateral resolution refers to the ability to resolve 2 adjacent objects perpendicular to the beam and improves with a narrow beamwidth and increases with the number of lines per frame

• Point objects within the beam are averaged over the effective beam diameter
• Best lateral resolution occurs at the focal zone/distance
• Lateral resolution worsens after the focal zone away from the transducer
• Focused transducers produce the focal zone by beam steering and transmit/receive focusing

Axial resolution refers to the ability to separate two objects parallel to the beam. This improves with shorter pulse length (most important)

• Spatial pulse length = number of cycles emitted per pulse (usually 3 cycles) x wavelength
• Axial resolution = spatial pulse length / 2

• Velocity through tissue = 1540m/sec = frequency x wavelength
• If frequency is 5MHz
• 1540 x 10^3 mm/sec = 5 MHz x wave length
• wavelength = speed of through tissue ultrasound / frequency
• 31mm = 1540 x 10^3 mm/sec / 5 x 10^6 Hz
• wavelength = 0.31mm
• SPL = 3 x 0.31mm = 0.93mm!

• Axial resolution = SPL / 2 = 0.47 mm!
• Shorter pulse length achieved by:
  • dampening the transducer element (usually 3 cycles!)
  • increasing frequency (reduces penetration)
• 2MHz ~ axial resolution is 1mm (ie. half of the pulse length of ~2mm)
• 4MHz ~ axial resolution is 0.5mm (ie half of pulse length of 1mm)
• axial resolution same at all depths
• Breast ultrasound high frequency 8-10MHz – axial resolution 0.23mm

Describe two ways in which the axial resolution can be improved.

As axial resolution = SPL/2 and SPL = cycle number x wavelength , improve axial resolution by:

• Reducing wavelength by increasing the frequency
  • But imaging depth reduced, better axial resolution with the shallow probes vs deep abdominal probes
• Reduce cycles per pulse (often 3 so reduce to 2 cycles per pulse by increasing the backing /damping material)
  • reduce cycles per pulse by increasing dampening of transducer element and low Q

Lateral resolution in real-time ultrasound imaging can be improved over a range of depths by using multiple focal zones. If 2 focal zones are set and the line density and maximum depth of imaging are unchanged, explain what will happen to the frame rate and why.

Using 2 focal zones will reduce the frame rate

• Focusing to each focal zone is achieved by beam steering and transmit/receive focusing
• Each focal zone needs a separate pulse-echo sequence to acquire data
• Acquire data along one beam direction according to the number of focal zones by only accepting the echoes with each focal zone to build a single line of in-focus zones by meshing the information together

Briefly explain the principles of operation behind Duplex scanning.  

Duplex scan = real-time 2D B-mode imaging + Pulsed doppler

• Electronic array transducers which switch between a group of transducers used to create a B-mode image and one or more transducers for the Doppler information
• Duplex system allows for the calculation of velocity as the doppler angle is able to be estimated
• Doppler and 2D B-mode must be obtained separately – decreases temporal resolution
• Allows real-time adjustment of the cursor while observing the cursor

B-mode (Brightness)

• The brightness of each dot is proportional to the echo signal amplitude
• Electronic conversation of A mode (amplitude) into bright-modulated dots along the A-line trajectory
• Need the real-time B mode image to visually guide the placement of the “doppler gate” over the vessel of interest

Describe the basic differences in operation between pulsed and continuous wave Doppler units.

Continuous-wave Doppler

• Two transducers continuously work: transmitting and detecting
• Oscillator produces a resonant frequency to drive the transmits transducer and provides same frequency signal to the demodulator which compares the returning frequency to the incident frequency
• Amplifies the returning signal and extracts residual information containing Doppler shift frequency by using a “low-pass” filter that removes the high-frequency oscillations
• Wall filter removes very low-frequency signals from vessel walls and other moving specular reflection
• Audio amplifier amplifies Doppler signal to audible sound level

 Pulsed wave Doppler

• Transmitting pulse wave sent out, wait for echoes in the time gate return before sending another pulse
• Combines the velocity determination of continuous-wave Doppler + range discrimination of pulse-echo imaging
• SPL is longer (5-25 cycles per pulse) to provide a higher Q factor and improve the accuracy of frequency shift (reduced axial resolution)

• Isolation of a selected area is achieved by gating the time of echo return and analysing only those echoes that fall within the time window

Discuss the pros and cons of Continuous Wave Doppler

Pros:

• High accuracy of high velocity due to narrow frequency bandwidth,
• No aliasing for high velocity
• Simplest and least expensive
• High sample volume
• Better sensitivity

Cons:

• No depth resolution – multiple overlying vessels will cause superimposition and make it difficult to distinguish a specific Doppler signal
• Poor range resolution

Discuss the pros and cons of Pulsed Wave Doppler. 

Pros:

• Better accuracy of frequency shift (but not at high frequency)
• Higher Q factor (AS LONGER SPL)
• Allows depth selection due to time gate circuit to reject echoes falling outside the gate window

Cons:

• Lower axial resolution – due to increased SPL (5-25 pulses/cycle)
• Limit on maximum velocity (PRF (pulse repetition frequency = sampling frequency) must be AT LEAST twice max Doppler frequency shift otherwise aliasing occurs)
• Small sample volume
• Low sensitivity

Define lateral spatial resolution and why would focusing the scan plane at the level of interest improve spatial resolution?

• Lateral spatial resolution is the ability to distinguish two separate points in a direction perpendicular to the direction of the ultrasound beam
• Lateral spatial resolution is directly proportional to the beam width
• Focusing the beam in the scan plane at the depth means the beam width is the narrower there and hence has the best lateral spatial resolution

Describe the two components of electronic (phased) focusing for a linear array transducer.

• Beam focusing refers to creating a narrow point in the cross-section of the ultrasound beam called the focal point.
• This is achieved using electronic (phased) focusing which involves transmit focusing (electronic focusing) and dynamic receive focusing. No mechanical focusing is used. 
• Transmit focusing (Electronic focusing):
  • This is achieved by applying progressively delayed electrical excitation pulses to individual array piezoelectric elements
  • Transmit focusing is achieved by firing the outer elements first and applying progressively increasing delays toward the centre of the array, with the central elements fired last.
  • These delays create a curved (parabolic) wavefront 
  • Wavefront converges at a focal point, improving lateral resolution
  • Focal depth can be varied electronically, which determines the time delay between pulses
    • Greater focal depths are achieved by reducing the difference in the time delay between the elements result in more beam divergence and greater depths
    • Shallower focal depths increase the difference in the time delay between the elements
  • Multiple focal zones can be used (multi-zone transmit focusing)

• Dynamic receive focusing:
  • On receive, the system applies dynamic time delays to echoes arriving at each element 
  • Echoes received at the outermost elements of the array travel a longer distance and hence arriver later than those at the centre of the array
  • Re-phasing is needed to prevent a loss of resolution 
  • Dynamic receiving focusing re-phases the signal by introducing electronic delays as a function of depth
  • A smaller time delay is needed for echoes returning from a greater depth and a larger time delay is needed for echoes returning from a shallower depth, so all the echoes can be analysed together
  • This creates continuous focusing at all depths during echo reception
  • Dynamic = focus changes continually with depth
  • Improves lateral resolution at greater depths

Both linear and phased arrays use electronic focusing. Describe one main difference in terms of the number of elements used for each beam line. 

• Linear array only uses/fires a subset (“sub-aperture”) of contiguous elements to generate each beam, producing a beam perpendicular to the array face with curved delayed patterns, so focusing is performed without beam steering (unless special linear-steered modes used)

• Phased array uses all (or nearly all) elements for every beam, and focusing is combined with beam steering: a linear delayed steers the beam and a curved delay profile superimposed  on this provides transmit focus. Phased arrays allow wide sector scanning. 

Define pulse repetition frequency. Explain why a lower pulse repetition frequency used at a maximum depth of 10cm compared to imaging with a maximum depth of 5cm.

Pulse repetition frequency is the number of pulses per second released from the ultrasound probe. It is not related to frequency. It is determined by depth:

• Shallow image = high PRF
• Deep image = low PRF

An ultrasound probe must wait until the echoes return from a pulse before sending another pulse.

Ultrasound of an object at 10cm would result in the ultrasound signal and returning echo travelling twice the distance and hence, take twice as long compared to an object at 5m depth. Hence the pulse repetition period must be longer for deeper lesions in order to avoid artefacts.

Explain why ultrasound probes used for abdominal imaging have a frequency range of 2-5MHz but the probes used for superficial skin have a frequency range of 5-15MHz.

The main reason is due to the principle of how ultrasound works: the emitted ultrasound pulse needs to return before sending out the next ultrasound pulse in order to reduce artefact.

Wave length x frequency = velocity of ultrasound (approx 1540m/sec in tissue).

Abdominal Ultrasound probes are lower frequencies (2-5mHz) in order to have longer wavelengths, which are less likely to be attenuated by tissue and hence more able to penetrate the body’s tissues and image deeper structures (abdomen approx 20cm thick). The probe is also a curvilinear array to allow a large FOV of the abdomen – trapezoidal format with a wide FOV. 

Superficial ultrasound probes have higher frequencies (5-15mHz) to have shorter wavelengths which are more likely to be attenuated and hence only used to image superficial structures (arm is much thinner 10cm), often needs 4cm penetration. The higher frequency also improves spatial resolution.

Explain the Doppler Frequency Shift

Doppler Frequency Shift refers to the difference between the incident frequency and the reflected frequency.

• RBC/reflector is moving toward the transducer produces higher frequency echoes
• RBC/reflector moving away from the transducer produces lower frequency echoes

During Colour Doppler scan of a carotid artery the entire artery is red (blood flowing away) however you notice in the centre there is blue (suggesting blood coming toward the probe). Assuming the flow is not turbulent, explain why this has occurred.  

The blue in the artery (which is otherwise red) is due to aliasing that is seen with pulsed Doppler whereby the Pulsed Repetitive Frequency (sampling frequency) is less than double the maximal Doppler frequency shift. For example, a 1.6kHz doppler shift needs a minimum PRF of 2×1.6kHz = 3.2kHz

In aliasing, higher frequencies in the spectrum will be misinterpreted as lower frequency signals with a 180-degree phase shift, such that the highest blood velocities in the centre of the vessel are measured as having reverse flow (and hence blue)

How would you reduce aliasing artefact during a Colour Doppler?

• Drop the baseline
• Increase the available velocity range
• Increase the Doppler frequency shift by using a lower isolating frequency
• Increase the Doppler angle (eg 60 degrees reduces the Doppler shift)

Define Mechanical Index and why it is important to monitor during an ultrasound.

The mechanical index is a measure of the relative risk of inducing cavitation and is based on a formula (MI = max peak rarefaction pressure after correction for attenuation divided by the square root of frequency).

• Small volumes of gas that can coalesce to form cavitation nuclei when exposed to ultrasound
• Ultrasound can produce inertial cavitation in which the rarefractional phase of the pressure wave EXPANDS the bubble to greater than its maximum stable volume, resulting in a sudden collapse of the bubble which produces local heating 1000 to 10000^oC

There is a higher risk of MI at lower frequencies (as wavelengths are longer and hence can expand the bubble larger)

Important to monitor Mechanical index where gaseous bodies are expected within the beam paths (chest and abdominal scanning especially in neonatal) (not as important during obstetric scans as no gas in the fetus!)

MI < 1.9 considered safe

MI can be breached during the use of spectral Doppler (fetal echo, tricuspid regurgitation or ductus venous flow characteristics, petal echocardiography) in paediatric abdominal and chest scanning.

Define Thermal Index of Bone.

Thermal Index Bone = Thermal index of bone at the focus = an indicator of the temperature increase at soft tissue bone interfaces. A TIB of 0.8 means that the temperature would increase by 0.8^oC at soft-tissue bone interfaces.

Define Thermal Index and the safety cut-offs for adults, prenatal and post-natal.

TI = thermal index = relevant acoustic power at the depth of interest divided by the estimated power necessary to raise the tissue equilibrium temperature by 1^oC.

It is a RELATIVE risk of heating, not an absolute scale of heating.

= device emitted acoustic power divided by the acoustic power that would increase the temperature by 1^oC. under conditions of minimum heat loss (without perfusion)

For adults:

TI < 1.5 – safe for extended period except for the eye

TI 1.5-4.0 – limit exposure time if possible

TI > 4 – UNSAFE for any length of time

For prenatal:

TI < 0.5  = unlimited time

TI 0.5-1  = <30 minutes scanning of prenatal

TI 2.5 = < 1 minute of scanning prenatal

TI > 3 – UNSAFE for any length of time

General rule: TI < 1.0 for first-trimester screening + exposure to little Doppler as possible, preferably less then 5-10minutes

For postnatal:

TI < 2 for extended periods of scanning

TI 2-6 – <30 minutes for postnatal

TI > 6 – < 1 minute for postnatal

For the three types of Thermal Index (TI), explain when they should be monitored during an ultrasound scan (ie. which patients and scanning body location).

• TI Soft tissue = TIs = 0-8 weeks gestation age + soft tissue scanning
• TI Bone = TIb = after 8 weeks gestational age + where bony-soft tissue interfaces are expected
• TI Cranium = TIc = where the bone is close to the transducer surface such as neonatal bone scanning

Explain what you would do if you saw the TIB was 4.0 during an ultrasound scan.

A TIB of 3 is above the recommended guidelines (TIB < 1.5 for adults, <0.5 for prenatal) – hence I would

• Minimise the scan to as short as possible
• Avoid scanning areas with bone-soft tissue interfaces
• Avoid the use of Colour and spectral Doppler (< 5-10 minutes recommended)
• Avoid harmonics
• Minimise the acoustic power (decrease frequency)

Explain why pulsed Doppler has the greatest likelihood for inducing thermal bioeffects in tissues.

Pulsed doppler has a higher energy acoustic output because of 2 reasons:

• High pulse repetition frequency PRF than B mode- The PRF must be at least twice the maximal Doppler frequency shift to avoid aliasing.
• Longer spatial pulse length (15mm, 5-25 cycles per pulse) for more accurate frequency shift- to provide a higher Q factor and improve the measurement accuracy of the frequency shift (although at the expense of axial resolution)

Depth selection in pulsed doppler is achieved with an electronic time gate circuit to reject all echo signals except those falling within the gate window, ad determined by the operator. Repeated echoes from the active gate are analysed and a Doppler signal is repeatedly built up.

Explain how harmonic ultrasound works:

Harmonic imaging provides better image quality as compared with conventional ultrasound technique as it utilizes the non-linear propagation of ultrasound through the body tissues. The shape of the ultrasound wave is distorted as the high-pressure portion of the wave travels faster than the low-pressure portion. This change in waveform leads to the generation of harmonics (multiples of the transmitted frequency) from a tissue. At present, the 2nd harmonic is being used to produce the image because the subsequent harmonics are of decreasing amplitude and insufficient to generate a proper image. 

Explain the advantages of harmonic ultrasound over conventional ultrasound

1. Improved axial and lateral resolution
2. Increased signal to noise ratio.
3. Decreased artefact: side lobe and reverberation.
4. improved resolution with large body habitus