Final Practice exam

Question 1

(a) On a typical spectral Doppler pulse as a function of time, (i) what is a typical number of cycles in this type of pulse?

Answer 1

(i) The Doppler pulse should have 10 to 20 fairly uniform cycles. If the bandwidth is too large the frequency shift is less easy to determine in the Doppler ultrasound electronic circuitry. To solve for this in pulse Doppler diagnostic ultrasound, the SPL is increased to about 10 to 20 wavelengths with a resulting significant decrease in the bandwidth, and unfortunately some loss in axial resolution that goes with the larger value of the spatial pulse length. Axial resolution = SPL/2

Question 2

On a typical spectral Doppler pulse as a function of time, (ii) Describe the pulse duration as it would appear on the diagram of the pulse.

Answer 2

(ii) The pulse duration is the full length, in time, of the whole pulse, that is all 10 or 20 cycles. Starting at the start of the first cycle and ending at the end of the last cycle.

Question 3

On a typical spectral Doppler pulse as a function of time (iii) How does the number of cycles in a spectral Doppler pulse differ from a B-mode pulse.

Answer 3

(iii) A B-mode pulse has only 3 to 5 cycles compared with 10 to 20 in a Doppler pulse. The b-mode pulse is highly damped and thus has a large bandwidth increasing it’s spatial resolution. The Doppler pulse is less damped with a long SPL resulting in a more narrow bandwidth, that is a more narrow range of frequencies. This makes it easier for the machine to detect frequency shifts required for an accurate Doppler trace but comes at the cost of axial and thus spatial resolution as represented in the formula axial resolution = SPL/2.

Question 4

(b) Explain the factors that affect spatial pulse length and, in turn, how this affects spatial resolution.

Answer 4

The spatial pulse length is the product of the wavelength and the number of cycles. Thus, spatial pulse length is inversely proportional to the frequency.
So, increasing frequency will reduce the spatial pulse length which, in turn, improves spatial resolution as defined in the formula Axial resolution = SPL/2.
In reality, low frequencies are used in spectral Doppler ultrasound so that an image may be formed. As Doppler imaging focuses on the imaging of red blood cells that scatter the beam in all directions, only a fraction of the beam will echo back to the transducer. Low frequency transducers are used to lessen the attenuation of ultrasound into soft tissue and maximize the return echoes and this comes at the cost of axial and thus spatial resolution.
Similarly, the longer SPL is a product of a pulse that is less damped than that used in b mode imaging. This means there is a far narrower range of frequencies used. This makes it easier for the machine to determine the frequency shift caused by the Doppler effect but also comes at the loss of axial and thus spatial resolution.

Question 5

(c) (c) Discuss the concept of dynamic range in the following terms. An ultrasound signal processor has an output voltage range of 0.0 to 1.0 volts with a voltage precision of ΔV = 0.001 V. (i) State the first three and the last three voltage steps in this arrangement.

Answer 5

The signal processor output comprises of voltage pulses of height that provide the necessary information about the reflecting interfaces in the patient at given depths. If, after amplification, the peak voltage values are in the range 0.0 - 1.0 V, and a given voltage is precise to 0.001 V (1 mV) then The signal processor output comprises of voltage pulses of height that provide the necessary information about the reflecting interfaces in the patient at given depths. If, after amplification, the peak voltage values are in the range 0.0 - 1.0 V, and a given voltage is precise to 0.001 V (1 mV) then The first three voltage steps will be 0.000 V, 0.001 V, 0.002V, and the last three voltage steps will be 0.998 V, 0.999V and 1.000V

Question 6

(c) (c) Discuss the concept of dynamic range in the following terms. An ultrasound signal processor has an output voltage range of 0.0 to 1.0 volts with a voltage precision of ΔV = 0.001 V. (ii) State the number of discrete voltages that can be recognized by the system?

Answer 6

The signal processor output comprises of voltage pulses of height that provide the necessary information about the reflecting interfaces in the patient at given depths. If, after amplification, the peak voltage values are in the range 0.0 - 1.0 V, and a given voltage is precise to 0.001 V (1 mV) then there are a total of (ii) 1,001 discrete voltage values that the electronics can recognize.

Question 7

(c) (c) Discuss the concept of dynamic range in the following terms. An ultrasound signal processor has an output voltage range of 0.0 to 1.0 volts with a voltage precision of ΔV = 0.001 V. (iii) Give the defining equation for dynamic range.

Answer 7

(iii) The defining equation for dynamic range is DR = (Vmax -Vmin)/ ΔV and in this instance is (1000-0) / 0.001 = 1000/1

Question 8

(c) (c) Discuss the concept of dynamic range in the following terms. An ultrasound signal processor has an output voltage range of 0.0 to 1.0 volts with a voltage precision of ΔV = 0.001 V. (iv) Express the dynamic range in decibels (dB).

Answer 8

(iv) Another way to express this is to take the dynamic range ratio and turn this into a dB value. In this case we have the DR = 1,000:1, or
20 log10(1.000/0.001) = 20 log10(1,000) = 60 dB
This number of 60 dB is reasonable for an electronic amplifier. The reason that 20 log10 is used, and not 10 log10, is that these voltages are considered as magnitudes and not power (magnitude squared).

Question 9

(a) List the features found in a typical arterial spectral Doppler display

Answer 9

B-mode image
Scan line 
Sample volume
Doppler angle cursor
Spectral broadening 
Spectral display
Intrinsic spectral broadening
Spurious spectral broadening

Question 10

Give a full description of B-mode image

Answer 10

obtained by scanning the area of interest and acquiring an appropriate frame from which the spectral Doppler display will be acquired. This image allows the user to adjust the Doppler angle cursor and define the location and size of the sample volume

Question 11

Give a full description of scan line

Answer 11

The single scan line on the display down which the pulses are sent to acquire the echo spectrum

Question 12

Give a full description of Sample volume

Answer 12

The sample volume of a Doppler spectral display is determined by slice thickness, lateral resolution and range gate duration. It is generally shown by a square box placed within the blood vessel along the scan line. Only the Doppler shifted echoes from the soft tissues and blood within the sample volume contribute to the spectral display, all other echoes are ignored. Range gating is when the machine only detects return echoes for a short amount of time (usually one or two microseconds) and it determines the axial limit of the sample volume. Increasing the axial limit of the sample volume will increase the range gate and vice versa. The time elapsed between the transmission of the Doppler pulse and the range gate is such that signals received only come from the required sample volume depth and any echoes that arrive before or after the range gate are ignored by the machine

Question 13

Give a full description of spectral broadening

Answer 13

Spectral broadening is a Doppler artefact signified by a variance in the Doppler signal and it can be either intrinsic or spurious. Spectral broadening manifests as widening of the trace associated with lengthening of the vertical spectral lines and ‘filling’ of the spectral window that should otherwise be dark in appearance.

Question 14

Give a full description of intrinsic spectral broadening

Answer 14

caused by the way the machine processes the data it receives, that is, as if there were a single line of sight, a single point of origin and a single Doppler angle. As the Doppler signal received is on a significant area of the transducer, from multiple paths, all with different Doppler angles, there will be an intrinsic range of Doppler shifts in the overall Doppler signal.

Question 15

Give a full description of spurious spectral broadening

Answer 15

caused by: setting a large sample volume, a large Doppler angle or placing a sample volume too close to a vessel wall where blood flow varies in speed. It may also be real and caused by pathology. If vessel disease causes the velocity profile to change and blood flow becomes turbulent the spectrum will have a larger variance in Doppler

Question 16

Give a full description of the doppler angle cursor

Answer 16

After defining the sample volume the Doppler angle is calculated by manipulating the Doppler angle cursor to be aligned with the direction of blood flow. Because the machine already knows the direction of the Doppler beam return echo it can now calculate the angle between it and the direction of blood flow, also known as the Doppler angle.
If the Doppler angle has been correctly measured, Doppler shift can be calculated and an accurate blood velocity is acquired from the Doppler spectral display

Question 17

Give a full description of the spectral doppler display

Answer 17

Portrays a range of frequency shifts, or velocities, in centimetres per second (cm/s), on the vertical (y) axis, over time (s), on the horizontal (x) axis

Question 18

(b) Describe why ‘aliasing’ cannot occur in continuous wave Doppler imaging

Answer 18

There is a fixed limit on the range of Doppler shifts that can be correctly displayed and this is called the Nyquist limit and is equal to one half of the Doppler PRF. Aliasing is due to an ambiguity arising from the pulse repetition frequency, PRF, being set inappropriately. If the Doppler shifts exceed the limits of less than -1/2 PRF or greater than +1/2PRF the display is aliased and the Doppler information appears to suddenly change its direction and velocity.
This artefact can arise in all pulse (PW) Doppler modes, but manifests itself in different ways. Aliasing can occur in any system (electronic or otherwise) where the information is sampled; not received and processed continuously. Which is why aliasing cannot occur in CW Doppler as there is no pulses that are used to sample information.

Question 19

Why is this phenomenon called ‘aliasing’?

Answer 19

It is called aliasing because the true signal appears as something else, having a frequency lower than the true frequency hence the name ‘alias’.

Question 20

Apart from lack of aliasing, what is the most important advantage of CW imaging compared with pulsed imaging

Answer 20

The main advantage of CW Doppler over PW Doppler is that as there is no pulsing of ultrasound there is no restriction associated with a PRR, so in principle there is no restriction to the flow velocity that can be measured. This means that very high blood velocities can be easily seen and measured unlike spectral Doppler.

Question 21

What is the main effect of aliasing and explain why this is a problem in diagnostic imaging?

Answer 21

Aliasing has the “top” of the spectral trace chopped off and simply translated downwards (or upwards) to the other half of the display region. There is a clear discontinuity on the trace across the zero frequency/velocity axis which causes blood velocities to be misrepresented which is a big problem as the purpose of spectral Doppler imaging is to asses blood flow.

Question 22

(c) A colour Doppler examination is performed using a phased array transducer. Describe the shape of the display. Describe the appearance of a blood vessel on the display including appropriate colours for the flow direction in question. Make sure you label your diagram clearly. Describe the use of appropriate (red/blue) colours in the display.

Answer 22

The shape of a display for a phased array transducer will be sector-shaped with a pointed top.
In colour Doppler, Doppler shift information is colour coded and overlaid on the grey scale image. Colour is displayed only within the colour box and only where flowing blood is found. Standard grey scale information is displayed everywhere else. The shape of the box will be approximately trapezoidal with the top and bottom curved like the sector display.
A blood vessel within the colour box will be coloured depending on its shape and its orientation relative to the transducer. It will remain grey if there is no velocity flow component, red with flow toward the transducer and blue with flow away from the transducer. If a blood vessel is parallel to the phased array transducer face and blood flow is moving from right to left. The blood flowing toward the transducer, forming an acute Doppler angle with the transducer, will be represented as red. While the blood flowing away from the transducer, forming an obtuse Doppler angle, will be represented as blue.

Question 23

(a) Explain the appearance of mirror artefact in B-Mode Imaging, and why it occurs

Answer 23

In the presence of a strongly reflective interface, tissues can be imaged by reflected ultrasound. The transmitted pulse will hit the interface and be reflected off then hitting another structure. This structure also reflects echoes back to the highly reflective interface and then back to the probe. Since the machine assumes the ultrasound has travelled in a straight line, the tissues will be displayed behind the highly reflective interface and also in their true location creating a mirror image. Sometimes their true location is outside the scan plane so they will not be shown and only the mirror image displayed. Gas-filled regions provide perfect reflectors for this to occur. That is to say interfaces between two tissues with significantly different acoustic impendences such as the diaphragm which is the interface between the liver and the air-filled lungs.

Question 24

(a) Explain the appearance of reverberation artefact in B-Mode Imaging, and why it occurs

Answer 24

Reverberation artefacts cause tissue structures to be displayed multiple times at equally spaced depths in the image. This artefact arises as the ultrasound bounces (reflects) back and forth between the interface and the transducer face-plate with each repeated echo appearing twice, three times and so on the depth of the original echo. This only happens if the surfaces have large reflection coefficients. Some examples include transducer/skin, tissue/bone, tissue/calcified structure, tissue/fibrous structure, tissue/air-filled structure.
Ring down artifact – When the reflectors are small gas bubbles ultrasounds reflects between the multiple gas bubbles, creating a continuous series of echoes returning to the transducer. If the gas bubbles are in liquid there will be very little attenuation, so the echoes returning to the transducer remain strong and because the reflection of US by gas bubbles is highly efficient the result is a bright line of echoes deep to the bubbles.
Comet tail artefact – Similar to ring down artefact but shorter lived and so a relatively ‘short ‘tail is seen rather than the long bright line as seen in ring down artefact. It is generally caused by small calcifications. They are strong reflectors, like gas bubbles, and so the ultrasound reverberates between the calcifications producing a series of echo as with ring down artifact. The major difference, and what causes the comet tail appearance, is that reflection of US by calcium causes significant loss of energy and hence the reverberation of echoes quickly reduce in amplitude and the artifact fades from the image with depth.

Question 25

(a) Explain the appearance of side lobe artefact in B-Mode Imaging, and why it occurs

Answer 25

Not all energy goes into the main lobe of the beam. Side lobes are additional beams that occur either side of the desired ultrasound beam. They are generally weaker than the main beams by a factor of 100 (20dB) they are still capable of contributing echoes and can therefore cause artifacts. Because they carry far lower energy than the main beam, the artefacts often appear as ghosting “copies” of the main feature. Since a machine cannot know that the echo from the object comes from a sidelobe it displays it across the midline of the beam. As the beam is scanned the object will be displayed in its correct location by the main beam and also to the left and right of it’s true location by the sidelobes. This is referred to as side lobe artifact.

Question 26

(b) Give a full description of the Doppler angle and how is it used in spectral Doppler imaging.

Answer 26

The Doppler angle (θ) is the angle between the forward blood flow direction and the returning Doppler echo. If the Doppler angle has been correctly measured, Doppler shift can be calculated and an accurate blood velocity is acquired from the Doppler spectral display. The Doppler angle is routinely set less than 60° to limit the error in the blood velocity calculation, when the angle is 60° a 5° error will mean a 15% error when calculating blood velocity.
Setting a Doppler angle allows the component of blood flow velocity vector directly toward or away from the transducer to be converted into blood flow velocity in the direction of the vessel. It does not affect the Doppler shifts detected rather the calculation of blood velocity.

Question 27

(c) Define the temperature index, TI, and briefly outline its role in diagnostic ultrasound in terms of its purpose and how it is used in practical ultrasound scanning.

Answer 27

The estimated temperature increase is displayed by the ultrasound machine as the thermal index. It is essentially a measure of the likely maximum temperature rise in the patients tissues in degrees Celsius. The thermal index is defined as the ratio of the acoustic power used, W0, to that power required to raise the tissue temperature by 1 Celsius degree, W1°C. For example, of TI of 0.8 means that like likely temperature rise due to US exposure will be 0.8 degrees Celsius. Because there is a large variety of tissue absorption coefficients, there are 3 further thermal indices to indicate the tissue type/position:
- TIS - soft tissue index; only soft tissue is in the beam
- TIB - bone index; bone in the focal zone
- TIC - cranial bone index; scanning through superficial bone.
In practice, the thermal index provides guidance as to whether the amount of heating is within safe limits or whether it should be of concern. It is widely agreed that a temperature increase of up 1.5 degrees Celsius above normal body temperature (37) can be tolerated by a fetus without harm which is why some machines won’t display the TI unless it is greater than 1. The user does at times, need to take additional factors into account. If a patient has a fever of 37.5 then only a 1 degree increase in temperature would be acceptable.
If a machine does not display a thermal index it likely indicates the manufacturers have limited its output to levels that are considered completely safe.