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Flashcards in Module 3 questions Deck (21)
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1

What is meant by the following instrument/system terms: (ii) gain

(ii) Gain is the control that adjusts the electronic amplification of all the echo signals returned the scanner. All echo voltage signals are amplified by the same amount using the gain control.

2

What is meant by the following instrument/system terms: (i) power

Power refers to the system control that allows the user to change the intensity of the ultrasound beams transmitted into the patient. Changing the power control effectively changes the magnitude of the voltage pulses that are used to excite the transducer elements.

3

What is meant by the following instrument/system terms: (iii) dynamic range

(iii) Dynamic range refers to the number of distinguishable signal magnitudes of the echo voltages. This ultimately provides the determining factor for the contrast available in the image.

4

What is meant by the following instrument/system terms: (iv) time gain compensation

(iv)Time gain compensation is a set of slide controls (usually down the righthand side of the image screen) that allow the user to adjust the gain of the echo voltages (in addition to the gain control) to compensate for the weaker echoes from deeper tissue interfaces.

5

explain the processing that takes place to the voltage signal caused by the return echo at the face-plate to the point where a digital signal is generated to represent that echo strength.

• Pre-amplification in the transducer housing – a gain for all echo voltages, done virtually at the point of reception to minimise noise/interference.
• Amplification of echo voltages in the scanner system that involves both gain and time-gain compensation.
• Rectification and demodulation of the amplified echo voltages to produce a positive signal with the high frequencies removed.

6

Outline the purpose of the scan converter in an ultrasound scanning system. In your answer define what is meant by an analog and a digital signal, and give examples of each in the processing that takes place in such a system

The scan converter is the electronic system that takes the analog echo voltage signals, after amplification, and turns them into digital signals and then arranges these echo signals and other signals that relate to the echo scan line position, orientation and echo depth (all in digital form these days) so that they can be used by the video input to form the display image of pixels.

An analog signal/voltage is one that is continuous in voltage (vertically) and time (horizontally) and, in the context of diagnostic B-mode ultrasound, looks a bit like that shown in Figure 1.1 on page 112 for a given echo. The scanner electronics tunes the echo strength (essentially the area under the rectified signal) into a digital signal/voltage; this involves forming a rectangular voltage pulse where the height is proportional to the echo strength, and then coding this pulse height into, say one of 256 integer values in an 8-bit system, ready for further processing and display.

7

Clearly differentiate between a pixel and pixel grey value. Relate these two terms to information about the patient being scanned. What do we mean by an 8-bit imaging system, and how many unique values are thus involved?

A pixel is a 2-D image area that corresponds to an actual area within the patient. Thus the pixel dimensions are always with reference to the patient (in a diagnostic medical image), not to the display screen.
A pixel grey value is a single integer number that represents the strength of the patient information that is derived from that pixel area. In diagnostic ultrasound the pixel grey value represents echo strength (in B-mode). In an 8-bit imaging system the pixel grey values are represented by an 8-bit code/number. As there are a total of 2^8 such codes, and 2^8 = 256, these codes are usually represented by the integers 0 through 255.

8

Differentiate between the operations of read-zoom and write-zoom in ultrasound imaging. In your answer define pre- and post-processing in general and state which of the two operations refer to write-zoom

Read zoom is a magnification process where a smaller region of the original image is displayed over the same original frame field of view. This increases the size of the pixels (so there are fewer of them in the frame). This is an example of post-processing as no new echo data is processed. This may make a region of the image clearer but does not actually improve the spatial resolution.
Write zoom is a magnification process where the echo data for a smaller region of the original image is pre-processed so there are essentially the same number of pixels (the pixels are thus smaller than in the original image). This actually improves the spatial resolution.
So post-processing is a process that uses the original echo data and changes scale or grey level display (contrast) etc. Pre-processing is the action of taking echo data and re-using it within the scan converter – it is truly a new image.

9

Define the term spatial resolution, and for an ultrasound imaging system state what three components need to be considered for spatial resolution. In each instance state what determines the spatial resolution.

Spatial resolution defines the smallest, high contrast, feature that can be reliably imaged. In diagnostic ultrasound there are three components (it is a 3-D imaging situation). The three components are:
- Axial – determined by half the spatial pulse length: ½ SPL.
- Lateral – determined by the beam width: BW.
- Elevation plane – determined by the slice thickness: ST.
As a general rule, the best you can do with all three components is about 1 mm.

10

Explain the difference between contrast and spatial resolution, giving typical values for an aspect of an ultrasound image.

Spatial resolution defines the smallest, high contrast, feature that can be reliably imaged. In diagnostic ultrasound there are three components (it is a 3-D imaging situation). The three components are:
- Axial – determined by half the spatial pulse length: ½ SPL.
- Lateral – determined by the beam width: BW.
- Elevation plane – determined by the slice thickness: ST.
As a general rule, the best you can do with all three components is about 1 mm.

Contrast, or contrast resolution is the smallest change in echo strength that can be visualised for two broad adjacent regions of the patient in the scanned image. If contrast resolution is quantified it is normally expressed as a percentage of the grey level difference between the two regions divided by the grey level average, multiplied by 100.
As a general rule, in B-mode diagnostic ultrasound, you might expect the best spatial resolution to be about 1 mm, and the best contrast resolution to be about 5%.

11

List the assumptions/approximations that arise in B-mode diagnostic scanning that give rise to image artefacts. For each item in the list give one example of a specific artefact and how it is caused.

The assumptions are:
(i) The speed of ultrasound is constant at 1540 m/s
(ii) All incident beams and echoes are infinitely narrow and travel in a direction that coincides with the central ray
(iii) All incident beams and echoes travel in straight lines from the transducer face-plate to the interface/scatterer and back to the transducer
(iv) Attenuation of the beam is constant at, for example, 0.7 dB/cm/Mz
These give rise to:
(i) Depth of origin artefacts (an example is reverberation)
(ii) Beam dimension artefacts (an example is grating lobe artefacts)
(iii) Beam path artefacts (an example is the mirror artefact)
(iv) Attenuation artefacts (an example is shadowing)

12

Explain the essential differences in cause and appearance between shadowing and enhancement artefacts.

Distal to a region of high attenuation (from high reflection or high absorption, or both) there is a low intensity transmitted signal. So echoes from this lower region are going to be very weak, giving rise to the appearance of a dark region – shadowing.
On the other hand, distal to a region of low attenuation (from low reflection or low absorption, or both) there is a high intensity transmitted signal. So, echoes from this lower region are going to be very strong, giving rise to the appearance of a bright region – enhancement.

13

Indicate how each of the listed setting changes will affect the image frame rate; Decreasing the number of focal zones. Increasing the width of the field of view and decreasing the line density.

Decreasing the number of focal zones will reduce the time required to acquire each frame of information, and hence increase the frame rate.
Increasing the width of the FOV will increase the time required to acquire each frame, decreasing the frame rate, and decreasing the line density will decrease the time to acquire each frame, increasing the frame rate. There is insufficient information to determine whether there will be a net increase or decrease in frame rate.

14

What is involved in the process of demodulation of the detected ultrasound signal?

The detected signal comprises a short burst of AC voltage. This is first rectified so that all of the voltages are positive. This may be achieved by inverting the negative components. The rapid fluctuations in voltages are then removed by filtering, taking out the high frequency component and leaving a smoothed amplitude pulse.

15

What is a pixel?

A pixel is a picture element and is the smallest element of a picture, and the image is formed from a grid matrix of pixels combined that correspond to areas within the patient

16

What sort of RGB code would be required for an RGB 8-bit system to represent the colour light grey?

Light grey would require 191, 191, 191. Grey shades are obtained using equal components of red, green and blue, and 191 is half way between mid grey (127) and white (256).

17

Write a brief explanation of the process of image smoothing.

Image smoothing reassigns pixel values after averaging the original pixel value with the pixel values in the adjacent pixel locations. This averaging is performed for all pixels, forming a new image with less noise and more gradual changes in contrast, blurring the edges of high contrast features.

18

Explain the cause of slice thickness artefact

Slice thickness artefact occurs due to the finite thickness of the scan plane, when the scan plane at a given location encompasses both anechoic tissue and some other more echogenic tissue. The average echo level is therefore higher than when just anechoic tissue is insonated, and this is evident as areas of increased grey level in the predominant black of an anechoic region.

19

A structure is displayed only once, and in the correct line of sight, but at a greater depth than that at which the structure is actually located. What artefact is responsible for this?

Velocity artefact can cause this when the velocity of sound in overlying structures is less than 1540 m/s. The sound wave takes longer to reach the structure and return so the resulting echo is displayed at a greater depth.

20

There are similarities between side lobe artefact and grating lobe artefact. There are also two notable differences. What are they?

1) Grating lobes can be of higher intensity than side lobes, hence producing higher intensity artefacts.
2) Grating lobes only occur within the scan plane, whereas sidelobes occur all around the main beam, so side lobe artefact may be produced by structures from outside of the scan plane.

21

List four depth of origin artefacts

There are five depth of origin artefacts, so any four of the following:-Reverberation artefact, comet tail artefact, ring down artefact, velocity artefact and range ambiguity artefact.