Factors Affecting Contrast In An Xray Image Health Essay

2037 words (8 pages) Essay

1st Jan 1970 Health Reference this

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To test resolution, a line pair guage is used. To measure MTF in an x-ray system, the sine wave equivalent of a line pair gauge is used. The closest distinguishable pair of cycles determines the best MTF, it is quoted in cycles per mm [2].

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A common way to express the system resolution is to quote the frequency where the MTF is reduced to either 3%, 5% or 10% of the original height [3]. MTF and spatial frequency are related by MTF curves. Examples of MTF curves are given below;

Mammography System:

Fig 2: MTF curve for a mammography system [4].

Chest Radiography System:

Fig 3: MTF curve for a typical chest X-ray. Three different detector types are indicated on the plot [5].

Fluoroscopy System:

Fig 4: MTF curve for a Fluoroscopy system with plots shown for individual components of the imaging process. The film and the optics have excellent resolution. The MTF of the imge intensifier is shown to have a limiting resolution of approximately 4.8 cycles/mm. The TV camera is the worst in the series, it limits the MTF of the overall image during live fluoroscopy and videotaped imaging. [6]

Question 2:

Contrast is the variation in brightness or optical density across an image. Factors affecting contrast in an x-ray image include the tube output, or the kVp. This is a measure of the energy of the x-ray beam leaving the x-ray tube and passing through the patient to form an image. X-rays with higher kVp can penetrate deeper into materials. In an image with the correct kVp bone is white and soft tissues and air are gray/black. If the kVp is too high, the x-rays will pass through even dense bone, creating an image that is mostly black with indistinguishable features [7]. The opposite occurs when a kVp which is too low is used. The most suitable kVp depends on the feature under investigation. Also among the factors affecting the image contrast is the patient. The density, the atomic number Z and the thickness of the part of the patient being imaged. Denser tissue, tissue with higher Z or tissue of a greater thickness results in lighter areas on the image because they have blocked the x-ray from exposing the image receptor. Variation in contrast occurs because tissues in the body attenuate x-rays differently. The human eye can percieve a difference of approximately 2% in contrast between adjacent areas [8]. The final influence on image contrast to be discussed here is the image receptor. In film imaging, the contrast of the resultant image depends on the sensitivity of the film used. To produce an image with the correct contrast, a film with corrresponding sensitivity must be chosen before imaging. In digital imaging, there is no fixed sensitivity. It has the advantage of being able to record the full range of exposures and digital processing after imaging can be used to improve the contrast in the image.

Image contrast can be evaluated using a densiometer. This device emits light of a known energy. The light is reflected back from the image and detected by the densitometer. The difference in energy between emitted and detected light is used to compute the optical density (blackness) in that area. Since contrast is the variation in optical density, this method can be used to survey the contrast in the image.

Question 3:

The following description is based on an article from the NDT database [9].

Spatial resolution of an x-ray system is limitied by the size of the focal spot. Fourier analysis can be used to calculate the focal spot size. X-rays are passed through a test object with a known pattern. This test object is placed between the x-ray source and detector, the arrangement is shown in the figure below. The focal spot of the x-ray is not assumed to be point-like, as the detector is moved away from the source, the detected focal spot appears larger. Blurring of the image by the detector is included, this blurring is related to the point spread funtion (psf) of the detector. Otherwise, an ideal detector is assumed. Image deterioration due to noise is also factored into the description.

Fig 5: Setup for determing the focal spot size. The X-ray source, the flat object, and the intensity distribution measured at the detector system lie in different planes for which different coordinate systems with the variables (x, y), (x’, y’) and (x”, y”) respectively, are used. This is done in order to include magnification effects in the calculations.

The measurement of the x-ray transmission, t, is derived mathematically. This is done by convolving the intensity distribution of the focal spot f with the transmission profile of the flat object g and the detector point spread function d. Also, t is deteriorated by noise, which is taken into consideration by addition of a noise term n to the result of the convolution. In order to take into account the geometrical magnification, V, of the setup, these functions are represented in one of these planes (here the plane of the detector), whereby the physical magnification effects of the setup were observed before the convolution is accomplished, this is shown in the second part of the equation below. The magnification is the distance between the source and the detector system divided by the distance between the source and the object.

The Convolution Theorem states that the Fourier transform of a convolution is the product of the Fourier transforms. Conversely, the Fourier transform of a product is the convolution of the Fourier transforms. Using the above equation, a deconvolution of t with g€ (€ d yields an estimate of f.

In a technique like this, a suitable test object is measured. The resulting image corresponds to a convolution of the test object with the intensity distribution of the focal spot and other factors.

Information on the focal spot is derived from this measurement using knowledge on the test

object and other influencing values which means that the convolution process is undone to

a certain extent. Also, with the presented method an arbitrary two dimensional intensity distribution can be measured, regardless of shape.

According to the convolution theorem, a convolution in the spatial domain corresponds to a point-by-point multiplication in the corresponding Fourier domain. Furthermore, according to the addition theorem, an addition in the spatial domain corresponds to an addition in the corresponding Fourier domain. (Note: lower case letters represent functions and upper case letters represent the Fourier transforms of the equivalent functions.) The initial equation now becomes;

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At certain spatial frequencies | N | can be significantly higher than| F ƒ-€ P |. At these spatial frequencies division of T by P mainly increases noise and deteriorates the image quality. This is due to the fact, that information on F is lost at these spatial frequencies. For that reason, independently of the deconvolution method applied, all spatial frequencies which are contained with high intensity in | F | should be contained with high intensity in |P| in order that | Fƒ- P | is significantly larger than | N |. This means that the test object (in combination with the detector imaging properties) should contain the major spatial frequencies which are required to describe the focal spot with sufficient intensity. In this case F can be restored well at these spatial frequencies, which yields a good estimate of f.

Question 4:

Using a bar phantom similar to that used for determining resolution can lead to an error determining the focal spot size. This is because the line pairs are aligned in one direction only. For accurate measurement of the focal spot size, many images with the bar phantom at different angles would be necessary [10]. To overcome this problem, a star phantom is used. This is a disc of alternating Lead spokes and x-ray transparent material. At a particular diameter of the focal spot the image of the spokes blurs, i.e., adjacent spokes cannot be distinguished from each other. The diameter of the blur is an indication of the focal spot size [11].

Fig 6: Star pattern for testing focal spot size [12]

Question 5:

5a.

The most obvious parts of a CT scanner are the moving patient table and the gantry or ‘tube’. Conventional projection radiography is limited because it collapses 3D objects onto 2D images. CT has a rotating system of emission and detection and so it can give accurate 3D diagnostic information about the distribution of structures inside the body. Inside the gantry there is the X-ray tube, x-ray detectors and slip-rings. The X-ray beam is collimated and radiates in a ‘fan-beam’ shape. The x-ray emitter and detectors rotate in the gantry to measure projections that form an image that is a ‘slice’ though the body. There are brushes around the rotating slip-rings to transmit signals. In CT, the linear attenuation coefficient, μ is measured. This tells how much intensity is lost as the beam travels through the medium. This distribution of μ is the basis of image formation. There are two distinct motions of the x-ray beam relative to the patient’s body during CT imaging. One motion is the scanning of the beam around the body. The other motion is the movement of the beam along the length of the body.  This is achieved by moving the body through the beam as it is rotating around

Fig 7: External appearance of a CT scanner. [13]

Fig 8: Basic schematic of the construction of a CT scanner.

Fig 9: CT image quality and electromechanical acceptance tests. The ‘Priority’ column indicates which of the tests are the most important. [14]

5b.

CT images are formed by multiple intersecting projections. This is illustrated in the figure on the left. In the bottom right section, it is seen that the combination of the projections causes blurring in the final image. The blurring goes as 1/r, i.e., it is proportional to the distance from the centre point. The 2D Fourier transform of 1/r is 1/ρ. Since the inverse of 1/ρ is |ρ|, it is possible to compute the 2D Fourier Transform of the blurred image, multiply the Fourier transform of the resultant image by |ρ| and the calculate the inverse Fourier transform to obtain a sharper image [15].

Fig 10: On the left, unfiltered back projections and on the right, filtered back projections. The conversion is done in the spatial domain by convolving the projection with the IFT of |ρ|.

Fig 11: The mathematics of the image reconstruction process, can be expressed compactly in the above equation, where the terms have been grouped to reflect the “filtered-back-projection” approach [15].

5c.

The fundamental system performance indicators are CT number, resolution, noise, and patient dose.26 The accuracy of CT numbers is measured by scanning a water-filled phantom at least monthly. The CT number for water should be zero over a 20-cm-diameter phantom, with a variation of less than 1 CT number. Deviation from the expected CT number of 0 for water at any energy is

adjusted by applying a correction factor for the pixel value. Constancy of the value

should be monitored with a daily scan. An overall check of system performance is obtained from semiannual measurements of CT image noise, defined as the standard deviation of CT numbers in a region of interest. Constancy of performance is checked by evaluation of the standard deviation in the daily water scan mentioned previously. Resolution is measured by scanning phantoms on a monthly basis. Of particular importance is low contrast resolution, which is a sensitive indicator of changes in component performance as they affect noise. Patient dose is evaluated semiannually. Specially designed ionization chambers provide measurements from which the dose may be calculated for the exposure conditions (narrow beam, variable slice thickness) used in CT. The values should agree with manufacturer’s specifications to within 20%. [hendee]

To test resolution, a line pair guage is used. To measure MTF in an x-ray system, the sine wave equivalent of a line pair gauge is used. The closest distinguishable pair of cycles determines the best MTF, it is quoted in cycles per mm [2].

A common way to express the system resolution is to quote the frequency where the MTF is reduced to either 3%, 5% or 10% of the original height [3]. MTF and spatial frequency are related by MTF curves. Examples of MTF curves are given below;

Mammography System:

Fig 2: MTF curve for a mammography system [4].

Chest Radiography System:

Fig 3: MTF curve for a typical chest X-ray. Three different detector types are indicated on the plot [5].

Fluoroscopy System:

Fig 4: MTF curve for a Fluoroscopy system with plots shown for individual components of the imaging process. The film and the optics have excellent resolution. The MTF of the imge intensifier is shown to have a limiting resolution of approximately 4.8 cycles/mm. The TV camera is the worst in the series, it limits the MTF of the overall image during live fluoroscopy and videotaped imaging. [6]

Question 2:

Contrast is the variation in brightness or optical density across an image. Factors affecting contrast in an x-ray image include the tube output, or the kVp. This is a measure of the energy of the x-ray beam leaving the x-ray tube and passing through the patient to form an image. X-rays with higher kVp can penetrate deeper into materials. In an image with the correct kVp bone is white and soft tissues and air are gray/black. If the kVp is too high, the x-rays will pass through even dense bone, creating an image that is mostly black with indistinguishable features [7]. The opposite occurs when a kVp which is too low is used. The most suitable kVp depends on the feature under investigation. Also among the factors affecting the image contrast is the patient. The density, the atomic number Z and the thickness of the part of the patient being imaged. Denser tissue, tissue with higher Z or tissue of a greater thickness results in lighter areas on the image because they have blocked the x-ray from exposing the image receptor. Variation in contrast occurs because tissues in the body attenuate x-rays differently. The human eye can percieve a difference of approximately 2% in contrast between adjacent areas [8]. The final influence on image contrast to be discussed here is the image receptor. In film imaging, the contrast of the resultant image depends on the sensitivity of the film used. To produce an image with the correct contrast, a film with corrresponding sensitivity must be chosen before imaging. In digital imaging, there is no fixed sensitivity. It has the advantage of being able to record the full range of exposures and digital processing after imaging can be used to improve the contrast in the image.

Image contrast can be evaluated using a densiometer. This device emits light of a known energy. The light is reflected back from the image and detected by the densitometer. The difference in energy between emitted and detected light is used to compute the optical density (blackness) in that area. Since contrast is the variation in optical density, this method can be used to survey the contrast in the image.

Question 3:

The following description is based on an article from the NDT database [9].

Spatial resolution of an x-ray system is limitied by the size of the focal spot. Fourier analysis can be used to calculate the focal spot size. X-rays are passed through a test object with a known pattern. This test object is placed between the x-ray source and detector, the arrangement is shown in the figure below. The focal spot of the x-ray is not assumed to be point-like, as the detector is moved away from the source, the detected focal spot appears larger. Blurring of the image by the detector is included, this blurring is related to the point spread funtion (psf) of the detector. Otherwise, an ideal detector is assumed. Image deterioration due to noise is also factored into the description.

Fig 5: Setup for determing the focal spot size. The X-ray source, the flat object, and the intensity distribution measured at the detector system lie in different planes for which different coordinate systems with the variables (x, y), (x’, y’) and (x”, y”) respectively, are used. This is done in order to include magnification effects in the calculations.

The measurement of the x-ray transmission, t, is derived mathematically. This is done by convolving the intensity distribution of the focal spot f with the transmission profile of the flat object g and the detector point spread function d. Also, t is deteriorated by noise, which is taken into consideration by addition of a noise term n to the result of the convolution. In order to take into account the geometrical magnification, V, of the setup, these functions are represented in one of these planes (here the plane of the detector), whereby the physical magnification effects of the setup were observed before the convolution is accomplished, this is shown in the second part of the equation below. The magnification is the distance between the source and the detector system divided by the distance between the source and the object.

The Convolution Theorem states that the Fourier transform of a convolution is the product of the Fourier transforms. Conversely, the Fourier transform of a product is the convolution of the Fourier transforms. Using the above equation, a deconvolution of t with g€ (€ d yields an estimate of f.

In a technique like this, a suitable test object is measured. The resulting image corresponds to a convolution of the test object with the intensity distribution of the focal spot and other factors.

Information on the focal spot is derived from this measurement using knowledge on the test

object and other influencing values which means that the convolution process is undone to

a certain extent. Also, with the presented method an arbitrary two dimensional intensity distribution can be measured, regardless of shape.

According to the convolution theorem, a convolution in the spatial domain corresponds to a point-by-point multiplication in the corresponding Fourier domain. Furthermore, according to the addition theorem, an addition in the spatial domain corresponds to an addition in the corresponding Fourier domain. (Note: lower case letters represent functions and upper case letters represent the Fourier transforms of the equivalent functions.) The initial equation now becomes;

At certain spatial frequencies | N | can be significantly higher than| F ƒ-€ P |. At these spatial frequencies division of T by P mainly increases noise and deteriorates the image quality. This is due to the fact, that information on F is lost at these spatial frequencies. For that reason, independently of the deconvolution method applied, all spatial frequencies which are contained with high intensity in | F | should be contained with high intensity in |P| in order that | Fƒ- P | is significantly larger than | N |. This means that the test object (in combination with the detector imaging properties) should contain the major spatial frequencies which are required to describe the focal spot with sufficient intensity. In this case F can be restored well at these spatial frequencies, which yields a good estimate of f.

Question 4:

Using a bar phantom similar to that used for determining resolution can lead to an error determining the focal spot size. This is because the line pairs are aligned in one direction only. For accurate measurement of the focal spot size, many images with the bar phantom at different angles would be necessary [10]. To overcome this problem, a star phantom is used. This is a disc of alternating Lead spokes and x-ray transparent material. At a particular diameter of the focal spot the image of the spokes blurs, i.e., adjacent spokes cannot be distinguished from each other. The diameter of the blur is an indication of the focal spot size [11].

Fig 6: Star pattern for testing focal spot size [12]

Question 5:

5a.

The most obvious parts of a CT scanner are the moving patient table and the gantry or ‘tube’. Conventional projection radiography is limited because it collapses 3D objects onto 2D images. CT has a rotating system of emission and detection and so it can give accurate 3D diagnostic information about the distribution of structures inside the body. Inside the gantry there is the X-ray tube, x-ray detectors and slip-rings. The X-ray beam is collimated and radiates in a ‘fan-beam’ shape. The x-ray emitter and detectors rotate in the gantry to measure projections that form an image that is a ‘slice’ though the body. There are brushes around the rotating slip-rings to transmit signals. In CT, the linear attenuation coefficient, μ is measured. This tells how much intensity is lost as the beam travels through the medium. This distribution of μ is the basis of image formation. There are two distinct motions of the x-ray beam relative to the patient’s body during CT imaging. One motion is the scanning of the beam around the body. The other motion is the movement of the beam along the length of the body.  This is achieved by moving the body through the beam as it is rotating around

Fig 7: External appearance of a CT scanner. [13]

Fig 8: Basic schematic of the construction of a CT scanner.

Fig 9: CT image quality and electromechanical acceptance tests. The ‘Priority’ column indicates which of the tests are the most important. [14]

5b.

CT images are formed by multiple intersecting projections. This is illustrated in the figure on the left. In the bottom right section, it is seen that the combination of the projections causes blurring in the final image. The blurring goes as 1/r, i.e., it is proportional to the distance from the centre point. The 2D Fourier transform of 1/r is 1/ρ. Since the inverse of 1/ρ is |ρ|, it is possible to compute the 2D Fourier Transform of the blurred image, multiply the Fourier transform of the resultant image by |ρ| and the calculate the inverse Fourier transform to obtain a sharper image [15].

Fig 10: On the left, unfiltered back projections and on the right, filtered back projections. The conversion is done in the spatial domain by convolving the projection with the IFT of |ρ|.

Fig 11: The mathematics of the image reconstruction process, can be expressed compactly in the above equation, where the terms have been grouped to reflect the “filtered-back-projection” approach [15].

5c.

The fundamental system performance indicators are CT number, resolution, noise, and patient dose.26 The accuracy of CT numbers is measured by scanning a water-filled phantom at least monthly. The CT number for water should be zero over a 20-cm-diameter phantom, with a variation of less than 1 CT number. Deviation from the expected CT number of 0 for water at any energy is

adjusted by applying a correction factor for the pixel value. Constancy of the value

should be monitored with a daily scan. An overall check of system performance is obtained from semiannual measurements of CT image noise, defined as the standard deviation of CT numbers in a region of interest. Constancy of performance is checked by evaluation of the standard deviation in the daily water scan mentioned previously. Resolution is measured by scanning phantoms on a monthly basis. Of particular importance is low contrast resolution, which is a sensitive indicator of changes in component performance as they affect noise. Patient dose is evaluated semiannually. Specially designed ionization chambers provide measurements from which the dose may be calculated for the exposure conditions (narrow beam, variable slice thickness) used in CT. The values should agree with manufacturer’s specifications to within 20%. [hendee]

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