Download Free Medical X-RAY DIAGNOSTICS AND IMAGING PowerPoint Presentation

X-RAY DIAGNOSTICS AND IMAGING PowerPoint Presentation

DOWNLOAD MEDICAL POWERPOINT PRESENTATION

About This Presentation

Description : Check out this medical presentation on Radiology and Imaging, which is titled "X-RAY DIAGNOSTICS AND IMAGING", to know about X-RAY DIAGNOSTICS AND IMAGING.

Tags : Radiology, Imaging, X-RAY DIAGNOSTICS

Published on : Aug 02, 2021 | Views : 2717 | Downloads : 0

Shareable Url :

Embed Code :

Comment

Transcript

Slide 1-

8. Diagnostic Radiology 8.2 X-RAY DIAGNOSTICS AND IMAGING Fluoroscope
Slide 2-

Direct-Exposure Film Direct exposure film has a relatively low absorption efficiency  for photons in the diagnostic range, however it is still used in many combinations of image receptor systems. Direct exposure film material has special design of two photographic emulsions with protective layers between to optimize the absorption efficiency  . Correct exposure is important to produce a reliable image on the film. Over- or underexposure will result in loss of contrast and therefore possibly in loss of diagnostic information. Many different types image receptors are used in modern diagnostic radiology. They all have in common that they form an image by absorption of energy from the X-ray beam (after transmitting through the body). The main characteristics will be discussed on the example of the direct exposure film.
Slide 3-

The proper film exposure can be obtained from the so-called characteristic curve of the film material. The blackening of the film after X-ray exposure is expressed in terms of its optical density: D = log10(I0/I) where I0 and I is the light intensities before and after passing through the exposed film material.
Slide 4-

The objective of this section is to correlate the optical density D (amount of blackening) with the received X-ray, exposure. This can be obtained with a simple model for the absorption process. A single emulsion of film material initially contains G silver-bromine grains per unit area, the average cross section of the grain is b. After irradiation with a flux of N X-ray photons per unit area a total number of g grains per area are sensitized. The number of sensitized grains per incoming X-ray photon is: with X as the received X-ray dose and k a conversion constant.
Slide 5-

The sensitized grains develop into a silver speck with an average cross section a after the film development. If light hits one of the silver specks in the developed film, it is completely absorbed (black spot).
Slide 6-

To relate the number of sensitized grains to the optical density D the absorption of the light in the film material of thickness t can be described as: Using this relation the optical density can be calculated to: This relation is known as Nuttings Law ! Maximum optical density for an area on the film is obtained when all grains are sensitized: g = G
Slide 7-

The correlation between the optical density D and the maximum number of sensitized grains results in a relation between the optical density D and the received dose X : This is displayed in the figure. The curve relating the optical density to the film exposure dose is called the characteristic curve of the film material. In the center part of the curve the relation between optical density and the logarithm of the dose is approximately linear:
Slide 8-

For a small contrast in dose DX= X1-X2 the associated change in optical density is: The constant F is known as the film-gamma and ranges between 2-3. It corresponds directly to the slope of the linear section of the characteristic curve. For low exposure or high exposure the characteristic curve levels out, exposure differences do not translate into differences in optical density (blackening).
Slide 9-

A film with an optical density of D  0.5 appears overall light, a film with optical density D  2 appears overall black. To achieve best contrast the film must be exposed hi such a way that the region of interest in the patient cause film doses which are in the center part of the characteristic curve.
Slide 10-

ppt slide no 10 content not found
Slide 11-

Additionally the contrast can be affected by the energy absorption efficiency of the image receptor material which in general decreases with energy. Efficiency and / or sensitivity of film material The sensitivity of film material depends on size and density of grains, emulsion thickness and X-ray absorption efficiency.
Slide 12-

The figure shows a typical X-ray absorption efficiency e for a double emulsion film as a function of energy. The efficiency drops rather rapidly with increasing energy and is mainly determined by the interaction probability of the photons with the film material (attenuation coefficient m ) and the thickness of the material t. An important goal is to maximize the efficiency of the image receptor material.
Slide 13-

The noise in the image may limit the contrast. The noise in the receptor image arises from several sources: fluctuations in the number of absorbed X-ray photons per unit area fluctuations in the absorbed photon energy fluctuations in the number of silver halide per unit area of emulsion The first and the last are the main sources for noise (quantum mottle and random darkening).
Slide 14-

To calculate the effect of quantum mottle we replace: with A as area, e as interaction efficiency, and N as number of incident photons. The resulting expression for the noise in the optical density of the image is: The noise due to quantum mottle is proportional to the slope G of the characteristic curve of the receptor material!
Slide 15-

To determine the noise due to random darkening the influence of granularity fluctuation (g  A)1/2 in the number of developed grains (g  A) in area A for the fluctuation in optical density DDG needs to be calculated: after substituting for g. Fluctuations therefore depend directly on fluctuations in grain size!
Slide 16-

The figure shows that the quantum mottle corresponds directly to the film-gamma and the random darkening due to the granularity distribution directly to the characteristic curve.
Slide 17-

ppt slide no 17 content not found
Slide 18-

Alternative Image Receptors Intensifying Screens in front of the photo emulsion convert X-rays into visible light. The film material is more sensitive to light photons than to X-ray photons. This increases the energy absorption efficiency e by more than one order of magnitude, but the resolution decreases due to additional noise components.
Slide 19-

Image intensifiers convert X-ray photons to electrons by photoelectric effect on a photocathode. The electrons are focused onto by electrical fields on a fluorescent screen where they form an intensified image which can be recorded on film or viewed with a TV camera.
Slide 20-

Xeroradiography is a dry non-silver photographic system, which produces images on paper (Xerox copies). It is slower than standard film receptors (which may result in higher doses) but has better resolution and better energy absorption efficiency e
Slide 21-

lonography replaces the photographic film by a position sensitive ion chamber. The X-rays induce by ionization of the gas electron clouds which can be detected by an electrode with good spatial resolution.
Slide 22-

X-Ray Transmission Computed Tomography Several problems exist with conventional radiography techniques: inability to distinguish soft body tissue because of limited contrast (see example blood-muscle); this can be fixed by the use of liquid contrast medium which has to be injected. inability to resolve spatially structures along the X-ray propagation axis resulting in loss of depth information (flat picture), because the three-dimensional body is projected on to a two-dimensional receptor.
Slide 23-

Computed tomography (CT) techniques allows sectional imaging . It is based on the principle that an image of an unknown object can be obtained if one has an infinite number of projections through the object. Two scans through the body gives an image in x- and y-direction (side view and front view). With increasing number of scans over a 360° angle range provides a set of images which allow to construct the three dimensional structure of the body.
Slide 24-

For X-ray tomography a planar slice of the body is defined and X-rays are passed along this plane in all directions. To produce a tomographic image typically 100 to 1000 scans are required. To store the multitude of images and process the data requires computer. The two-dimensional image corresponds to a three dimensional section of the patient with the third dimension being the slice thickness which is typically a few millimeter thick. The resulting spatial resolution is 1 mm, a density discrimination (contrast) of better than 1% can be obtained with this technique.
Slide 25-

Present CT machines have an rotatable X-ray source. This allows to scan the patient who is located along the rotational axis from all sides and angles. The image receptor system is designed as a stationary ring of detectors around the patient which receive the portions of the fanned X-ray beam.
Slide 26-

Each data point acquired by the detector array is a transmission measurement through the patient along a given line x between source and detector pixel. Each data point therefore follows the basic equation: The attenuation coefficient p represents the sum of all attenuation coefficients along the line x:
Slide 27-

From the image information the projection P(x) of the image is calculated: Because the attenuation coefficient m corresponds directly to the density of the body tissue along the projection axis x, the projection corresponds to the density. From a complete scan along different axes (x, f) the cross sectional density along the slice can be constructed using Fourier analysis methods.
Slide 28-

From the image information the projection P(x) of the image is calculated : The projection is directly proportional to the summed attenuation coefficient and to the length x through the body. Because the attenuation coefficient m corresponds directly to the density of the body tissue along the projection axis x, the projection corresponds to the density. From a complete scan along different axes (x, f) the cross sectional density along the slice can be constructed using Fourier analysis methods.
Slide 29-

The resulting CT image is a two-dimensional matrix of numbers with each number corresponding to a spatial location in image and patient along the plane of the slice. The matrix is constructed from 512x512 pixels. Each pixel has a value (up to 4096=12 bits) which corresponds to the level of gray (darkness, attenuation). This number is called CT number and contains the physical information ( attenuation  density ) about the corresponding body section. As the attenuation coefficient ranges between m=0 (air) and m=1 (metal inlet) the attenuation is scaled to a maximum of 4096. The CT therefore corresponds to a two-dimensional map of attenuation across the body slice. The CT number CT is normalized to the attenuation of water: If the attenuation coefficient for a given pixel is equal to water, the CT number=0, soft tissue material has CT number in the range of -100 to +100, dense tissues like bones have high CT numbers from 300 to 3000. CT numbers are rescaled attenuation coefficients!
Slide 30-

ppt slide no 30 content not found
Slide 31-

Typical Radiation Doses in Radiographic and in CT Examinations The radiation dose received in CT is considerably higher than that of conventional screen radiography. To compare the two types of X-ray exposures the received dose D of X-ray radiation is converted to integral dose DI, which corresponds to the total amount of energy deposited in the body tissue of mass m:
Slide 32-

ppt slide no 32 content not found
Slide 33-

In radiographic examinations the dose is not distributed evenly but drops from the entrance dose at the skin D0 towards deeper layers. For the skull m  0.33 cm1, A  256 cm2, the typical entrance dose is D0  4.8 mGy. This yields for the integral dose: The integral dose in a seven slice head CT is about twenty times higher than the integral dose of a single X-ray exposure. The integral dose is described by: DI = 3.7J
Slide 34-

ppt slide no 34 content not found
Slide 35-

ppt slide no 35 content not found