PRH353_Topic3.pdf Compilation of fluoroscopy notes

Fluoroscopy notes

1. Fluoroscopy
•References:
•Medical Imaging Physics by Hendee W. R. and
Ritenour E. R. (2002)
•The Physics of Diagnostic Imaging by Dowsett
D. J., Kenny P. A. and Johnson R. E. (1998)

2. •Fluoroscopic Imaging
•Fluoroscopy refers to the use of an X ray beam and a suitable image
receptor for viewing images of processes or instruments in the body in
real time.
•It is a dynamic imaging which allows radiologist to view the internal
structures of the patient in action.
•Image receptors in early fluoroscopic systems is an intensifying screen
similar to that used in radiographic screen film imaging.
•During imaging the screen would be viewed directly by the radiologist.
•The disadvantages of the direct view (early fluoroscopic) systems include:
•* Production of dim images that required the radiologist’s eyes to be
dark adapted.
•* High patient dose
•* High radiologist dose
•Modern fluoroscopic systems use X-Ray Image Intensifier (XRII).

3. •A typical XRII system is shown below








•Stages of images formation by XRII
•1. The x-rays shown are those that exit the patient.
•2. The antiscatter grid removes contrast degrading scattered
radiation from the X ray beam.
•3. X rays that have passed through the patient, antiscatter grid and
the metal support structure of the XRII are incident on an input
phosphor

4. •4. These light photons strike a very thin bi- or multi-alkali photocathode,
where electrons are released through photoelectric effect.
•5. The released electron beam is accelerated (to energies between 25 and
30 keV) and is focused on the positively charged anode at the output
phosphor by a series of electrodes located at the sides of the XRII.
•6. The output phosphor converts the electrons to light.
•7. The optical image from the output of the XRII is routed to other
components for display on a viewing monitor or for recording of the
images.
•The optical system consists of three elements:
•(i) a collimating lens to shape the divergent light from the output
phosphor into an almost parallel beam.
•(ii) an aperture to limit the amount of light reaching a video (or TV)
camera.
•(iii) a lens to focus the image on to the video camera

5. •8. The optical image is captured by the camera and converted to an
analogue electrical signal that conforms to a recognized video format
•The size of aperture in the optical system of XRII is measured in f stops. A
decrease in aperture size by one f stop allows double the amount of light
through, while an increase by one f stop allows half the amount of light
through.
•f stop is the focal length of the lens divided by the diameter of the
aperture.
•Intensification of x-ray image
•This occur through two mechanisms, namely electronic (or flux) gain and
minification gain.
•Flux gain is a result of the kinetic energy gained by electrons as they are
accelerated between the photocathode and the output phosphor (anode).
•The amount of light produced at the output phosphor depends on the
kinetic deposited by the electrons on the phosphor.
•??????
�=
??????���� ������� ���� ������ ��������
??????���� ������� �� ������������

•A typical image intensifier has a flux gain of at least 50 i.e. g
f ≥ 50.

6. •Example
•Determine the flux gain of the image intensifier of the fluoroscopic
system with the following parameters:
* X-ray flux at input phosphor = 3 x 10
5
* Effective energy of the x-ray = 60 keV
* Input phosphor has 50% x-ray to light conversion
efficiency
* Mean energy of light produced from input phosphor is
3eV.
* Photocathode light to electron conversion
efficiency is 3%.
* Output phosphor produced 1 light for every
15.87 eV electron energy.
* The image intensifier was operated at 30 kV)

7. •Solution
•??????
�=
??????���� ������� ���� ������ ��������
??????���� ������� �� ������������

•Hence we need to calculate (i) Light photons at photocathode and (ii) Light
photons from output phosphor
•(i) Light photons at photocathode = light photons from input phosphor
•This is calculated following the steps listed below:
•The number of x-rays that will produce light photons in the input
phosphor = 3 x 10
5
x 50% = 1.5 x 10
5

•Each of these 1.5 x 10
5
has an energy of 60 keV = 60000 eV.
•Since it requires 3 eV of the x-ray photon to a light photon, the number of
light photons that will be produced by each x-ray photon = 60000/3 =
20000.
•Therefore the total light photons produced by the input phosphor = 1.5 x
10
5
x 20000 = 3 x 10
9

•This implies, Light photons that reach the photocathode = 3 x 10
9

8. •(ii) Light photons from output phosphor is calculated following the steps
listed below:
•The number of electrons produced from the photocathode = number of
light photons that reach the photocathode x light-electron conversion
efficiency of the photocathode = 3 x 10
9
x 3% = 9 x 10
7

•Since the voltage applied is 30 kV each electron on reaching the output
phosphor will possess energy 30 keV = 30000 eV.
•The electron energy required to produce 1 light in the output phosphor =
15.87 eV, hence the number of light photons that each electron will
produce in the output phosphor = 30000/15.87 = 1890
•Hence the total light photons from the output phosphor = 9 x 10
7
x 1890 =
1.7 x 10
11

•Therefore the light photons from output phosphor = 1.7 x 10
11

•??????
�=
??????���� ������� ���� ������ ��������
??????���� ������� ���� ����� ��� ������������
=
1.7×10
11
3×10
9
=56.6

9. •Minification gain is a result of the reduction of a large X ray
image at the input phosphor on to the smaller diameter
output phosphor.
•The reduction means that the amount of light per unit area
from the output phosphor is greater than that from the
input phosphor. This implies increased image brightness.
•Hence if the brightness of input screen is B
1 ( =
�
�1
) and the
brightness of output screen is B
2 ( =
�
�2
), then the
minification gain is
•??????
�=
�2
�1
=
�1
�2
=
�1
2
�2
2

•A
1 and A
2 are the areas of the input and output phosphors
respectively. d
1 and d
2 are the diameters of the input and
output phosphors respectively.

10. •For example, the minification gain of an XRII using an FOV with diameter
50 cm and output phosphor with diameter 5 cm would be 50
2
/5
2
or 100.
•Brightness gain is the product of the electronic gain and minification gain
and is a measure of the overall system gain. i.e
•??????= ??????
�×??????
�
•Conversion factor (CF) is another measure of the gain of an XRII. It is used
to specify XRII performance and hence can be used to compare two image
intensifiers.
•CF is the ratio of the luminance in units of candela per square metre
(cd/m
2
), measured at the output phosphor to the X ray Incident Air Kerma
Rate (IAKR), in units of mGy/s, measured at the input phosphor using
standardized beam conditions. i.e.
•????????????=
??????�������� �� ������ ������ (
��
??????
2)
������� ���� �� ����� ������ (
??????????????????
�
)

•CF decreases over time and may ultimately fall to a level that requires
replacement of the XRII.

11. •Dual- and Triple-Field Image intensifiers
•Some image intensifiers allow magnified viewing of the central region of the input
screen (phosphor).






•When only mode of magnified viewing is possible, the intensifier is referred to as
dual-field intensifier.
•When two magnified viewing are possible, the intensifier is referred to as triple-
field intensifier.
•In the magnified viewing mode, the image on the output screen is produced by
electrons from only the central portion of the input screen.
•Fewer electrons, than in the normal mode, are used to produce the image. Hence,
the brightness of the image would be reduced unless the exposure rate to the
input screen is increased.
•The increase in exposure rate is achieved automatically when the mode is changed
from normal to magnified.
•The exposure rate increases with decreasing area of the active input phosphor.

12. •Note: exposure rate from the input phosphor inversely proportional
to the area of the phosphor ∝
1
�

•Example
•Estimate the increase in exposure rate required to maintain image
brightness when an image intensifier field of view is decreased
from 3 cm to 1.5 cm in diameter.
•Solution
•To get the same brightness when the FOV is decreased to 1.5 cm,
the exposure rate needs to be multiplied by the factor which is
equivalent to
•
�������� ���� �� ��?????? �� 1.5 ��
�������� ���� �� ��?????? �� 3 ��
=
1
??????1.5
2
1
??????3
2
=
3
2
1.5
2
=4

13. •Size of Image intensifier







•If the distance from the region of interest to the x-ray tube is d
r and
d
i is the distance from x-ray tube to the input phosphor, then
•�=�
�
??????
��

•a is the actual length of the region of interest and b is the length of
the region of interest on the input phosphor


a
b
X-ray
tube Patient
Input phosphor
Region of interest in patient

14. •Example
•An x-ray tube is located 30 cm below a fluoroscopic table. The input
screen of the image intensifier is 15 cm in diameter and 35 cm
above the table. What is the maximum length of an object that is
included completely on the screen. What is the magnification of
the image? The object is 10 cm above the table.
•Solution
•Hence we need to calculate a when b = 15 cm.
•di = 30 cm + 35 cm = 65 cm
•dr = 30 cm + 10 cm = 40 cm
•�=�
�
??????
��
i.e �=�
��
�
??????
=15×
40
65
=9.23 cm
•Magnification = 15/9.23 = 1.6

15. •Flat panel detectors
•Nowadays, fluoroscopic systems incorporate flat panel
detectors.
•No lens, aperture or camera system is needed for flat plate
detector systems.
•Advantages of flat panel detectors over XRII
•These include:
•(1) their larger size
•(2) less bulky profile
•(3) absence of image distortions
•(4) a higher Quantum Detection Efficiency (QDE) at
moderate to high IAKRs to the receptor.
•(5) The made possible new applications such as rotational
angiography and cone beam CT (CBCT).

16. •Disadvantages of flat panel detectors over XRII
•(1) They suffer from additive noise sources, including read noise, and
therefore perform poorly compared with XRIIs at low IAKR.
•Automatic Exposure Control for Fluoroscopic system
•Automatic exposure control (AEC) are devices that automatically adjust
radiographic technique factors (most often the mAs) to deliver a constant signal
intensity at the image receptor in response to differences in patient thickness, X
ray tube energy, focus to detector distance and other technical factors.
•In fluoroscopic systems, the AEC controls the IAKR to the XRII, to prevent
fluctuation in image brightness and SNR that would make diagnosis or navigation
of instruments difficult.
•Flux Magnification
•Flux magnification refers to the use of a focusing electrode to deminify the
fluoroscopic image by selecting a smaller portion of the input phosphor to project
on to the output phosphor.
•F magnification improves the image MTF but also decreases minification gain and
decreases the sampling pitch of the input phosphor, increasing noise.

17. •Television display of the fluoroscopic image
•The image from the output screen of the image intensifier is collected by a
television camera connected to the back of the intensifier.
•The signal from the camera is transmitted by cable to a television monitor.
•Camera
•A typical analog camera has a thin layer of photoconducting material that is an
electrical insulator when not exposed to light.
•When light from the output screen of the image intensifier falls upon the
photoconducting layer, the resistance across the layer decreases.
•This makes the resistance across any region on the illuminated photoconducting
layer depends on the amount of light incident on that region.
•Electron gun is another component of a typical analog camera. Scanning electron
beam produced the electron gun deposits electrons on the photoconducting layer.
•Some of the electrons migrate towards the positive side of the layer and are
collected by the conducting layer.
•In any region, the number of electrons that migrate to the positive side depends
on the resistance and, consequently, the illumination of the region.
•Electrons collected by any region of the conducting layer generate electrical signal
that depends on the illumination of the region.

18. •Instead of the analog camera, some new fluoroscopic systems use charge couple
device cameras.
•In CCD cameras, the photoconducting and scanning are replaced with arrays of
CCDs that produce electrical signals in response to the amount of light falling on
them.
•Scanning of the photoconducting layer
•In scanning, the scanning electron beam moves horizontally across the
photoconducting layer of the camera (or the CCD elements are struck by photons).
•This motion of the scanning electron beam is referred to as active sweep.
•The horizontal scanning of the layer in the opposite direction is referred to as
retrace sweep, during which no information is transmitted to the television
monitor.
•The results of the next active sweep is displayed slightly below the preceeding
one.
•Number of active sweeps of the scanning electron beam across the image may be
525, 625, 837, 875, 945 or 1024.
•525 active sweeps means the image is divided into 525 horizontal lines on the
screen of the television monitor.
• Images with a larger number of sweeps provide better vertical resolution than
those formed with a smaller number of sweeps.

19. •To prevent brightness flickering in the image, the electron
beam first scans photoconducting layer along even-
numbered sweeps (2, 4, 6, …).
•The resulting set of horizontal lines or field is then
interlaced with sweeps of the electron beam along the odd-
numbered lines (1, 3, 5, …).
•The final image, produced by two interlaced fields, is
referred to as a frame.
•A frame is usually completed in 1/30 seconds. This means it
takes 1/60 seconds to generate each set of the even-
numbered and odd-number lines.
•If the scanning is done sequentially along all lines (1, 2, 3,
….), the image will appear at a rate of 30 times per second
and a sensation of flickering light would be perceived.

20. •Usually not all the available horizontal lines contribute to the image. Those
that contribute are the active sweeps.
•The vertical resolution of a television image is given by
•Vertical resolution = (number of active sweeps/frame)*(Kelly factor)
•Kelly factor is the fraction of the active sweeps that are actually effective
in preserving detail in the image (For most television the value is 0.7).
•Horizontal resolution depends on number of scan lines per frame and
upon the range of frequencies transmitted by the electronic circuitry
between the television camera and monitor.
•The highest frequency of signal modulation that is transmitted undistorted
from the television camera to the television monitor is referred to as
frequency badpass or bandwidth of the television system.
•Example
•Determine the vertical resolution of a 525-line television that provides
about 490 active sweeps per frame

21. •Solution
•Vertical resolution = (Number of active sweeps/frame)*(Kelly factor) = 490 x 0.7 =
343 lines
•Example
•What is the minimum bandwidth that is required to provide equal horizontal
resolution (343 resolution elements per line) and vertical (343 lines per frame)
resolution in the image?
•Solution
•Bandwidth (Number of resolution elements transmitted per second) = (Horizontal
resolution)*(vertical resolution)*(Number of frames per second) = (343 resolution
elements per line)*(343 line per frame)*(30 frames per second) = 3.53 x 10
6

resolution elements per second = 3.53 MHz.

•A more exact expression for horizontal resolution (in resolution elements per sec)
is
•??????��????????????����?????? �??????��??????��??????��=
2�������������??????����� �������������� �����
������ ��� ����������� ��� �����

22. •Horizontal fraction is the ratio of time required to complete one active
sweep and the one required to complete both the active and the retrace
sweeps.
•Aspect ratio is the ratio of the width of a television frame to its height.
•Example
•An image of light and dark vertical and horizontal lines is presented on the
output screen of an image intensifier. This image is transmitted to a
television monitor by a 525-line television camera. The bandwidth is 3.5
MHz for the television chain. The framing time of the camera is 1/30
second. What is the maximum number of light and dark lines on the
output screen that are duplicated without distortion on the television
monitor? Determine the horizontal resolution given that the aspect ratio is
1 and horizontal fraction is 0.83.
•Solution
•??????��????????????����?????? �??????��??????��??????��=
2�������������??????����� �������������� �����
������ ��� ����������� ��� �����

23. •??????��????????????����?????? �??????��??????��??????��=

23.5×10
6
0.831.0
30 ������ ��� ������525 ����� ��� �����
=368 lines.
•This means the image will be transmitted undistorted if
it is composed of no more than 184 dark lines
separated by 184 light lines.
•Digital Fluoroscopy
•One way by which digital fluoroscopy is done is by
digitizing the signal from the video camera.
•In other digital fluoroscopic systems, the video camera
is replaced with CCD camera or flat-panel detector.

24. •Image quality in Fluoroscopy
•Factors to consider regarding image quality in fluoroscopy include
contrast, noise, sharpness, temporal resolution and artefacts or
image distortions.
•Contrast
•Subject contrast is inherently poor in fluoroscopic imaging, mainly
because high kV values are used in order to maintain patient dose
at an acceptable level.
•To improve contrast radiopaque markers are used on catheters and
other instruments, and through the use of exogenous contrast
agents.
•Iodine and barium are two contrast agents commonly used in
fluoroscopic imaging.
•Gadolinium or carbon dioxide may be used when iodine contrast is
contraindicated owing to allergies or impaired renal function

25. •Noise
•The noise level in fluoroscopic images is necessarily high, as a low
IAKR is typically used to maintain the patient dose at an acceptable
level.
• XRII based fluoroscopic systems are also characterized by low
additive electronic noise levels. Therefore, the system is still
quantum limited at low IAKR values.
•Flat panel based fluoroscopic systems suffer from high levels of
electronic noise (read noise, specifically), and their imaging
performance is limited by this noise at low IAKR values.
•Therefore systems utilizing flat panels require a higher IAKR than
XRII based systems for fluoroscopic imaging. Conversely, flat panels
perform better than XRIIs at high IAKR, such as those used during
digital acquisition imaging

26. •Sharpness
•The sharpness of a fluoroscopic image is influenced by several factors, including:
•* the display matrix
•* FOV
•* video camera matrix
•* focal spot size
•* geometric magnification
•* image noise and motion.
•Resolution
•The resolution of an image intensifier depends on
(a) resolution of the input phosphor
(b) resolution of the output phosphor
(c) ability of the focusing electrodes to preserve the image as it is being
transferred.
•Contributions to resolution loss from outside image intensifier are:
(a) presence of scattered radiation in the x-ray beam received by the input
phosphor.
(b) unsharpness in the image conributed by patient motion
(c) finite size of the focal spot.

27. •Artefacts
•Artefacts in fluoroscopic imaging usually stem from
image distortions caused by the image chain
components.
•Flat panel image receptors are generally free from
image distortions.
•On the other hand, XRIIs suffer from several
common image distortions, including:
•* veiling glare
•* vignetting
•* blooming
•* pincushion
•* distortion and S distortion

28. •S distortion causes straight objects to appear curved and results
from the acceleration of electrons in the electron optical system of
the XRII in the presence of an external magnetic field.
•Pincushion distortion causes enlargement of the fluoroscopic
image near the edges and results from the curvature of the input
phosphor, which is required for proper electronic focusing and
structural support. Pincushion distortion is more severe for a large
FOV.
•Blooming is caused by the input of signals to the video camera that
exceed its dynamic range.
•Vignetting is an optical distortion that produces a fall off in light
intensity or darkening near the edges of an image. This may be
caused by a number of factors, including: deterioration of the video
camera, and is also inherent to multielement lenses. Vignetting can
be reduced in some cases by restricting the aperture size.

29. •Veiling glare is a contrast reducing ‘haze’ that results from the scattering
of information carriers within the XRII, including electrons within the
electron optical system and, most importantly light photons within the
glass output window.
•Fluoroscopic system Designs
•There are a number of fluoroscopic system designs, including:
•(1) Under-table x-ray tube fluoroscopy system
•In this configuration, the X ray tube is located under the patient table and
the XRII and auxiliary imaging equipment are placed on a movable ‘tower’
above the patient table.
• Lead curtains hang from the XRII tower and shield the operator from stray
radiation scattered from the patient.
•There is a control panel for local operation
attached to the image intensifier
•The table material usually have a high
transparency to x-rays.

30. (2) Over-table x-ray tube fluoroscopy system
•In this configuration, the X ray tube is located above the
patient table or couch and the XRII under the patient table.
•Exposures are controlled remotely from a shielded control
station.



•

•

31. •(3) Mobile C-arm Fluoroscopy system