Ch. 15-19 Physics questions Flashcards
Which electronic component is required for gray scale imaging?
A. VCR
B. non-interlaced display
C. computer memory
D. scan converter
Scan converter
Which of the following scan converters will provide the best spatial resolution?
A. 256 x 256 pixels
B. 512 x 512 pixels
C. 128 x 128 pixels
D. 1000 x 1000 pixels
1000 x 1000 pixels
Which of these statements regarding a bit of computer memory is false?
A. it is the smallest element of a image
B. it is bistable
C. Eight bits combine to make a byte
D. it is a component of a digital scan converter
it is the smallest element of a image
How many gray shades can be represented by a group of:
4 bits?
2 bits?
4 bits= 16
2 bits= 4
What is the number of shades that can be represented by 3 bits?
8
What is the number of shades that can be represented by 8 bits?
256
What is the number of shades that can be represented by 10 bits?
1024
What is the number of shades that can be represented by n bits?
2 is multiplied by itself n times. 2^n different gray shades can be stored using n bits.
How many bits are needed to store:
10 shades of gray?
11 shades of gray?
15 shades of gray?
10 shades= 4 bits
11 shades= 4 bits
15 shades = 4 bits
Which of the following statements regarding a pixel is false?
A. it is the smallest part of a digital image
B. a collection of bits, assigned to each pixel, stores the shade of gray.
C. it displays up to 3 gray shades simultaneously.
it displays up to 3 gray shades simultaneously.
Are the following procedures usually pre- or postprocessing?
A. Modifying a frozen image
B. read zoom
C. write zoom
D. adjusting the brightness on the monitor
E. increasing the receiver gain
Modifying a frozen image= postprocessing
read zoom= postprocessing
write zoom= preprocessing
adjusting the brightness on the monitor= postprocessing
increasing the receiver gain= preprocessing
What is the primary disadvantage of video tape and computer disc methods of image archiving?
A. these methods are very costly
B. the information is vulnerable to erasure when exposed to strong magnetic fields
C. only low resolution images are stored
D. only black and white images are stored
the information is vulnerable to erasure when exposed to strong magnetic fields
Where does coded excitation take place?
A. in the transducer
B. in the receiver
C. in the pulser
D. in the PACS system
in the pulser
Which of the following is not true about spatial compounding?
A. data is collected from different viewing angles
B. frame rates are reduced
C. shadowing artifact can be eliminated
D. can be performed with a mechanical transducer
can be performed with a mechanical transducer
All of the following are advantages of PACS systems except:
A. data will not deteriorate over time
B. images can be delivered to distant locations on the imaging network
C. older, archived studies and reports can be reviewed easily
D. images have higher resolution than what appears on the system’s display
images have higher resolution than what appears on the system’s display
What is DICOM?
A. standards for transducer design
B. standards for patient exposure
C. protocols for two- dimensional imaging
D. protocols for medical image data
protocols for medical image data
What is the primary storage device on a PACS system?
A. Magneto-optical storage
B. hard disc drives
C. solid state memory
D. compact discs
hard disc drives
Which of the following technologies uses long sound pulses to create an image?
A. coded excitation
B. harmonics
C. pulse inversion
D. rendering
coded excitation
With frequency compounding, how many pulses are transmitted down each scan line?
A. 1
B. 2
C. 4
D. cannot be determined
1
Which of the following techniques is used to create image data where non really exists?
A. edge enhancement
B. temporal compounding
C. fill in interpolation
D. spatial compounding
fill in interpolation
In modern elastography, how are the tissues deformed?
A. by pressing the transducer into the skin
B. with the use of a second probe
C. automatically
D. As a result of respiration
automatically
An uncompressed signal within the receiver of an ultrasound system has a dynamic range of 110dB. The signal undergoes 40dB of compression. What is the dynamic range of the compressed signal?
110dB - 40dB = 70dB
An uncompressed signal has a dynamic range of 85dB. The signal undergoes 30dB of compression. What is the dynamic range of the compressed signal?
85dB - 30dB = 55dB
After compression, a signal within the ultrasound system has a dynamic range of 70dB. The original signal was compressed by 40dB. What is the dynamic range of the original, uncompressed signal?
70dB - 40dB = 110dB