Scan Principles Image Basics - Calculations* Flashcards
Assuming all other parameters are the same, an echo time of ___________ ms provides the best Signal-to-Noise Ratio
15
Fat has a _______ T1 relaxation time and a ________ T2 relaxation time.
short; short
Water has a __________ T1 relaxation time and a ________ T2 relaxation time.
long; long
If the repetition time is doubled, the scan time will:
Double
TR is directly proportional to scan time (TR x Phase Matrix x NEX)
Which of the following is NOT affected by changing the slice thickness?
The T1 contrast
Repetition Time (TR) is the controlling factor of T1 weighting/contrast in an image; slice thickness does nothing to alter the T1 information.
Which of the following would NOT compensate for the change in signal-to-noise as a result of decreasing the FOV?
Reducing the number of excitations
A reduction in excitations (NEX/NSA) would reduce SNR, which has the same effect as decreasing the field of view
Reducing the phase matrix from 256 to 192 will
All of the above
Reducing the phase encoding matrix yields higher SNR, shorter scan times, but decreased spatial resolution due to the increased pixel size.
SNR increases with all the following adjustments, EXCEPT:
Decreasing pixel size
Smaller pixel size (higher phase matrix) leads to improved resolution, but decreased SNR.
Another name for a picture element is a:
Pixel
What formula is used to calculate the size of the pixel?
FOV ÷ matrix
Which parameter will provide the better spatial resolution?
4mm slice thickness; 256 x 256 matrix
Thinner slices yield improved spatial resolution, in addition to smaller pixel size (higher matrix)
If a sequence with a matrix of 256 x 512 is adjusted to 512 x 512,
The SNR will decrease and the scan time will increase
In a fast spin echo sequence with a 28 cm FOV, 256 x 512 matrix, TR 2200, 90 TE, 5mm slice thickness, 8 ETL and 3 NEX, the scan time would be:
3 minutes 31 seconds
Scan time formula: TR x Phase Matrix x NEX ÷ ETL = total in milliseconds, then ÷ 1000 to achieve scan time in seconds. Be careful if you divide by 60 to achieve the scan time in decimal format (This example computes to 211 seconds, if you divide by 60 you get 3.52 seconds, which is 3 minutes 31 seconds, not 3 minutes 52 seconds)
Decreasing the TR will:
Both A and B
With regards to k-space, the data containing high resolution are located along:
The outer lines
Referring to k-space, the data containing high resolution are located along the outer lines
With regards to k-space, the data containing signal and contrast information are located along:
The central lines
Referring to k-space, the data containing signal and contrast information are located along the central lines.
As the number of excitations (NEX) is increased from 1 to 3, which of the following does NOT occur?
The pixel size is increased
Changes in NEX have no effect on the resolution of an image.
Reducing the FOV (field of view) only will result in:
Decreased signal-to-noise
A reduction in FOV will yield a decrease in SNR, but an increase in spatial resolution.
The scan time of a fast spin echo sequence utilizing a TR of 3500ms, a TE of 90ms, a 256x256 matrix, 1NSA, a 220mm FOV and an echo train length of 5 is ______ minutes, approximately.
3
TR (3500) x Phase Matrix (256) x NSA (1) ÷ ETL (5) = 179200 ms, then divide by 1000 to get 179.2 seconds, or nearly 3 minutes.