1. The DF increases: DF = PW/PRF.
2. The PRF increases: PRF = 1/PRP.
4. When WL decreases, the PRF increases.
The increase in PRF decreases the DF.
PRF = V/WL
Calculate the PRP, PRF, PD, DF, SPL, wavelength, period, and Axial Resolution if a 2 cycle pulse emitted from a 10 MHz transducer if the echo time received was 0.26 ms in soft tissue.
Part II - Using the information above:
1. If the PRF decreases what happens the DF?
2. If PRP decreases what happens to the PRF?
3. If the frequency increases what happens to the axial resolution (improve or degrade)?
4. If wavelength decreases what happens to the DF?
2 answers
Part 1:
distance:
Distance - d=ct
Echo time- 10 ms
1540 m/s x .01s = 15.4 m
15.4/2
= 7.7m
PRF:
PRF = c/2D
1540 m/s / (2 x 7.7 m)
=1540 m/s / 15.4 m
=100 s
PRP:
PRP = 1/PRF
1/100 HZ
= .01s
Period:
Period - T= 1/f
1/5 MHZ
=.2 us
PD:
PD= nT
3 cycles x .2 us
= .6 us
DF:
DF= PD/PRP
.6us = .0000006 s
.0000006 s / .01s
=.00006
= .006%
wavelength:
Wavelength=c/f
1540m/s / 5 MHZ
1540 m/s / 5,000,000 1/s
.000308
= .308 mm
SPL:
SPL=nw
3 x .308 mm
= .92 mm
distance:
Distance - d=ct
Echo time- 10 ms
1540 m/s x .01s = 15.4 m
15.4/2
= 7.7m
PRF:
PRF = c/2D
1540 m/s / (2 x 7.7 m)
=1540 m/s / 15.4 m
=100 s
PRP:
PRP = 1/PRF
1/100 HZ
= .01s
Period:
Period - T= 1/f
1/5 MHZ
=.2 us
PD:
PD= nT
3 cycles x .2 us
= .6 us
DF:
DF= PD/PRP
.6us = .0000006 s
.0000006 s / .01s
=.00006
= .006%
wavelength:
Wavelength=c/f
1540m/s / 5 MHZ
1540 m/s / 5,000,000 1/s
.000308
= .308 mm
SPL:
SPL=nw
3 x .308 mm
= .92 mm
0 answers