CVTE 110 Doppler, Pulses Exam #1 Flashcards
(158 cards)
1
Q
What is DOPPLER ULTRASOUND?
A
Shift happens. This is good—it’s how we make our living,saving lives and stamping out disease.
2
Q
Who is Christian Andreas Doppler? When was he alive?
A
1803 – 1853 Movement of stars toward or away, trumpet players on train, etc.
3
Q
What is the Doppler effect?
A
Frequency shift due to motion Motion can be on the part of the source, the receiver, or both.Or a reflector (RBCs).In that case, the reflectors become secondary sources.
4
Q
What happens when ultrasound is sent into tissue? How is backscattered ultrasound received from what went out?
A
Ultrasound is sent into tissue,gets scattered bymoving RBCs. Backscattered ultrasound from RBCs is received at a different frequency from what went out: ∆f
5
Q
Very convenient feature:The change of frequency is?
A
The change of frequency is proportional to the velocity. Faster flow shifts frequency more. Slower flow shifts frequency less.
6
Q
Faster flow shifts frequency ______.
A
More
7
Q
Slower flow shifts frequency ______.
A
Less
8
Q
How is the frequency shifted if the flow is toward the beam?
A
higher
9
Q
Flow toward beam: frequency shifted ______
A
higher
10
Q
How is the frequency shifted if the flow is away from the beam?
A
lower
11
Q
Flow away beam: frequency shifted ______
A
lower
12
Q
Trick question:
You get a stronger frequency shift when flow is toward the beam than when it’s away from the beam, right?
Toward the beam is better?
A
Nope.
A shift is a shift.
A plus or a minus doesn’t matter.
They’re both proportional to velocity.
13
Q
Basic functions of Doppler ultrasound in medical diagnostic work:
A
- Estimate arterial stenosis
- Estimate cardiac valve pressure gradient 4 (V2)
- Evaluate flow character qualitatively (e.g., LE veins)
14
Q
What influences the frequency shift?
A
- If the velocity of blood flow increases, the frequency shift increases. Faster flow makes bigger shift.
- If the operating frequency of the transducer increases,the frequency shift increases.Higher operating frequency makes bigger shift.
- If the angle of the Doppler beam relative to the direction of flow increases, the frequency shift decreases.Higher angle makessmaller shift.(This is the tricky relationship.)
15
Q
On the left: 4 HHz
On the right: 8.2 MHz
(Same velocity, same angle)
A
On the left: 4 HHz
On the right: 8.2 MHz
(Same velocity, same angle)
16
Q
If the angle of the doppler beam relative to the direction of flow increases, the frequency shift ________.
Higher ange makes _______ shift.
A
If the angle of the Doppler beam relative to the direction of flow increases, the frequency shift decreases.
Higher angle makes smaller shift.
(This is the tricky relationship.)
17
Q
If the operating frequency of the transducer increases,the frequency shift ___________.
Higher operating frequency makes _________ shift.
A
If the operating frequency of the transducer increases,the frequency shift increases.Higher operating frequency makes bigger shift.
18
Q
If the velocity of blood flow increases, the frequency shift _________.
Faster flow makes _______ shift.
A
If the velocity of blood flow increases, the frequency shift increases.
Faster flow makes bigger shift.
19
Q
What happens to ∆f with different angles?
A
20
Q
Radial artery with CW Doppler
A
21
Q
The simple Doppler equation:
A
∆f: frequency shift ∆f = 2 fo V
fo: operating frequency (Hz) ___________
V: velocity (m/sec) c
c: speed of sound (a constant)
22
Q
Velocity through a mitral valve is1.0 m/sec.
You’re using 2.5 MHz operating frequency.
What frequency shift will this create?
A
23
Q
Change of operating frequency?
4 MHz vs. 8 MHz
(Same velocity in radial artery for bothsame angle of about 45°)
A
24
Q
This equation assumes that the motion (i.e., the flow) is directly toward the receiveror directly away—
a 0° or 180° angle.
If the angle is other than 0° or 180°, you have to compensate.
How?
A
With a cosine.
25
What on earth is a cosine?
Don’t worry. It’s just a trig function of a given angle.It has a value between 0 and 1.
26
Angle Cosine Angle Cosine (Worth remembering the bold ones)
0° 1.000 10° 0.985
20° 0.940 30° 0.866
45° 0.707 60° 0.500
70° 0.342 90° 0.000
Angle Cosine
0° \_\_\_\_\_\_\_
45° \_\_\_\_\_\_\_
60° \_\_\_\_\_\_\_
90° \_\_\_\_\_\_\_
27
Notice how there’s little change for the _______ angles,
and rapid change for the _______ angles.
Much more error with _______ angles.
Notice how there’s little change for the lower angles,
and rapid change for the higher angles.
Much more error with higher angles.
28
The worst Doppler angle possible?
The worst doppler angle possible is 90°
Cosine of 90° is 0,
and 0 times anything is 0:
no frequency shift,so no waveform.
(Actually you get just junk along the baseline.)
29
The Doppler angle that gives the maximum frequency shift?
0°.
The cosine of 0° is 1.00, the maximum possible.
Echo uses 0°.
30
Vascular work—
can’t usually get a 0° angle in vessels,
so we use an intermediate angle, usually \_\_\_\_\_\_\_
50–60°.
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the more complicated Doppler equation:
∆f = 2 fo V (cos 0 )
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
c
0 : angle of beam incidence relative to flow direction
cos: cosine function of that angle
32
Be careful to make the distinction between the angle ( 0 )
and the cosine of that angle.
Theta refers to the ______ \_\_\_ \_\_\_\_\_\_\_\_;
the cosine is a _________ \_\_ ____ \_\_\_\_\_.
Theta refers to the angle of incidence;
the cosine is a function of that angle.
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Where’s the flow going? Where’s the beam going?
34
Angles: acute and obtuseHow many of each are there?
35
So answer these, then.
1. If operating frequency increases, does ∆f increase or decrease?
2. If velocity of blood flow increases, does ∆f increase or decrease?
3. If angle q increases, does ∆f increase or decrease? (Watch it.)
4. Is the speed of sound in tissue likely to change appreciably?
36
37
Velocity in CCA is 100 cm/sec,and operating frequency is 5 MHz. The angle of incidence is 60°.
What is the frequency shift?
(Be sure to use whole units...)
Note the difference between this and the echo result for the same velocity. Why?
38
We don’t use ∆f to grade stenosis or valve area.
What do we use it for?
We get the ∆f back from the moving blood, then use another form of the Doppler equation to estimate the velocity that created that ∆f.
39
The actual information that the scanner gets is ________ \_\_\_\_\_\_\_\_\_.
What are they caused by?
What is your job when using doppler and why?
The actual information that the scanner gets is frequency shifts
(caused by given velocities)
The meaning of those shifts can change depending on the variables in the Doppler equation.Your job is to use the Doppler intelligently (and humbly) so that the raw ∆f information really does have meaning.
40
Solving for velocity instead of frequency:
∆f c
Ve = \_\_\_\_\_\_\_\_\_\_
2 fo (cos 0 )
[Ridgway uses Ve, meaning velocity estimate, to keep in mind the limitations of our accuracy.]
41
Now answer these questions:
1. If ∆f increases, does Ve increase or decrease?
2. If operating frequency increases, does Ve increase or decrease?
3. If angle q increases, does Ve increase or decrease? (Really watch it.)
42
You use 5 MHz Doppler and a 60° angle to get a 4 KHz peak frequency shift from an ICA.
What is the velocity?
(The answer will be in m/secif you calculate with whole units.)
43
Doppler processing:
* Audio output
* Zero-crossing detector
* Spectral analysis (FFT)
* Color flow
* Power Doppler
44
Zero-crossing detector:
Detects crossings of zero!(Counts zero-crossings of signal to estimate return frequencies)
Single-line tracing resulting from average of shifts
Also known as “analog Doppler”
45
Spectral analysis:
Fast Fourier Transform
FFT
Analyzes (breaks down) the complex signalinto component frequencies
Unlike analog Doppler, this allows all the frequency shifts to be displayed rather than just an average.
46
Note: Page 336 in Edelman—bottom waveform is overgained to simulate spectral broadening.
Not true turbulence.
You want true turbulence?
This is true turbulence.Note the high amplitude(bright pixels)nearer to and underthe baseline.
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Here you can see individual pixels.
49
Elements of the FFT display:
X:
Y:
Z:
X = time
Y = ∆f (Ve)
Z = echo strength at a given ∆f at a given moment
(i.e., how much blood moving at that speed...) Indicated by pixel brightness
50
Pixel brightness suggests echo strength at that ∆f.
(Doppler gain should show gray scale, not just bleached-out white.
Overgained Doppler is a common mistake.)
Which pixels suggest more blood at a given speed? Picture on front or back of note card?
51
Another trick question:
Can you use spectral analysis with CW Doppler?
It’s only for PW Doppler, right?
No, CW frequency shifts can be processed with FFT too, but you’d expect a wider range of frequency shifts, since you get all the flow along the beam,across the entire lumen (not just center stream).
Echo often uses CW with spectral analysis.
52
Aliasing:
Display shows cut-off waveforms with the peaks “wrapped around” on other side of display.
Cause:∆f exceeds the Nyquist limit.
53
Aliasing has to do with _____ \_\_\_\_
Aliasing has to do with sampling rate
You must sample each cycle at least twice to yield a reasonable estimate of frequency.
Usual example: The wagon wheel in “Stagecoach” (1939, John Wayne, John Carradine, Thomas Mitchell, et al.)
54
Sample rate:
Think of movie frames acting like a strobe light, catching action at intervals.Wagon wheel spoke,rolling to the right:
The wheel doesn’t get all the way aroundbefore the next capture…
55
The Nyquist–Shannon sampling theorem is a
fundamental result in the field of information theory, in particular telecommunications and signal processing. The theorem is commonly called the Shannon sampling theorem, and is also known as Nyquist–Shannon–Kotelnikov, Whittaker–Shannon–Kotelnikov, Whittaker–Nyquist–Kotelnikov–Shannon, WKS, etc., sampling theorem, as well as the Cardinal Theorem of Interpolation Theory. It is often referred to as simply the sampling theorem. Sampling is the process of converting a signal (for example, a function of continuous time or space) into a numeric sequence (a function of discrete time or space). The theorem states that:
Exact reconstruction of a continuous-time base band signal from its samples is possible if the signal is band limited and the sampling frequency is greater than twice the signal bandwidth.
56
Nyquist limit (Harry Nyquist in 1923) represents:
represents the upper limit of frequencies that can be processed normally.
We need to sample often enough to get accurate ∆f processing:
at least TWICE the peak frequency.
57
Nyquist limit =
And PRF =
Nyquist limit = 1/2 the PRF
(pulse repetition frequency:pulses of ultrasound per second)
And PRF = twice the Nyquist limit
The Nyquist limit means sampling at a rate at least TWO TIMES the highest frequency in the signal.
58
59
The Nyquist limit is a result of the \_\_\_\_.
PRF (the sampling rate).
If you sample 10,000 times per second, the Nyquist limit is automatically half of that.
Again, that’s because you need two samples per Hz of signal that you want to display.
60
If you have Doppler running with a 12,000 Hz PRF, what is the upper limit of ∆f that can be processed without aliasing?
12,000/2 = 6,000 HzAnything higher than 6 KHz will alias: wrap around from the bottom of the display.
Note that this phenomenon happens only with PW Doppler.
61
CW has NO \_\_\_
CW has no PRF:
it’s always sending and always receiving, so no aliasing with CW.
(In other words, CW samples continuously instead of intermittently.)
62
What influences the likelihood of aliasing?
Anything that changes PRF or frequency shift.
63
Lower PRF:
lower Nyquist limit
64
From Edelman p. 318
65
What can you do to reduce the likelihood of aliasing?
(And why does each work?)
* Raise the angle of incidence
* Lower the operating frequency
* Find an approach that lessens depth
* Cut the Gordian knot: Change to CW!
Also: use maximum scale, high-PRF mode, low baseline
66
Left waveform is aliasing:
PRF is low (small scale).
Fixed on right by increasing scale
(increasing PRF, therefore raising the Nyquist limit).
67
Echo techs must switch to CW Doppler at velocities above ~2 m/sec,
while vascular techs can use PW to assess over 600 cm/sec.
How can this be?
Echo: 12-18 cm depth
Vascular:3-8 cm depth
PRF will be lower for echodue to much more depth.
Why? Travel time…
68
Echo: 0°
Vascular: 50-60°
∆f c
Ve = \_\_\_\_\_\_\_\_\_
2 fo (cos q)
69
Another way to deal with aliasing:
“High PRF” mode
Send a pulse before the first one gets back.Mild range ambiguity...
70
Why all the fuss about aliasing?
Why not just use CW all the timeand forget about aliasing?
Because, when possible, it’s preferable to have a discrete sample volume,
so you can assess flow in one spot.
CW Doppler can’t do that.
71
Angle correction:
(Vascular issue, not echo)
Line the angle-correct cursor parallel to the wall
to approximate the flow direction.
72
Another form of Doppler display:
Color flow
Many sample volumes
Process for direction and shift
Display as red or blue (direction)
Display as brighter or darker hue
(shift/velocity)
73
Differences between spectral analysis and color flow:
Spectral analysis Color flow
* One sample volume \*Many sample volumes
* Samples MANY times per second (256) over large area of image
* Gives ALL frequencies \*Samples 3-8 times per sample volume
(velocities) \*Gives MEAN frequency
(velocity) for each sample
| (FFT) (Autocorrelation)
74
Color flow processing:
Autocorrelation
Each sample is compared to the previous oneto arrive at a rough estimateof mean velocity(not peak).
75
The usual color assignment for echo:
For vascular work:
The usual color assignment for echo:
Red toward, blue away
For vascular work:
Make arteries red, veins blue(usually)
(not always possible)
76
The color bar at the side of the display tells you the color assignment.
Toward the beam is red
Toward the beam is blue
77
Brighter color suggests higher velocity
(It really suggests higher ∆f, for whatever reason:higher velocity, lower angle, etc.)
78
Aliasing due to stenosis?Or due to flow going faster downhill?
79
| (Edelman also talks about the “variance” color map,showing range of velocitiesas well as fast/slow velocities. Not really used that much.You may ignore that onefor this class.)
80
Toward or away from beam?
Is flow toward or away from the beam?
81
VERY severe mitral valve regurge
Severe ICA stenosis, with color bruit in tissue
82
Color aliasing:
Same phenomenon as with FFT.
Higher shift than Nyquist limit “wraps around” into opposite color.
83
Tricky aspect of color flow: Those darned angles.
You must constantly be aware offlow direction relative to beam direction.
Femoral artery (top) and vein—flow in artery reversed due to stenosis?
No; that was early diastole.This is systole.
84
Color flow requires lots of processing time, so there is a constant tension between enhancing color flow and keeping a high-enough frame rate.
If you improve color resolution or have a bigger color box, for example,you slow down the frame rate.
85
Important control for Doppler:
Wall filter
(a.k.a. high pass filter)
Takes out low-frequency,high-amplitude signals that can interfere with the information you care about(usually the higher frequencies).But must lower this to find slow flow.
86
Another form of color flow display:
Power Doppler
(a.k.a. Color angio, etc.)
Puts color on screen if there is a shift, but no directionality.Just yes or no to flow.
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START PULSES PP REMIX
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The 7 parameters of ultrasound:
1. Frequency
2. Period
3. Wavelength
4. Propagation speed .
5. Amplitude
6. Power
7. Intensity
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90
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Period =
Frequency =
Wavelength:
Speed of sound in average soft tissue:
Period = 1/f
Frequency = 1/P
Wavelength: λ = c/f
Speed of sound in average soft tissue: 1540 m/sec
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c
\_\_\_\_\_\_\_\_\_\_
λ f
93
Pulses in Ultrasound
Pulse:
Pulse:
A group of cycles;a short burst of ultrasound(“Pulse train”)
94
For imaging and most Doppler modes, you need pulses of ultrasound so the scanner knows distance.A pulse is sent, and echoes come back at different times depending on how far they have traveled. (In ultrasound, time is distance, distance is time.)
NOTE:No matter how short the pulse(usually a tiny fraction of a second), the ultrasound frequency is still expressed in cycles per second (Hz).
95
PRF:
pulse repetition frequency
The number of pulses per second
Usually in kHz
Note that this is pulses per second (i.e., groups of cycles per second) not cycles per second.
96
PRP:
pulse repetition period
Length of time from the beginning of a pulse to the beginning of the next pulse
97
PRF and PRP are inverse of each other:
PRF = 1/PRP
PRP = 1/PRF
(Similar to relationship between period and frequency with single cycles)
98
PD:
Pulse duration:
The length of time of the pulse itself from the beginning to the end of the group of cycles
PD = number of cycles in the pulse x time of one cycle (period)
99
Pulse duration is determined by the construction of the transducer:
frequency of transducer anddamping of vibration(more on this later)
100
Typical PD for diagnostic ultrasound transducers:
0.5 to 3 μsec
101
What makes PRP different from PD?
How long is the PRP?
102
Example:
If frequency is 10 Hz, and you create 2 pulses per second with 2 cycles per pulse,
what is the PDand the PRFand the PRP?
Period = 1/f = 1/10 Hz = 0.1 sec
PD = cycles per pulse x period =.
2 cycles x 0.1 sec = 0.2 sec
PRF = pulses per second = 2 Hz
PRP = 1/PRF = 1/2 Hz = 0.5 sec
103
If operating frequency is 2 MHz, and each pulse has 200 cycles,what is the pulse duration?
104
Duty factor:
Proportion of time the transducer is actually vibrating to produce ultrasound
(i.e., percent of time the transducer actually works)
105
DF is usually very tiny for diagnostic ultrasound:
1% or less.The rest of the time, the transducer is waiting for echoes to return.
No units - can be expressed as decimal or %
106
107
108
If the frequency is 20 Hz,there are 2 pulses per second, and there are 2 cycles per pulse, what is the duty factor?
NOTE! The frequency is 20 Hz,but only when sound is being generated.There aren’t actually 20 cycles happening each second...It’s just a rate that tells us how long each cycle lasts.
Pulse duration = period x cycles per pulse =
0.05 sec/cycle x 2 cycles/pulse = 0.1 sec/pulse
PRP = 1/2Hz PRF = 0.5 sec
DF = PD/PRP = 0.1 sec/0.5 sec =
0.2, or 20%
(No units for DF, just %)
109
Note that the frequency of ultrasound is NOT related to PRF.
You could have a PRF of 10,000 Hz
(i.e., 10,000 pulses per second)
with an operating frequency of
2 MHz, 5 MHz, 7.5 MHz, etc.
110
PD = 1.4 µsec
PRP = 0.5 msec
What is the duty factor?
DF = PD/PRP = 0.0000014 sec
\_\_\_\_\_\_\_\_\_\_\_\_\_
0.0005 sec
= 0.0028 or 0.28% duty factor
111
112
113
If the duty factor is 0.28%, how many cycles per second in a beam using 3.5 MHz frequency?
0.0028 x 3,500,000 = 9,800 cycles per second
(3,500,000 is the number of cycles per second if the duty factor were 100%—but it’s a smaller proportion than that.How much smaller? That’s the DF.)
This doesn’t tell you how many pulses there are, or how many cycles per pulse. If you knew the PRF, you could divide the number of cycles in a second by the number of pulses in a second to get the number of cycles per pulse, and from there get the PD.
114
Let’s say PRF is 4 kHz:
4,000 pulses per second.
9,800 cycles/sec
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
4,000 pulses/sec
= 2.45 cycles/pulse
THEN:PD = # of cycles in a pulse times the period of one cycle
PD = 2.45 cycles/pulse x (1/3,500,000) sec/cycle (period) = 2.45 x 0.00000029 =
0.00000071 sec/pulse or 0.71 µsec/pulse
115
Duty factor of 100% =
Duty factor of 0% =
Duty factor of 100% =always creating sound:continuous wave.
Duty factor of 0% =never creating sound:it’s off.
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117
Spatial pulse length:
distance of one pulse
SPL = cycles per pulse x λ
Remember: λ = c
\_\_
f
118
High frequency —\> ________ wavelength
Low frequency —\> _______ wavelength
High frequency —\> shorter wavelength
Low frequency —\> longer wavelength
119
Medium with faster propagation speed —\>
__________ wavelength
Medium with slower propagation speed —\>
\_\_\_\_\_\_\_\_\_\_wavelength
Medium with faster propagation speed —\>
longer wavelength
Medium with slower propagation speed —\>
shorter wavelength
120
Wavelength is to SPL
as Period is to PD
121
122
What is wavelength if frequency = 5 MHz?
λ = 1,540 m/sec
5,000,000 cycles/sec
= 0.000308 m/cycleor 308 µm per cycle
123
If f = 2 MHz, and there are 25 cycles per pulse, what is SPL?
First get the wavelength.
λ = 1540/2,000,000 =
0.00077 m/cycle
Then:
0. 00077 m/cycle x 25 cycles/pulse =
0. 01925 or 0.02 m/pulse SPL
124
Leaving Theoryland, pulses aren’t made up of equal-frequency cycles.
There’s more than one frequency present, due to damping of the transducer (more on this later).
Instead, you have a center frequencyalong with a number of other frequencies above and below the center—in other words, there is bandwidth above and below the center frequency (operating frequency, resonant frequency)
125
The bandwidth represents the range of frequencies in the signal created by the transducer,the number of frequencies produced.
126
As we’ll see later, in the transducer section, damping produces a shorter pulse.
This is good for resolution, but produces wider bandwidth.
(Ring bell for clear tone, then lightly hold the edge and ring it to get a clunk.)
Imaging transducers have wide bandwidths—you want short SPL for good resolution.
127
Quality factor:
Index of how clean and efficientthe signal is—
Ratio of center frequency to bandwidth
Operating frequency
QF = -------------------------------
bandwidth
fo
QF = -------
BW
128
A transducer has 5 MHz operating frequency.
Quality factor with BW of 2 MHz?
5/2 = 2.5 (no units)
Quality factor with BW of 5 MHz?
5/5 = 1
(Better Q factor with a high or low number?)
129
Wider BW —\> _______ Q factor
Wider BW —\> lower Q factor
Therefore, the transducers we use have low quality factors.
(Higher Q factor is more efficient,but we need short pulses for good resolution, so we live with low Q transducers.)
130
Which has the low Q factor?
Which has the high Q factor?
131
REVIEW FOR 6-WEEK EXAM
132
DOPPLER STUFF
Equations solving forfrequency shift or velocity estimate
What happens to ∆f with changes in velocity of blood flow, angle, or operating frequency?
What happens to Ve with changes in ∆f, operating frequency, or angle?
133
Toward vs. away—how do you tell?
What is the result on the spectral display or color flow display?
CW vs. pulsed Doppler
134
Processing for different kinds of Doppler:
analog (zero-crossing detector)
spectral (FFT)
color flow (autocorrelation)
135
Spectral display: 3 axes of information
X: Time
Y: Frequency shift (velocity)
Z: Amplitude of shifted echo
indicated by pixel brightness suggesting...what?
136
Angles and cosines
0° 45° 60° 90°
Which angles used for echo vs. vascular
Influence on frequency shift and velocity estimate (and likelihood of aliasing)
137
Nyquist limit:
What is it? (It’s about sampling rate...)
Why do we care?
Relation to PRF (and what is PRF?)
Influences on Nyquist limit and aliasing:
Depth, angle, operating frequency
138
Color flow:
conventions for color assignmentecho vs. vascular application
139
The 7 parameters of ultrasound:
1. Frequency
2. Period
3. Wavelength
4. Propagation speed
5. Amplitude
6. Power
7. Intensity
140
Imaging physics concepts:
PRF
PRP
DF
PD
SPL
141
Duty factor:
possible range
units
interaction with PD and PRP
142
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Wavelength (refers to one cycle)
Spatial pulse length (refers to group of cycles)
Units:
λ = c
-----
f
SPL = λ x cycles per pulse
144
Relationship between
amplitude and frequency:
145
Finding SPL:
1. Find wavelength (you need frequency and speed of sound)
2. Multiply by number of cycles per pulse (i.e., number of wavelengths)
146
SPL = wavelength X cycles/pulse
(It’s about distance)
PD = period X cycles/pulse
(It’s about time)
147
Can you draw a duty factor of 100%?
Can you draw a duty factor of 0%?
148
MATH
1 point for formula
1 point for correct work
1 point for correct answer
1 point for correct units
149
Math problems:
Circle what you’re solving for
Draw pictures
Remember to write formula
Watch for non-essential information
Check by working it backward
Don’t forget to double-check units
150
Duty factor problems:
How many cycles in a second if
DF = 0.2% and frequency is 5 MHz?
DF = PD PD = DF x PRP etc
------- c
PRP ----------------
(Make a magic triangle out of that.) λ f
151
Spatial pulse length:
Wavelength times cycles per pulse
Wavelength = speed of sound
-----------------------
frequency
What is SPL if frequency = 7.5 MHz, and there are 10 cycles in a pulse,and it’s soft tissue?
(Start by finding the length of one cycle)
152
Again:
PRF = pulses per second
PRP = time from start of one pulse to start of next
PRF = 1/PRP
PRP = 1/PRF
153
If PRF = 50 kHz,
then PRP = 1/50 k—\> 0.02 msec
(kHz usually inverts to msec,but be careful.)
Or (safer)
1/50,000 = 0.00002 sec =
20 µsec
154
If the frequency of the ultrasound is 5 MHz, and there are 4 cycles per pulse,what is the pulse duration?
P = 1/f = 1/5,000,000 =
0.0000002 sec = 0.2 µsec
PD = P x #cycles = 0.2 µsec x 4 =
0.8 µsec pulse duration
155
Another way to get pulse duration:
PD = # cycles per pulse
---------------------------
frequency
PD = # cycles per pulse x period
period = 1/frequency, so
PD = # cpp x 1
----
f
156
If frequency is 2.5 MHz, and there are 10 cycles per pulse, what is the pulse duration?
PD = cpp
---------
frequency
= 10
-----
2,500,000
= 0.000004 sec/pulse
= 4 µsec/pulse
157
158
If duty factor = 0.25%, and frequency = 3.5 MHz, How many cycles per second are being produced?
(Convert % to decimal: 0.0025)
If DF were 100%, there would be 3,500,000 cycles produced per second.
But cycles are being produced only 0.0025 of the time.
So:0.0025 x 3,500,000 = 8750 cycles/sec
