CVTE 110 Doppler, Pulses Exam #1

Question 1

What is DOPPLER ULTRASOUND?

Answer 1

Shift happens. This is good—it’s how we make our living,saving lives and stamping out disease.

Question 2

Who is Christian Andreas Doppler? When was he alive?

Answer 2

1803 – 1853 Movement of stars toward or away, trumpet players on train, etc.

Question 3

What is the Doppler effect?

Answer 3

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.

Question 4

What happens when ultrasound is sent into tissue? How is backscattered ultrasound received from what went out?

Answer 4

Ultrasound is sent into tissue,gets scattered bymoving RBCs. Backscattered ultrasound from RBCs is received at a different frequency from what went out: ∆f

Question 5

Very convenient feature:The change of frequency is?

Answer 5

The change of frequency is proportional to the velocity. Faster flow shifts frequency more. Slower flow shifts frequency less.

Question 6

Faster flow shifts frequency ______.

Answer 6

More

Question 7

Slower flow shifts frequency ______.

Answer 7

Less

Question 8

How is the frequency shifted if the flow is toward the beam?

Answer 8

higher

Question 9

Flow toward beam: frequency shifted ______

Answer 9

higher

Question 10

How is the frequency shifted if the flow is away from the beam?

Answer 10

lower

Question 11

Flow away beam: frequency shifted ______

Answer 11

lower

Question 12

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?

Answer 12

Nope.

A shift is a shift.

A plus or a minus doesn’t matter.

They’re both proportional to velocity.

Question 13

Basic functions of Doppler ultrasound in medical diagnostic work:

Answer 13

1. Estimate arterial stenosis
2. Estimate cardiac valve pressure gradient 4 (V2)
3. Evaluate flow character qualitatively (e.g., LE veins)

Question 14

What influences the frequency shift?

Answer 14

1. If the velocity of blood flow increases, the frequency shift increases. Faster flow makes bigger shift.
2. If the operating frequency of the transducer increases,the frequency shift increases.Higher operating frequency makes bigger shift.
3. 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.)

Question 15

On the left: 4 HHz

On the right: 8.2 MHz

(Same velocity, same angle)

Answer 15

On the left: 4 HHz

On the right: 8.2 MHz

(Same velocity, same angle)

Question 16

If the angle of the doppler beam relative to the direction of flow increases, the frequency shift ________.

Higher ange makes _______ shift.

Answer 16

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.)

Question 17

If the operating frequency of the transducer increases,the frequency shift ___________.

Higher operating frequency makes _________ shift.

Answer 17

If the operating frequency of the transducer increases,the frequency shift increases.Higher operating frequency makes bigger shift.

Question 18

If the velocity of blood flow increases, the frequency shift _________.

Faster flow makes _______ shift.

Answer 18

If the velocity of blood flow increases, the frequency shift increases.

Faster flow makes bigger shift.

Question 19

What happens to ∆f with different angles?

Answer 19

Question 20

Radial artery with CW Doppler

Answer 20

Question 21

The simple Doppler equation:

Answer 21

∆f: frequency shift ∆f = 2 fo V
fo: operating frequency (Hz) ___________
V: velocity (m/sec) c
c: speed of sound (a constant)

Question 22

Velocity through a mitral valve is1.0 m/sec.

You’re using 2.5 MHz operating frequency.

What frequency shift will this create?

Question 23

Change of operating frequency?

4 MHz vs. 8 MHz

(Same velocity in radial artery for bothsame angle of about 45°)

Question 24

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?

Answer 24

With a cosine.

Question 25

What on earth is a cosine?

Answer 25

Don’t worry. It’s just a trig function of a given angle.It has a value between 0 and 1.

Question 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

Answer 26

Angle Cosine

0° _______

45° _______

60° _______

90° _______

Question 27

Notice how there’s little change for the _______ angles,

and rapid change for the _______ angles.

Much more error with _______ angles.

Answer 27

Notice how there’s little change for the lower angles,

and rapid change for the higher angles.

Much more error with higher angles.

Question 28

The worst Doppler angle possible?

Answer 28

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.)

Question 29

The Doppler angle that gives the maximum frequency shift?

Answer 29

0°.

The cosine of 0° is 1.00, the maximum possible.

Echo uses 0°.

Question 30

Vascular work—

can’t usually get a 0° angle in vessels,

so we use an intermediate angle, usually _______

Answer 30

50–60°.

Question 31

the more complicated Doppler equation:

Answer 31

∆f = 2 fo V (cos 0 )

_______________

c

0 : angle of beam incidence relative to flow direction

cos: cosine function of that angle

Question 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 _________ __ ____ _____.

Answer 32

Theta refers to the angle of incidence;

the cosine is a function of that angle.

Question 33

Where’s the flow going? Where’s the beam going?

Question 34

Angles: acute and obtuseHow many of each are there?

Question 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?

Question 36

Question 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?

Question 38

We don’t use ∆f to grade stenosis or valve area.

What do we use it for?

Answer 38

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.

Question 39

The actual information that the scanner gets is ________ _________.

What are they caused by?

What is your job when using doppler and why?

Answer 39

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.

Question 40

Solving for velocity instead of frequency:

Answer 40

∆f c
Ve = __________

      2 fo (cos 0 )

[Ridgway uses Ve, meaning velocity estimate, to keep in mind the limitations of our accuracy.]

Question 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.)

Question 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.)

Question 43

Doppler processing:

Answer 43

• Audio output
• Zero-crossing detector
• Spectral analysis (FFT)
• Color flow
• Power Doppler

Question 44

Zero-crossing detector:

Answer 44

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”

Question 45

Spectral analysis:

Fast Fourier Transform

FFT

Answer 45

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.

Question 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.

Answer 46

Question 47

Question 48

Here you can see individual pixels.

Question 49

Elements of the FFT display:

X:

Y:

Z:

Answer 49

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

Question 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?

Answer 50

Question 51

Another trick question:

Can you use spectral analysis with CW Doppler?

It’s only for PW Doppler, right?

Answer 51

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.

Question 52

Aliasing:

Answer 52

Display shows cut-off waveforms with the peaks “wrapped around” on other side of display.

Cause:∆f exceeds the Nyquist limit.

Question 53

Aliasing has to do with _____ ____

Answer 53

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.)

Question 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…

Question 55

The Nyquist–Shannon sampling theorem is a

Answer 55

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.

Question 56

Nyquist limit (Harry Nyquist in 1923) represents:

Answer 56

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.

Question 57

Nyquist limit =

And PRF =

Answer 57

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.

Question 59

The Nyquist limit is a result of the ____.

Answer 59

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.

Question 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?

Answer 60

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.

Question 61

CW has NO ___

Answer 61

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.)

Question 62

What influences the likelihood of aliasing?

Answer 62

Anything that changes PRF or frequency shift.

Question 63

Lower PRF:

Answer 63

lower Nyquist limit

Question 64

From Edelman p. 318

Question 65

What can you do to reduce the likelihood of aliasing?

(And why does each work?)

Answer 65

• 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

Question 66

Left waveform is aliasing:

PRF is low (small scale).

Fixed on right by increasing scale

(increasing PRF, therefore raising the Nyquist limit).

Answer 66

Question 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?

Answer 67

Echo: 12-18 cm depth

Vascular:3-8 cm depth

PRF will be lower for echodue to much more depth.

Why? Travel time…

Question 68

Echo: 0°

Vascular: 50-60°

∆f c
Ve = _________
2 fo (cos q)

Answer 68

Question 69

Another way to deal with aliasing:

Answer 69

“High PRF” mode

Send a pulse before the first one gets back.Mild range ambiguity…

Question 70

Why all the fuss about aliasing?

Why not just use CW all the timeand forget about aliasing?

Answer 70

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.

Question 71

Angle correction:

(Vascular issue, not echo)

Answer 71

Line the angle-correct cursor parallel to the wall

to approximate the flow direction.

Question 72

Another form of Doppler display:

Color flow

Answer 72

Many sample volumes

Process for direction and shift

Display as red or blue (direction)

Display as brighter or darker hue
(shift/velocity)

Question 73

Differences between spectral analysis and color flow:

Answer 73

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)

Question 74

Color flow processing:

Autocorrelation

Answer 74

Each sample is compared to the previous oneto arrive at a rough estimateof mean velocity(not peak).

Question 75

The usual color assignment for echo:

For vascular work:

Answer 75

The usual color assignment for echo:
Red toward, blue away

For vascular work:
Make arteries red, veins blue(usually)
(not always possible)

Question 76

The color bar at the side of the display tells you the color assignment.

Toward the beam is red

Answer 76

Toward the beam is blue

Question 77

Brighter color suggests higher velocity

(It really suggests higher ∆f, for whatever reason:higher velocity, lower angle, etc.)

Question 78

Aliasing due to stenosis?Or due to flow going faster downhill?

Question 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.)

Question 80

Toward or away from beam?

Answer 80

Is flow toward or away from the beam?

Question 81

VERY severe mitral valve regurge

Answer 81

Severe ICA stenosis, with color bruit in tissue

Question 82

Color aliasing:

Same phenomenon as with FFT.

Answer 82

Higher shift than Nyquist limit “wraps around” into opposite color.

Question 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?

Answer 83

No; that was early diastole.This is systole.

Question 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.

Question 85

Important control for Doppler:

Wall filter

(a.k.a. high pass filter)

Answer 85

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.

Question 86

Another form of color flow display:

Power Doppler

(a.k.a. Color angio, etc.)

Answer 86

Puts color on screen if there is a shift, but no directionality.Just yes or no to flow.

Question 87

START PULSES PP REMIX

Question 88

The 7 parameters of ultrasound:

Answer 88

1. Frequency
2. Period
3. Wavelength
4. Propagation speed .
5. Amplitude
6. Power
7. Intensity

Question 89

Answer 89

Question 90

Answer 90

Question 91

Period =

Frequency =

Wavelength:

Speed of sound in average soft tissue:

Answer 91

Period = 1/f

Frequency = 1/P

Wavelength: λ = c/f

Speed of sound in average soft tissue: 1540 m/sec

Question 92

c
__________

λ f

Question 93

Pulses in Ultrasound

Pulse:

Answer 93

Pulse:

A group of cycles;a short burst of ultrasound(“Pulse train”)

Question 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.)

Answer 94

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).

Question 95

PRF:

Answer 95

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.

Question 96

PRP:

Answer 96

pulse repetition period

Length of time from the beginning of a pulse to the beginning of the next pulse

Question 97

PRF and PRP are inverse of each other:

Answer 97

PRF = 1/PRP

PRP = 1/PRF

(Similar to relationship between period and frequency with single cycles)

Question 98

PD:

Answer 98

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)

Question 99

Pulse duration is determined by the construction of the transducer:

Answer 99

frequency of transducer anddamping of vibration(more on this later)

Question 100

Typical PD for diagnostic ultrasound transducers:

Answer 100

0.5 to 3 μsec

Question 101

What makes PRP different from PD?

How long is the PRP?

Question 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?

Answer 102

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

Question 103

If operating frequency is 2 MHz, and each pulse has 200 cycles,what is the pulse duration?

Answer 103

Question 104

Duty factor:

Answer 104

Proportion of time the transducer is actually vibrating to produce ultrasound

(i.e., percent of time the transducer actually works)

Question 105

DF is usually very tiny for diagnostic ultrasound:

Answer 105

1% or less.The rest of the time, the transducer is waiting for echoes to return.

No units - can be expressed as decimal or %

Question 106

Question 107

Question 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.

Answer 108

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 %)

Question 109

Note that the frequency of ultrasound is NOT related to PRF.

Answer 109

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.

Question 110

PD = 1.4 µsec

PRP = 0.5 msec

What is the duty factor?

Answer 110

DF = PD/PRP = 0.0000014 sec
_____________
0.0005 sec

= 0.0028 or 0.28% duty factor

Question 111

Answer 111

Question 112

Answer 112

Question 113

If the duty factor is 0.28%, how many cycles per second in a beam using 3.5 MHz frequency?

Answer 113

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.

Question 114

Let’s say PRF is 4 kHz:

4,000 pulses per second.

Answer 114

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

Question 115

Duty factor of 100% =

Duty factor of 0% =

Answer 115

Duty factor of 100% =always creating sound:continuous wave.

Duty factor of 0% =never creating sound:it’s off.

Question 116

Answer 116

Question 117

Spatial pulse length:

Answer 117

distance of one pulse

SPL = cycles per pulse x λ

Remember: λ = c
__
f

Question 118

High frequency —> ________ wavelength
Low frequency —> _______ wavelength

Answer 118

High frequency —> shorter wavelength
Low frequency —> longer wavelength

Question 119

Medium with faster propagation speed —>
__________ wavelength
Medium with slower propagation speed —>
__________wavelength

Answer 119

Medium with faster propagation speed —>
longer wavelength
Medium with slower propagation speed —>
shorter wavelength

Question 120

Wavelength is to SPL

as Period is to PD

Answer 120

Question 121

Answer 121

Question 122

What is wavelength if frequency = 5 MHz?

Answer 122

λ = 1,540 m/sec

5,000,000 cycles/sec

= 0.000308 m/cycleor 308 µm per cycle

Question 123

If f = 2 MHz, and there are 25 cycles per pulse, what is SPL?

Answer 123

First get the wavelength.

λ = 1540/2,000,000 =
0.00077 m/cycle

Then:

0. 00077 m/cycle x 25 cycles/pulse =
1. 01925 or 0.02 m/pulse SPL

Question 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).

Answer 124

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)

Question 125

The bandwidth represents the range of frequencies in the signal created by the transducer,the number of frequencies produced.

Answer 125

Question 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.)

Answer 126

Imaging transducers have wide bandwidths—you want short SPL for good resolution.

Question 127

Quality factor:

Answer 127

Index of how clean and efficientthe signal is—
Ratio of center frequency to bandwidth

   Operating frequency  QF =  -------------------------------
   bandwidth

    fo  QF =  -------
      BW

Question 128

A transducer has 5 MHz operating frequency.

Quality factor with BW of 2 MHz?

Answer 128

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?)

Question 129

Wider BW —> _______ Q factor

Answer 129

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.)

Question 130

Which has the low Q factor?
Which has the high Q factor?

Answer 130

Question 131

REVIEW FOR 6-WEEK EXAM

Question 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?

Question 133

Toward vs. away—how do you tell?

What is the result on the spectral display or color flow display?

CW vs. pulsed Doppler

Question 134

Processing for different kinds of Doppler:

analog (zero-crossing detector)

spectral (FFT)

color flow (autocorrelation)

Question 135

Spectral display: 3 axes of information

X: Time

Y: Frequency shift (velocity)

Z: Amplitude of shifted echo

indicated by pixel brightness suggesting…what?

Question 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)

Question 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

Question 138

Color flow:

conventions for color assignmentecho vs. vascular application

Question 139

The 7 parameters of ultrasound:

1. Frequency
2. Period
3. Wavelength
4. Propagation speed
5. Amplitude
6. Power
7. Intensity

Question 140

Imaging physics concepts:

PRF
PRP
DF
PD
SPL

Question 141

Duty factor:

possible range

units

interaction with PD and PRP

Question 142

Question 143

Wavelength (refers to one cycle)
Spatial pulse length (refers to group of cycles)

Units:
λ = c
—–
f

SPL = λ x cycles per pulse

Question 144

Relationship between

amplitude and frequency:

Question 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)

Question 146

SPL = wavelength X cycles/pulse

(It’s about distance)

PD = period X cycles/pulse

(It’s about time)

Question 147

Can you draw a duty factor of 100%?

Can you draw a duty factor of 0%?

Question 148

MATH

1 point for formula
1 point for correct work
1 point for correct answer
1 point for correct units

Question 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

Question 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

Question 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)

Question 152

Again:

PRF = pulses per second

PRP = time from start of one pulse to start of next

PRF = 1/PRP

PRP = 1/PRF

Question 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

Question 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

Question 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

Question 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

Question 157

Question 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