01:01:

Introduction Introduction to to UltrasoundUltrasound

George David, M.S.Associate Professor of Radiology

Speakertransmits sound pulses

Microphonereceives echoes

Acts as both speaker & microphone› Emits very short sound pulse› Listens a very long time for returning echoes

Can only do one at a time

Voltage generated when certain materials are deformed by pressure

Reverse also true!› Some materials change dimensions

when voltage applied dimensional change causes pressure

change

› when voltage polarity reversed, so is dimensional change

V

What does your scanner know about the sound echoes it hears?

AcmeUltra-Sound

Co.

I’m a scanner, Jim,

not a magician.

How loud is the echo?

inferred from intensity of electrical pulse from transducer

What was the time delay between sound broadcast

and the echo?

Direction sound was emitted

The sound’s pitch or frequency

Sound travels at 1540 m/s everywhere in body› average speed of sound in soft tissue

Sound travels in straight lines in direction transmitted

Sound attenuated equally by everything in body› (0.5 dB/cm/MHz, soft tissue average)

Sound travels at 1540 m/s everywhere in body› average speed of sound in soft tissue

Sound travels in straight lines in direction transmitted

Sound attenuated equally by everything in body› (0.5 dB/cm/MHz, soft tissue average)

Dot position ideally indicates source of echo

scanner has no way of knowing exact location› Infers location from

echo

?

Scanner aims sound when transmitting

echo assumed to originate from direction of scanner’s sound transmission

ain’t necessarily so

?

Time delay accurately measured by scanner

distance = time delay X speed of sound

distance

scanner assumes speed of sound is that of soft tissue› 1.54 mm/sec› 1540 m/sec› 13 usec required for echo object 1 cm from

transducer (2 cm round trip)

distance = time delay X speed of sound

1 cm13 sec

Handy rule

of thumb

Sometimes

?

soft tissue ==> 1.54 mm / sec

fat ==> 1.44 mm / sec

brain ==> 1.51 mm / sec

liver, kidney ==> 1.56 mm / sec

muscle ==> 1.57 mm / sec

•Luckily, the speed of sound is almost the same for most body parts

Ultrasound is gray shade modality

Gray shade should indicate echogeneity of object

? ?

Based upon intensity (volume, loudness) of echo

? ?

Loud echo = bright dot Soft echo = dim dot

Deep echoes are softer (lower volume) than surface echoes.

Correction needed to compensate for sound attenuation with distance

Otherwise dots close to transducer would be brighter

scanner assumes entire body has attenuation of soft tissue› actual attenuation

varies widely in body

• Fat 0.6

• Brain 0.6

• Liver 0.5

• Kidney 0.9

• Muscle 1.0

• Heart 1.1

Tissue Attenuation Coefficient (dB / cm / MHz)

Assumptions made by scanner cause artifacts when assumed conditions not true

All sound in body travels at same speed Sound travels only in straight lines Sound attenuated equally by everything

in body

All sound in body travels at same speed

Distance = Speed X Time Delay / 2

1380 m/s X 58usec / 2 = 4 cm

Actual Distance to interface

1540 m/s X 58usec / 2 = 4.47 cm

Calculated Distance to interface

AssumedAccurately Measured

Assumed speed

Actual speed

Actual object position

X Imaged object position

X

Incorrect dot placement can result in incorrect› Object placement› Object size› Object shape

Actual object position

X Imaged object position

X

Position of Object on Image

Actual Object Position

X

MultipathArtifact

X

Actual Object Position

X Position of Object on Image X

Refraction

Change in speed of sound causes beam to change direction

reflection from reflector “2” splits at “I”

some intensity re-reflected toward “2”

Result› later false echoes heard› scanner places dots

behind reflector “2”

1

2

Echo #1Echo #2Echo #3

Comet tail› dozens of multiple

reflections between transducer & reflector 2 reflectors

Comet tail› dozens of multiple

reflections between transducer & reflector 2 reflectors

Mirror Image› common around strong

reflectors Diaphragm Pleura

Scanner emits 2nd pulse before all reflections received from 1st pulse

scanner assumes echo from 2nd pulse

places echo too close & in wrong direction

X

Actual Object Position

X Position of Object on Image

Sound attenuated equally by everything in body

Scanner assumes soft tissue attenuation

0.5 dB/cm per MHz

Attenuates more than .5 dB/cm/MHz

ShadowedReflector

http://raddi.uah.ualberta.ca/~hennig/teach/cases/artifact/noframe/imag2-f2.htm

Attenuates less .5

dB/cm/MHz

Enhanced reflector

http://raddi.uah.ualberta.ca/~hennig/teach/cases/artifact/noframe/imag6-f1.htm

Results from random interference between scattered echoes from many reflectors

Ich heisse Johaan Christian Doppler

difference between received & transmitted frequency

caused by relative motion between sound source & receiver

Frequency shift indicative of reflector speed

IN

OUT

change in pitch of as object approaches & leaves observer› train› Ambulance siren

moving blood cells› motion can be presented as sound or as an image

Doppler spectrum speckle Cause

› same as acoustic speckle› random constructive &

destructive interference from sound scattered in blood

duplicate vessel image visible on opposite side of strong reflector

Analogous to mirror image artifact Doppler data also duplicated

Femoral vein duplication in region of adductor canal.

Duplication of left vertebral artery in Doppler ultrasonography: arrows duplicated left vertebral arteries, LC left common carotid artery

Sufficient Sampling

Insufficient Sampling

Results in detection of improper flow direction

occurs because sampling rate too slow

Similar to wagon wheels rotating backwards in movies

Which way is this shape turning?

#1 #2 #3

Does it help to sample more often?

#1 #2#1A

All else staying equal, larger reflector speed must produce larger Doppler shift

77 X fD (kHz)v (cm/s) = -------------------------- fo (MHz) X cos

Lower operating frequency results in lower Doppler shift

77 X fD (kHz)v (cm/s) = -------------------------- fo (MHz) X cos

Constant

larger Doppler angle results in› Lower cos()› Lower Doppler shift

77 X fD (kHz)v (cm/s) = -------------------------- fo (MHz) X cos(

Constant

decrease imaging depth increase pulse repetition

frequency› Increases sampling rate› Lessens aliasing

BUT

› increases likelihood of range ambiguity for pulsed instruments

RangeAmbiguityTrade-offTriangle

Depth

Lines / Frame Frames / sec(dynamics)

operator instructs scanner to assume aliasing occurring› scanner does calculations based on operator’s assumption

scanner cannot independently verify