2. 01: Introduction to Ultrasound George David, M.S. Associate Professor of Radiology

3. Speaker transmits sound pulses Microphone receives 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

4.  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

5.  What does your scanner know about the sound echoes it hears? Acme Ultra- Sound Co. I’m a scanner, Jim, not a magician.

6. How loud is the echo?  inferred from intensity of electrical pulse from transducer

7. What was the time delay between sound broadcast and the echo?

8.  Direction sound was emitted

9. The sound’s pitch or frequency

10.  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)

11.  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)

12.  Dot position ideally indicates source of echo  scanner has no way of knowing exact location › Infers location from echo ?

13.  Scanner aims sound when transmitting  echo assumed to originate from direction of scanner’s sound transmission  ain’t necessarily so ?

14.  Time delay accurately measured by scanner distance = time delay X speed of sound distance

15.  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 cm 13  sec Handy rule of thumb

16.  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

17.  Ultrasound is gray shade modality  Gray shade should indicate echogeneity of object ? ?

18.  Based upon intensity (volume, loudness) of echo ? ?

19.  Loud echo = bright dot  Soft echo = dim dot

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

21.  Correction needed to compensate for sound attenuation with distance  Otherwise dots close to transducer would be brighter

22.  scanner assumes entire body has attenuation of soft tissue › actual attenuation varies widely in body Fat0.6 Brain0.6 Liver0.5 Kidney0.9 Muscle1.0 Heart1.1 Tissue Attenuation Coefficient (dB / cm / MHz)

23.  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

24. 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 Assumed Accurately Measured Assumed speed Actual speed Actual object position X Imaged object position X

25.  Incorrect dot placement can result in incorrect › Object placement › Object size › Object shape Actual object position X Imaged object position X

26. Position of Object on Image Actual Object Position X Multipath Artifact X

28. Actual Object Position X Position of Object on Image X Refraction  Change in speed of sound causes beam to change direction

29.  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

30.  Comet tail › dozens of multiple reflections between  transducer & reflector  2 reflectors

31.  Comet tail › dozens of multiple reflections between  transducer & reflector  2 reflectors  Mirror Image › common around strong reflectors  Diaphragm  Pleura

32.  Scanner emits 2 nd 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

33.  Scanner assumes soft tissue attenuation  0.5 dB/cm per MHz

34. Attenuates more than.5 dB/cm/MHz Shadowed Reflector http://raddi.uah.ualberta.ca/~hennig/teach/cases/artifact/noframe/imag2-f2.htm

35. Attenuates less.5 dB/cm/MHz Enhanced reflector http://raddi.uah.ualberta.ca/~hennig/teach/cases/artifact/noframe/imag6-f1.htm

36.  Results from random interference between scattered echoes from many reflectors

37. Ich heisse Johaan Christian Doppler

38.  difference between received & transmitted frequency  caused by relative motion between sound source & receiver  Frequency shift indicative of reflector speed IN OUT

39.  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

40.  Doppler spectrum speckle  Cause › same as acoustic speckle › random constructive & destructive interference from sound scattered in blood

41.  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

42. Sufficient Sampling Insufficient Sampling

43.  Results in detection of improper flow direction  occurs because sampling rate too slow  Similar to wagon wheels rotating backwards in movies

44.  Which way is this shape turning? #1#2#3

45. Does it help to sample more often? #1 #2 #1A

46.  All else staying equal, larger reflector speed must produce larger Doppler shift  77 X f D (kHz) v (cm/s) = -------------------------- f o (MHz) X cos 

47.  Lower operating frequency results in lower Doppler shift  77 X f D (kHz) v (cm/s) = -------------------------- f o (MHz) X cos  Constant

48.  larger Doppler angle  results in › Lower cos(  ) › Lower Doppler shift  77 X f D (kHz) v (cm/s) = -------------------------- f o (MHz) X cos(  Constant

49.  decrease imaging depth  increase pulse repetition frequency › Increases sampling rate › Lessens aliasing BUT › increases likelihood of range ambiguity for pulsed instruments Range Ambiguity Trade-off Triangle Depth Lines / FrameFrames / sec (dynamics)

50.  operator instructs scanner to assume aliasing occurring › scanner does calculations based on operator’s assumption  scanner cannot independently verify