Bioengineering 280A Principles of Biomedical Imaging. Fall Quarter 2006 Ultrasound Lecture 1

Transcription

1 Bioengineering 280A Principles of Biomedical Imaging Fall Quarter 2006 Ultrasound Lecture 1 From Suetens

2 Basic System Echo occurs at t=2z/c where c is approximately 1500 m/s or 1.5 mm/µs Macovski 1983 Basic System Brunner

3 Transducer Prince and Links 2006 A-Mode (Amplitude) 3

4 M-Mode (Motion) B-Mode (Brightness) Brunner

5 B-Mode (Brightness) Brunner 2002 B-Mode 5

6 Harmonic Imaging CW Doppler Imaging 6

7 PW Doppler Imaging Color Doppler Imaging 7

8 Acoustic Waves Suetens 2002 c = 1 "# [m s-1 ] Speed of Sound " = compressibility [m s 2 kg -1 ] = [1/Pascal] # = density [kg m -3 ] Material Air Water Bone Fat Liver Density Speed m/s

9 Impedance Impedance Z = Pressure Velocity = P v = "c = " # density kg/m 3 speed of sound Brain 1541 m/s Liver 1549 Skull bone 4080 m/s Water 1480 m/s Note: particle velocity and speed of sound are not the same! Acoustic Wave Equation $ " 2 p = # 2 #x + # 2 2 #y + # 2 ' & ) p = 1 # 2 p % 2 #z 2 ( c 2 #t 2 Solutions are of the form p(x,t) = A 1 f 1 (x * ct) + A 2 f 2 (x + ct) 9

10 ( ) p(z,t) = cos k(z " ct) % = cos 2# ( ' (z " ct) * & $ ) % = cos 2#f ( ' (z " ct) * & c ) ( ) = cos 2#f (z /c " t) Plane Waves p(z,t) = exp( jk(z " ct) ) k = wavenumber = 2" # = 2"k z # = wavelength = c f f = frequency [cycles/sec] T = period = 1 f z " = wavelength T = period =1/ f t Outward wave Spherical Waves p(r,t) = 1 r "(t # r /c) + 1 "(t + r /c) r Inward wave Outward wave p(r,t) = 1 exp( j2"f (t # r /c)) r 10

11 Acoustic Intensity I = pv = p2 Z Also called acoustic energy flux. Analogous to electric power Echos 11

12 Specular Reflection P i,v i P t,v t Z 1 Z 2 P r,v r v i " v r = v t (velocity boundary condition) Material Brain-skull Fat-muscle Muscle-blood Soft-tissue-air Reflectivity P i Z 1 " P r Z 1 = P t Z 2 P i + P r = P t (pressure boundary condition) R = P r P i = Z 2 " Z 1 Z 2 + Z 1 # $Z Z 0 Reflection and Refraction Snell s Law sin" i c 1 = sin" r c 1 = sin" t c 2 12

13 Reflection and Refraction v i cos" i = v r cos" r + v t cos" t p i Z 1 cos" i = p r Z 1 cos" r + p t Z 2 cos" t p i + p r = p t Pressure Reflectivity Pressure Transmittivity R = p r = Z cos" # Z cos" 2 i 1 t p i Z 2 cos" i + Z 1 cos" t T = p t 2Z = 2 cos" i p i Z 2 cos" i + Z 1 cos" t Reflection and Refraction Intensity Reflectivity R I = I r = p 2 r I i p = $ Z cos" # Z cos" ' 2 i 1 t 2 & ) i % Z 2 cos" i + Z 1 cos" t ( 2 Intensity Transmittivity T I = I t = p 2 t Z 1 = I i p 2 i Z 2 4Z 1 Z 2 cos 2 " i ( ) 2 Z 2 cos" i + Z 1 cos" t 13

14 Example Example : Fat/liver interface at normal incidence Z fat =1.35 "10 #6 kg m -2 s -1 Z liver =1.66 "10 #6 kg m -2 s -1 $ R I = Z liver # Z ' fat & % Z liver + Z ) fat ( 2 = Scattering Point scatterers retransmit the incident wave equally in all direction (e.g. isotropic scattering). 14

15 Attenuation Loss of acoustic energy during propagation. Conversion of acoustic energy into heat. p(z,t) = A z f (t " c /z) µ a = " 1 z ln A z A 0 = A 0 exp("µ a z) f (t " c /z) Amplitude attenuation factor # = "20 1 z log 10 Attenuation coefficient : units = nepers/cm A z = 20µ a log 10 ( e) $ 8.7µ a : db/cm A 0 "( f ) = " 0 f n Attenuation For frequencies used in medical ultrasound, n #1. "( f ) # " 0 f Material fat liver Cardiac muscle bone α 0 [db/cm/mhz]

16 Example Example : Fat at 5 MHz Attenuation coefficient = 5MHz " 0.63 db/cm/mhz = 3.15dB/cm After 4 cm, attenuation = 4 * 3.15 = 12.6 db Relative amplitude is 10 (-12.6/20) = Recall db # 20log 10 ( A z / A 0 ) Received signal e "2#z e(t) = K $ $ $ R(x,y,z)s(x,y) p(t " 2z /c)dxdydz z Beam width Attenuation Reflection/Scattering Pulse 16

17 Received signal! Attenuation Correction! 17

18 Attenuation Correction and Signal Equation e "2#z e(t) = K $ $ $ R(x, y,z)s(x, y)p(t " 2z /c)dxdydz z % K e"#ct ct /2 $ $ $ R(x, y,z)s(x, y)p(t " 2z /c)dxdydz e c (t) = cte #ct e(t) $ $ $ % K R(x, y,z)s(x, y)p(t " 2z /c)dxdydz = c 2 $ $ $ R(x, y,c& /2)s(x, y)p(t " &)dxdyd& e c (t) = K # # # # # # Impulse Response Transducer centered at 0,0 (defines image at 0,0) R(x, y,z)s(x, y)p(t " 2z /c)dxdydz = K $(x, y,z " z 0 )s(x, y)p(t " 2z /c)dxdydz = Ks(0,0) p(t " 2z 0 /c) Transducer centered at x 0,y 0 (defines image at these coordinates) e c (t) = K # # # # # # R(x, y,z)s(x " x 0,y " y 0 )p(t " 2z /c)dxdydz = K $(x, y,z " z 0 )s(x " x 0,y " y 0 )p(t " 2z /c)dxdydz = Ks("x 0,"y 0 ) p(t " 2z 0 /c) Lateral Response is therefore s(-x,-y) 18

19 Impulse Response Depth response p(t " 2z 0 /c) = p(2z /c " 2z 0 /c) # = p 2(z " z ) & 0 % ( $ c ' Therefore impulse response is simply p(t) in the time domain or p(2z /c) in the spatial domain Signal Equation In general, we can write e c (t, x 0, y 0 ) = K # # # R(x, y,z)s(x " x 0, y " y 0 )p(t " 2z /c)dxdydz = K c [ R(x,y,ct /2) $$$ s("x,"y) p(t) ] 2 # # # x= x 0,y= y 0 e c ( z %, x 0, y 0 ) = K R(x, y,z)s(x " x 0,y " y 0 )p(2( z % " z) /c)dxdydz [ ] x= x0,y= y 0 = R(x, y, z %) $$$ s("x,"y)p(2 z %/c) 19

20 Depth Resolution p(t) = p(2z/c) determines the depth resolution Pulses are of the form a(t)cos(2"f 0 t +#) where a(t) is the envelope function and f 0 is the resonant frequency of the transducer. The duration of T of a(t) is typically chosen to be about 2 or 3 periods (e.g. T = 3/f 0 ). If the duration is too short, the bandwidth of the pulse will be very large and much of its power will be attenuated. The depth resolution is approximately $z = ct/2 %1.5c / f 0 =1.5&. Depth Resolution The depth resolution is approximately "z = ct/2 #1.5c / f 0 =1.5$. Example : For f 0 = 5 MHz, "z = (1.5)(1500m / s) /(5 %10 6 Hz) = 0.45 mm Trade -off Higher f 0 & Smaller "z & but more attenuation Example : Assume 1dB/cm/MHz For 10 cm depth, 20 cm roundtrip path length. At 1 MHz 20 db of attenuation & Attenuation = 0.1 At 10 MHz 200 db of attenuation & Attenuation = 1x