Basic System. Basic System. Sonosite 180. Acuson Sequoia. Echo occurs at t=2z/c where c is approximately 1500 m/s or 1.5 mm/µs

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1 Bioengineering 280A Principles of Biomedical Imaging Fall Quarter 2007 Ultrasound Lecture Sonosite 80 From Suetens 2002 Acuson Sequoia Basic System Basic System Echo occurs at t=2/c where c is approximately 500 m/s or.5 mm/µs Brunner 2002 Macovski 983

2 Transducer A-Mode (Amplitude) Prince and Links 2006 M-Mode (Motion) B-Mode (Brightness) Brunner

3 B-Mode (Brightness) B-Mode Brunner 2002 Mayo Clinic B-Mode CW Doppler Imaging Credit: Mayo Clinic 3

4 PW Doppler Imaging Color Doppler Imaging Acoustic Waves c = "# [m s- ] Speed of Sound " = compressibility [m s 2 kg - ] = [/Pascal] # = density [kg m -3 ] Material Air Water Bone Fat Liver Density Speed m/s Suetens

5 Acoustic Wave Equation $ " 2 p = # 2 #x + # 2 2 #y + # 2 ' & ) p = # 2 p % 2 # 2 ( c 2 #t 2 Solutions are of the form p(x,t) = A f (x * ct) + A 2 f 2 (x + ct) ( ) p(,t) = cos k( " ct) % = cos 2# ( ' ( " ct) * & $ ) % = cos 2#f ( ' ( " ct) * & c ) ( ) = cos 2#f ( /c " t) Plane Waves p(,t) = exp( jk( " ct) ) k = wavenumber = 2" # = 2"k # = wavelength = c f f = frequency [cycles/sec] T = period = f " = wavelength T = period =/ f t Outward wave Spherical Waves p(r,t) = r "(t # r /c) + "(t + r /c) r Inward wave Impedance Impedance Z = Pressure Velocity = P v = "c = " # Outward wave p(r,t) = exp( j2"f (t # r /c)) r density kg/m 3 speed of sound Brain 54 m/s Liver 549 Skull bone 4080 m/s Water 480 m/s Note: particle velocity and speed of sound are not the same! 5

6 Impedance Acoustic Intensity Z = "c = " # I = pv Material Air Water Bone Fat Liver Density Speed m/s Z (kg/m 2 /s) = p2 Z Also called acoustic energy flux. Analogous to electric power Echos Specular Reflection P i,v i P t,v t Z 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 " P r Z = P t P i + P r = P t (pressure boundary condition) R = P r P i = " Z + Z # $Z Z 0 6

7 Reflection and Refraction Reflection and Refraction v i = v r cos" r + v t cos" t Snell s Law p i Z = p r Z cos" r + p t cos" t p i + p r = p t sin" i c = sin" r c = sin" t c 2 Pressure Reflectivity R = p r p i = # Z cos" t + Z cos" t Pressure Transmittivity T = p t p i = 2 + Z cos" t Reflection and Refraction Example Example : Fat/liver interface at normal incidence Z fat =.35 "0 #6 kg m -2 s - Z liver =.66 "0 #6 kg m -2 s - Intensity Reflectivity Intensity Transmittivity R I = I r = p 2 r I i p = $ # Z cos" t ' 2 & ) i % + Z cos" t ( T I = I t = p 2 t Z = I i p 2 i 4Z cos 2 " i ( ) 2 + Z cos" t 2 R I = $ liver # Z ' fat & % Z liver + Z ) = 0.03 fat ( 7

8 Scattering Attenuation Loss of acoustic energy during propagation. Conversion of acoustic energy into heat. p(,t) = A f (t " c /) µ a = " ln A A 0 = A 0 exp("µ a ) f (t " c /) Amplitude attenuation factor : units = nepers/cm Point scatterers retransmit the incident wave equally in all direction (e.g. isotropic scattering). # = "20 log 0 Attenuation coefficient A = 20µ a log 0 ( e) $ 8.7µ a : db/cm A 0 Attenuation "( f ) = " 0 f n For frequencies used in medical ultrasound, n #. "( f ) # " 0 f Example Example : Fat at 5 MH Attenuation coefficient = 5MH " 0.63 db/cm/mh = 3.5dB/cm Material fat liver Cardiac muscle bone α 0 [db/cm/mh] After 4 cm, attenuation = 4 * 3.5 = 2.6 db Relative amplitude is 0 (-2.6/20) = Recall db # 20log 0 ( A / A 0 ) 8

9 Received signal Received signal e(t) = K $ $$! e"2# R( x, y,)s( x, y) p(t " 2 /c)dxdyd! Beam width Attenuation! Pulse Reflection/Scattering Attenuation Correction Attenuation Correction and Signal Equation e(t) = K $ %K $$ e"#ct ct /2 e"2# R(x, y,)s(x, y) p(t " 2 /c)dxdyd $ $ $ R(x, y,)s(x, y) p(t " 2 /c)dxdyd ec (t) = cte#ct e(t) %K $! = $ $ R(x, y,)s(x, y) p(t " 2 /c)dxdyd c! $$ 2 $ R(x, y,c& /2)s(x, y) p(t " & )dxdyd&! 9

10 Depth Response p(t) Depth Resolution Depth response p(t " 2 0 /c) = p(2 /c " 2 0 /c) # = p 2( " ) & 0 % ( $ c ' T 0 p(t-2 0 /c) Therefore impulse response is simply p(t) in the time domain or p(2 /c) in the spatial domain 2 0 /c (2 0 /c)+t P(2/c-2 0 /c) t 0 0 +ct/2 /f 0 Depth Resolution p(t) = p(2/c) determines the depth resolution Depth Resolution The depth resolution is approximately " = ct/2 #.5c / f 0 =.5$. T 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 $ = ct/2 %.5c / f 0 =.5&. Example : For f 0 = 5 MH, $ = c/f = (500m /s) /(5 %0 6 H) = 0.3mm " =.5$ = 0.45 mm Trade - off Higher f 0 & Smaller " & but more attenuation Example : Assume db/cm/mh For 0 cm depth, 20 cm roundtrip path length. At MH 20 db of attenuation & Attenuation = 0. At 0 MH 200 db of attenuation & Attenuation = x0-0 0

11 Depth of Penetration Assume system can handle L db of loss, then " A L = 20log % 0 $ ' # A 0 & We also have the definition ( = - 20log " A % 0$ ' # A 0 & and the approximation ( = ( 0 f Total range a wave can travel before attenuation L is = L ( 0 f Depth of penetration is d p = L 2( 0 f Depth of Penetration Assume L = 80 db; α 0 = db/cm/mh Frequency (MH) Depth of Penetration(cm) Pulse Repetition and Frame Rate Example Need to wait for echoes to return before transmitting new pulse Pulse repetition interval is T R " 2d p c Pulse repetition rate is f R = T R If N pulses are required to form an image, then the frame rate is F = NT R N = 256, L= 80dB, c =540m /s, " 0 =db /cm / MH What frequency should be used to achieve a frame rate of 5 frame/sec? T R = FN = 0.26ms T R # 2d p c = L " 0 fc f # L =.99MH act R

12 Single-slit Diffraction Square Aperture Diffraction Source:wikipedia Source:wikipedia Huygen s Principle Huygen s Principle r 0 Source:wikipedia Wavenumber k = 2" # Oliquity Factor Anderson and Trahey

13 Two-Slit Interference Plane Wave (Fraunhofer) Approximation Target x 0 r 0 Source d θ θ d = "x sin# x Assume overall distance for phase calculation is approximately +d Source:wikipedia sin# = x 0 $ x 0 r 0 d $ " x 0x r exp( jkr) " Plane Wave Approximation ( ) = exp j 2# ( $ % x 0x ( exp jk ( + d ), U(x 0 ) = - s(x ) %, r exp( jkr)dx, " - s(x ) %, exp & j 2# & $ % x 0x )) ( ( + + dx ' ' * * = & 2#), & exp( j + - s(x ' $ * ) exp % j2#x 0x ) ( + dx %, ' $ * = & 2#), exp( j + - s(x ' $ * ) exp (% j2#k %, x x )dx = & 2#) exp( j + F[ s(x) ] ' $ * k x = x 0 $ & ' & ' )) + + ** In general Plane Wave Approximation U(x 0,y 0 ) = $ 2"' exp& j ) F[ s(x, y) ] % # ( k x = x 0 #,k y = y 0 #, Example s(x, y) = rect(x /D)rect(y /D) U(x 0,y 0 ) = exp( jk)d2 sinc( Dk x )sinc( Dk y ) = exp( $ jk)d2 sinc D x ' $ 0 y & ) sinc Dk 0 ' & y ) % #( % #( Zeros occur at x 0 = n# D and y 0 = n# D Beamwidth of the sinc function is # D 3

14 Example D 2 " Sidelobes D " D 2 D 2 2 " D 2 " D " x %) " x % rect$ ' rect$ # D& # d& '( d comb " x %, + $ # d '. / Dsinc(Dk * & x ) ( d sinc(dk x )comb(dk x ) - [ ] Question: What should we do to reduce the sidelobes? Anderson and Trahey