1TT Liu, BE280A, UCSD, Fall 2006Bioengineering 280APrinciples of Biomedical ImagingFall Quarter 2006Ultrasound Lecture 1TT Liu, BE280A, UCSD, Fall 2006From Suetens 20022TT Liu, BE280A, UCSD, Fall 2006Basic SystemMacovski 1983Echo occurs at t=2z/c where c is approximately1500 m/s or 1.5 mm/µs TT Liu, BE280A, UCSD, Fall 2006Basic SystemBrunner 20023TT Liu, BE280A, UCSD, Fall 2006TransducerPrince and Links 2006TT Liu, BE280A, UCSD, Fall 2006A-Mode (Amplitude)Seutens 20024TT Liu, BE280A, UCSD, Fall 2006M-Mode (Motion)Seutens 2002TT Liu, BE280A, UCSD, Fall 2006B-Mode (Brightness)Brunner 2002Seutens 20025TT Liu, BE280A, UCSD, Fall 2006B-Mode (Brightness)Brunner 2002TT Liu, BE280A, UCSD, Fall 2006B-ModeSeutens 20026TT Liu, BE280A, UCSD, Fall 2006Seutens 2002Harmonic ImagingTT Liu, BE280A, UCSD, Fall 2006Seutens 2002CW Doppler Imaging7TT Liu, BE280A, UCSD, Fall 2006Seutens 2002PW Doppler ImagingTT Liu, BE280A, UCSD, Fall 2006Seutens 2002Color Doppler Imaging8TT Liu, BE280A, UCSD, Fall 2006Acoustic WavesSuetens 2002TT Liu, BE280A, UCSD, Fall 2006Speed of Sound! c =1"# [m s-1]"= compressibility [m s2 kg-1] = [1/Pascal]#= density [kg m-3]15701060Liver1450920Fat40801380-1810Bone14801000Water3301.2AirSpeed m/sDensityMaterial9TT Liu, BE280A, UCSD, Fall 2006Impedancedensity kg/m3speed of soundBrain 1541 m/sLiver 1549Skull bone 4080 m/sWater 1480 m/s! Impedance Z = PressureVelocity=Pv="c ="#Note: particle velocity and speed of sound are not the same!TT Liu, BE280A, UCSD, Fall 2006Acoustic Wave Equation! "2p =#2#x2+#2#y2+#2#z2$ % & ' ( ) p =1c2#2p#t2Solutions are of the formp(x,t) = A1f1( x * ct) + A2f2( x + ct)Seutens 2002Seutens 200210TT Liu, BE280A, UCSD, Fall 2006Plane Waves! p(z,t) = cos k(z " ct)( )= cos2#$(z " ct)% & ' ( ) * = cos2#fc(z " ct)% & ' ( ) * = cos 2#f (z /c " t)( )! p(z,t) = exp jk(z " ct)( )! k = wavenumber =2"#= 2"kz#= wavelength =cff = frequency [cycles/sec]T = period =1f! "= wavelength! T = period = 1/ fz tTT Liu, BE280A, UCSD, Fall 2006Spherical Waves! p(r,t) =1rexp( j2"f (t # r /c))! p(r,t) =1r"(t # r /c) +1r"(t + r /c)Outward waveInward waveOutward wave11TT Liu, BE280A, UCSD, Fall 2006Acoustic Intensity! I = pv=p2ZAlso called acoustic energy flux.Analogous to electric powerTT Liu, BE280A, UCSD, Fall 2006Echos12TT Liu, BE280A, UCSD, Fall 2006Specular Reflection! vi" vr= vt (velocity boundary condition)PiZ1"PrZ1=PtZ2Pi+ Pr= Pt (pressure boundary condition)R =PrPi=Z2" Z1Z2+ Z1#$ZZ0! Pi,vi! Pt,vt! Pr,vr! Z1! Z2.9995Soft-tissue-air0.03Muscle-blood0.10Fat-muscle0.66Brain-skullReflectivityMaterialTT Liu, BE280A, UCSD, Fall 2006Reflection and RefractionSeutens 2002! sin"ic1=sin"rc1=sin"tc2Snell’s Law13TT Liu, BE280A, UCSD, Fall 2006Reflection and RefractionSeutens 2002! vicos"i= vrcos"r+ vtcos"tpiZ1cos"i=prZ1cos"r+ptZ2cos"tpi+ pr= ptR =prpi=Z2cos"i# Z1cos"tZ2cos"i+ Z1cos"tT =ptpi=2Z2cos"iZ2cos"i+ Z1cos"tPressure ReflectivityPressure TransmittivityTT Liu, BE280A, UCSD, Fall 2006Reflection and RefractionSeutens 2002! RI=IrIi=pr2pi2=Z2cos"i# Z1cos"tZ2cos"i+ Z1cos"t$ % & ' ( ) 2TI=ItIi=pt2Z1pi2Z2=4Z1Z2cos2"iZ2cos"i+ Z1cos"t( )2Intensity ReflectivityIntensity Transmittivity14TT Liu, BE280A, UCSD, Fall 2006Example! ! Example : Fat/liver interface at normal incidenceZfat= 1.35 "10#6 kg m-2 s-1Zliver= 1.66 "10#6kg m-2 s-1RI=Zliver# ZfatZliver+ Zfat$ % & & ' ( ) ) 2= 0.103TT Liu, BE280A, UCSD, Fall 2006Scattering! Point scatterers retransmit the incident wave equallyin all direction (e.g. isotropic scattering).15TT Liu, BE280A, UCSD, Fall 2006Attenuation! Loss of acoustic energy during propagation. Conversion of acoustic energy into heat.p(z,t) = Azf (t " c /z) = A0exp("µaz) f (t " c / z)µa= "1zlnAzA0 : units = nepers/cm#= "201zlog10AzA0= 20µalog10e( )$ 8.7µa : dB/cmAttenuation coefficientAmplitude attenuation factorTT Liu, BE280A, UCSD, Fall 2006Attenuation! "( f ) ="0fnFor frequencies used in medical ultrasound, n # 1."( f ) #"0f20.0bone1.8Cardiac muscle0.94liver0.63fatα0 [dB/cm/MHz]Material16TT Liu, BE280A, UCSD, Fall 2006Example! ! Example : Fat at 5 MHzAttenuation coefficient = 5MHz " 0.63 dB/cm/MHz = 3.15dB/cmAfter 4 cm, attenuation = 4 * 3.15 = 12.6 dBRelative amplitude is 10(-12.6/20)= 0.2344Recall dB # 20log10Az/ A0( )TT Liu, BE280A, UCSD, Fall 2006Received signal! ! e(t) = Ke"2#zz$$$R(x, y,z)s( x, y) p(t " 2z /c)dxdydzAttenuationReflection/ScatteringBeam widthPulse17TT Liu, BE280A, UCSD, Fall 2006Received signal! http://radiographics.rsnajnls.org/content/vol23/issue4/images/large/g03jl25c1x.jpegTT Liu, BE280A, UCSD, Fall 2006Attenuation Correction!18TT Liu, BE280A, UCSD, Fall 2006Attenuation Correction and Signal Equation! ! e(t) = Ke"2#zz$$$R(x, y ,z)s(x, y) p(t " 2z /c)dxdydz% Ke"#ctct /2R(x, y ,z)s(x, y) p(t " 2z /c)dxdydz$$$ec(t) = cte#cte(t)% K R(x, y ,z)s(x, y) p(t " 2z /c)dxdydz$$$=c2R(x, y,c&/2)s(x, y) p(t "&)dxdyd&$$$TT Liu, BE280A, UCSD, Fall 2006Impulse Response! ! ec(t) = K R(x, y,z)s(x, y) p(t " 2z /c)dxdydz###= K$(x, y,z " z0)s(x, y) p(t " 2z /c)dxdydz###= Ks(0,0) p(t " 2z0/c)Transducer centered at 0,0 (defines image at 0,0)Transducer centered at x0,y0 (defines image at these coordinates)! ec(t) = K R(x, y,z)s(x " x0, y " y0) p(t " 2z /c)dxdydz###= K$(x, y,z " z0)s(x " x0, y " y0) p(t " 2z /c)dxdydz###= Ks("x0,"y0) p(t " 2z0/c)! Lateral Response is therefore s(-x,-y)19TT Liu, BE280A, UCSD, Fall 2006Impulse Response! ! p(t " 2z0/c) = p(2z /c " 2z0/c)= p2(z " z0)c# $ % & ' ( ! Depth response! Therefore impulse response is simplyp(t) in the time domain orp(2z /c) in the spatial domainTT Liu, BE280A, UCSD, Fall 2006Signal Equation! ! In general, we can write ec(t, x0, y0) = K R(x, y,z)s(x " x0, y " y0) p(t " 2z /c)dxdydz### = Kc2R(x, y,ct /2) $$$ s("x,"y) p(t)[ ]x= x0,y= y0 ec(% z , x0, y0) = K R(x, y,z)s(x " x0, y " y0) p(2(% z " z) /c)dxdydz### = R(x, y,% z )$$$ s("x,"y) p(2% z /c)[ ]x= x0,y= y020TT Liu, BE280A, UCSD, Fall 2006Depth Resolution! ! p(t) = p(2z/c) determines the depth resolution Pulses are of the form a(t) cos(2"f0t +#) where a(t) is the envelope function and f0 is the resonantfrequency of the transducer.The duration of T of a(t) is typically chosen to be about 2 or 3 periods (e.g. T = 3/f0). If the duration is too short, the bandwidth of thepulse will be very large and much of its power will be attenuated.The depth resolution is approximately $z = cT/2 %1.5c / f0=1.5&.TT Liu, BE280A, UCSD, Fall 2006Depth Resolution! ! The depth resolution is approximately "z = cT/2 #1.5c / f0=1.5$.Example : For f0= 5
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