1!Bioengineering 280A"Principles of Biomedical Imaging"Fall Quarter 2009"Ultrasound Lecture 1"From Suetens 2002!Sonosite 180!Acuson Sequoia!See also: http://www.youtube.com/watch?v=7gU1uSlxKDs!Basic System!Macovski 1983!Echo occurs at t=2z/c where c is approximately!1500 m/s or 1.5 mm/µs !Basic System!Brunner 2002!2!Transducer!Prince and Links 2006!A-Mode (Amplitude)!Seutens 2002!M-Mode (Motion)!Seutens 2002!B-Mode (Brightness)!Brunner 2002!Seutens 2002!3!B-Mode (Brightness)!Brunner 2002!B-Mode!Seutens 2002!Mayo Clinic!B-Mode!Credit: Mayo Clinic!Seutens 2002!CW Doppler Imaging!4!Seutens 2002!PW Doppler Imaging!Seutens 2002!Color Doppler Imaging!Acoustic Waves!Suetens 2002!Speed of Sound!€ c =1κρ [m s-1]κ= compressibility [m s2 kg-1] = [1/Pascal]ρ= density [kg m-3]Material! Density! Speed m/s!Air! 1.2! 330!Water! 1000! 1480!Bone! 1380-1810! 4080!Fat! 920! 1450!Liver! 1060! 1570!5!Acoustic 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 2002!Seutens 2002!Plane Waves!€ p( z,t) = cos k(z − ct)( )= cos2πλ( z − ct)⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = cos2πfc(z − ct)⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = cos 2πf (z /c − t)( )€ p( z,t) = exp jk(z − ct)( )€ k = wavenumber =2πλ= 2πkzλ= wavelength =cff = frequency [cycles/sec]T = period =1f€ λ= wavelength€ T = period = 1/ fz! t!Spherical Waves!€ p( r,t) =1rexp( j2πf (t − r /c))€ p( r,t) =1rφ(t − r /c) +1rφ( t + r /c)Outward wave!Inward wave!Outward wave!Note: The phase depends on both space and time. At a given time, wavefront occurs at r = ct. At a given location, wavefront arrives at t = r/c. !Impedance!density kg/m3!speed of sound !Brain 1541 m/s!Liver 1549!Skull bone 4080 m/s!Water 1480 m/s"€ Impedance Z = PressureVelocity=Pv=ρc =ρκNote: particle velocity and speed of sound are not the same!!6!Impedance!Material! Density! Speed m/s! Z (kg/m2/s)!Air! 1.2! 330! 0.0004!Water! 1000! 1480! 1.5!Bone! 1380-1810! 4080! 3.75-7.38!Fat! 920! 1450! 1.35!Liver! 1060! 1570! 1.64-1.68!€ Z =ρc =ρκAcoustic Intensity!€ I = pv=p2ZAlso called acoustic energy flux. Analogous to electric power !Echos!Specular Reflection!€ vi− vr= vt (velocity boundary condition)PiZ1−PrZ1=PtZ2Pi+ Pr= Pt (pressure boundary condition)R =PrPi=Z2− Z1Z2+ Z1≈ΔZZ0€ Pi,vi€ Pt,vt€ Pr,vr€ Z1€ Z2Material! Reflectivity!Brain-skull! 0.66!Fat-muscle! 0.10!Muscle-blood! 0.03!Soft-tissue-air! .9995!7!Reflection and Refraction!Seutens 2002!€ sinθic1=sinθrc1=sinθtc2Snell’s Law!Reflection and Refraction!Seutens 2002!€ vicosθi= vrcosθr+ vtcosθtpiZ1cosθi=prZ1cosθr+ptZ2cosθtpi+ pr= ptR =prpi=Z2cosθi− Z1cosθtZ2cosθi+ Z1cosθtT =ptpi=2Z2cosθiZ2cosθi+ Z1cosθtPressure Reflectivity!Pressure Transmittivity!Reflection and Refraction!Seutens 2002!€ RI=IrIi=pr2pi2=Z2cosθi− Z1cosθtZ2cosθi+ Z1cosθt⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2TI=ItIi=pt2Z1pi2Z2=4Z1Z2cos2θiZ2cosθi+ Z1cosθt( )2Intensity Reflectivity!Intensity Transmittivity!Example!€ € 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= 0.1038!Scattering!€ Point scatterers retransmit the incident wave equallyin all direction (e.g. isotropic scattering). Attenuation!€ 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/cmα= −201zlog10AzA0= 20µalog10e( )≈ 8.7µa : dB/cmAttenuation coefficient!Amplitude attenuation factor!Attenuation!€ α( f ) =α0fnFor frequencies used in medical ultrasound, n ≈1.α( f ) ≈α0fMaterial! α0 [dB/cm/MHz]!fat! 0.63!liver! 0.94!Cardiac muscle! 1.8!bone! 20.0!Example!€ € 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( )9!Received signal!€ http://radiographics.rsnajnls.org/content/vol23/issue4/images/large/g03jl25c1x.jpeg!Received signal!€ € e(t) = Ke−2αzz∫∫∫R( x, y,z)s(x, y) p(t − 2z / c)dxdydzAttenuation!Reflection/Scattering!Beam width!Pulse!Attenuation Correction!€ Attenuation Correction and Signal Equation!€ € e(t) = Ke−2αzz∫∫∫R(x, y,z)s(x, y) p(t − 2z /c)dxdydz≈ Ke−αctct /2R(x, y,z)s(x, y) p(t − 2z /c)dxdydz∫∫∫ec(t) = cteαcte(t)≈ K R(x, y,z)s(x, y) p(t − 2z /c)dxdydz∫∫∫=c2R(x, y,cτ/2)s(x, y) p(t −τ)dxdydτ∫∫∫10!Depth 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 domainDepth Resolution!€ z!p(t)!t!2z0/c! (2z0/c)+T!T!z0!p(t-2z0/c)!€ z!z0!z0+cT/2!P(2z/c-2z0/c)!Depth Resolution!€ € p(t) = p(2z/c) determines the depth resolution Pulses are of the form a(t)cos(2πf0t +θ) 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 Δz = cT/2 ≈ 1.5c / f0= 1.5λ.T!1/f0!Depth Resolution!€ € The depth resolution is approximately Δz = cT/2 ≈1.5c / f0= 1.5λ.Example : For f0= 5 MHz, λ= c/f = (1500m /s)/(5 ×106Hz) = 0.3mm Δz = 1.5λ= 0.45 mmTrade - offHigher f0⇒ Smaller Δz ⇒ but more attenuationExample : Assume 1dB/cm/MHzFor 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 = 1x10-1011!Depth of Penetration!€ Assume system can handle L dB of loss, thenL = 20log10AzA0⎛ ⎝ ⎜ ⎞ ⎠ ⎟ We also have the definition α= -1z20log10AzA0⎛ ⎝ ⎜ ⎞ ⎠ ⎟ and the approximationα=α0fTotal range a wave can travel before attenuation L isz = Lα0fDepth of penetration isdp=L2α0fDepth of Penetration!Frequency (MHz)!Depth of Penetration(cm)!1! 40!2! 20!3! 13!5! 8!10! 4!20! 2!Assume L = 80 dB; α0= 1dB/cm/MHz!Pulse Repetition and
View Full Document