1TT Liu, BE280A, UCSD, Fall 2004Bioengineering 280APrinciples of Biomedical ImagingFall Quarter 2004Ultrasound Lecture 1TT Liu, BE280A, UCSD, Fall 2004From Suetens 2002TT Liu, BE280A, UCSD, Fall 2004Basic SystemMacovski 1983Echo occurs at t=2z/c where c is approximately1500 m/s or 1.5 mm/µs2TT Liu, BE280A, UCSD, Fall 2004Basic SystemMacovski 1983Brunner 2002TT Liu, BE280A, UCSD, Fall 2004TransducerSeutens 2002www.engineering.uiowa.edu/~bme_285/ Lecture/Notes%20on%20Ultrasound.pdfTT Liu, BE280A, UCSD, Fall 2004A-Mode (Amplitude)Seutens 20023TT Liu, BE280A, UCSD, Fall 2004M-Mode (Motion)Seutens 2002TT Liu, BE280A, UCSD, Fall 2004B-Mode (Brightness)Brunner 2002Seutens 2002TT Liu, BE280A, UCSD, Fall 2004B-Mode (Brightness)Brunner 20024TT Liu, BE280A, UCSD, Fall 2004B-ModeSeutens 2002TT Liu, BE280A, UCSD, Fall 2004Seutens 2002Harmonic ImagingTT Liu, BE280A, UCSD, Fall 2004Seutens 2002CW Doppler Imaging5TT Liu, BE280A, UCSD, Fall 2004Seutens 2002PW Doppler ImagingTT Liu, BE280A, UCSD, Fall 2004Seutens 2002Color Doppler ImagingTT Liu, BE280A, UCSD, Fall 2004Acoustic WavesSuetens 20026TT Liu, BE280A, UCSD, Fall 2004Acoustic 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 2002TT Liu, BE280A, UCSD, Fall 2004Acoustic Wave Equation€ Solutions of the wave equationPlane wavep(z,t) = exp( j2πf (t − z /c))Superposition of plane wavesp(z,t) = P( f )−∞∞∫exp( j2πf (t − z /c))dfAt z = 0 : p(0,t) ≡ p(t) = P( f )−∞∞∫exp( j2πft)df = F−1P( f )( )p(z,t) = p(0,t − z /c) = p(t − z /c)Spherical Wavep(r,t) =1rexp( j2πf (t − r /c))TT Liu, BE280A, UCSD, Fall 2004Impedancedensity kg/m3speed of soundBrain 1541 m/sLiver 1549Skull bone 4080 m/sWater 1480 m/s€ Impedance Z = PressureVelocity=Pv=ρc7TT Liu, BE280A, UCSD, Fall 2004EchosTT Liu, BE280A, UCSD, Fall 2004Specular Reflection€ vi− vr= vt (velocity boundary condition)PiZ1−PrZ1=PtZ2Pi+ Pr= Pt (pressure boundary condition)R =PrPi=Z2− Z1Z2+ Z1≈ΔZZ0€ Pi,vi€ Pi,vi€ Pi,vi€ Z1€ Z2.9995Soft-tissue-air0.03Muscle-blood0.10Fat-muscle0.66Brain-skullReflectivityMaterialTT Liu, BE280A, UCSD, Fall 2004Reflection and RefractionSeutens 2002€ sinθic1=sinθrc1=sinθtc2R =PrPi=Z2cosθi− Z1cosθtZ2cosθi+ Z1cosθ8TT Liu, BE280A, UCSD, Fall 2004Scattering€ Point scatterers retransmit the incident wave equallyin all direction (e.g. isotropic scattering). TT Liu, BE280A, UCSD, Fall 2004Attenuation€ Loss of acoustic energy during propagation. Conversion of acoustic energy into heat.H( f ,z) = exp(-α(f)z) ≈ exp(-α0fnz)For frequencies used in medical ultrasound, n ≈1.In liver, typical value is α0= 0.5 db/cm/MHzFor tissues in general, α0≈1.0 db/cm/MHzExample : Liver at 2 MHz, attenuation 1 dB/cm.After the 6 cm, 6 dB of attenuation.10^(-3/20) = 0.5Recall dB ≡ 20log10Az/ A0( )TT Liu, BE280A, UCSD, Fall 2004Received signal€ € e(t) = Ke−2αzz∫∫∫R(x, y,z)s( x,y) p(t − 2z /c)dxdydzAttenuationReflection/ScatteringBeam widthPulse9TT Liu, BE280A, UCSD, Fall 2004Received signal€ http://radiographics. rsnajnls .org/content/vol23/issue4/images/large/g03jl25c1x.jpegTT Liu, BE280A, UCSD, Fall 2004Attenuation Correction€ TT Liu, BE280A, UCSD, Fall 2004Attenuation 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∫∫∫= Kc2R(x, y, ct /2)∗ ∗∗ s(−x,−y) p(t)[ ]x=0,y=010TT Liu, BE280A, UCSD, Fall 2004Signal Equation Example€ € ec(t) ≈ K R(x,y,z)s( x, y) p(t − 2z / c)dxdydz∫∫∫ = Kc2R(x,y,ct /2) ∗∗ ∗ s(− x,−y) p(t)[ ]x=0,y=0Let R(x,y,z) =δ( x)δ( y)δ(z − z0) + δ( x)δ( y)δ(z − z1)and s( x, y) = rect(x / L)rect( y / L)ec(t) = Kδ( x)δ( y)δ(z − z0) + δ( x)δ( y)δ(z − z1)[ ]rect( x / L) rect( y / L) p(t − 2z / c)dxdydz∫∫∫ = K p(t − 2z0/c) + p(t − 2z1/c)[ ]TT Liu, BE280A, UCSD, Fall 2004Signal Equation Example€ € ec(t) ≈ K R(x,y,z)s( x, y) p(t − 2z /c)dxdydz∫∫∫ = K R( x, y,ct / 2)∗ ∗∗ s(−x,−y) p(t)[ ]x=0,y=0Let R(x,y,z) =δ(x)δ(y)δ(z − z0) + δ(x)δ(y)δ(z − z1)What happens to the signal as we move the transducerto some arbitrary position x0, y0?ec(t, x0, y0) = Kδ(x)δ(y)δ(z − z0) + δ(x)δ(y)δ(z − z1)[ ]s( x − x0, y − y0) p(t − 2z / c)dxdydz∫∫∫ = K p(t − 2z0/c)s(−x0,−y0) + p(t − 2z1/c)s(−x0,−y0)[ ]TT Liu, BE280A, UCSD, Fall 2004Signal Equation Summary€ € 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=y0Response to a point target δ(x - x1)δ(y - y1)δ(z - z1) iss( x1− x0, y1− y0) p(2(′ z − z1) /c)11TT Liu, BE280A, UCSD, Fall 2004Signal Equation Summary€ € 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=y0Response to a point target δ(x - x1)δ(y - y1)δ(z - z1) iss( x1− x0, y1− y0) p(2(′ z − z1) /c)TT Liu, BE280A, UCSD, Fall 2004Signal Equation Summary€ € Response to a point target δ(x - x1)δ(y - y1)δ(z - z1) iss( x1− x0, y1− y0) p(2( ′ z − z1) /c)Thus, s( x, y) determine the lateral response and p(t) determines the depth response. TT Liu, BE280A, UCSD, Fall 2004Depth 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.12TT Liu, BE280A, UCSD, Fall 2004Depth Resolution€ € The depth resolution is approximately Δz =
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