941733 IIIIIIIIIIIII1IIIZIILIIIaIIIIIIOIOIIIIGIIIII9I4IIIIIIIIIIIIIIIIIIIIIII I, _‘..., THESIS .. V . II éfi-IL ~- "5' Q .in- I hammers finite i Ué’z‘é'fifeé «TEL-rt; —- w" v ‘- .r-=Fr .1-v .m_ -“MJ This is to certify that the thesis entitled ACOUSTIC ATTENUATION AND IMPEDANCE CHARACTERIZATION BY BI-DIRECTIONAL IMPULSE RESPONSE TECHNIQUE presented by M.A. Anura P. Jayasumana has been accepted towards fulfillment of the requirements for MASTERS degree in SCIENCE 66%? Ma' professor Date 12 th May 1982 0-7 639 MS U is an Affirmative Action/Equal Opportunity Institution MSU LIBRARIES 4—;—_ RETURNING MATERIALS: Place in book drop to remove this checkout from your record. FINES wiII be charged if book is returned after the date stamped below. AUG 0 5 100: é //74 lggufli;odong.the_be§gggxis. ’However, the range can be increased by increasing the amount of incident signal power. The first section assumes the object to be a layered structure. An inhomogeneous_object, where the acoustical impedance is a continuous function of the distanee_j§_eonsidered_next. The impulse response of the object when interrogated from the opposite sides are given by h1(t) and h2(t). 3.1 RELATIONSHIP OF ATTENUATION T0 IMPULSE RESPONSE FUNCTION OF A LAYERED OBJECT Consider a layered structure with N+1 layers - N interfaces - as shown in Figure 3.2. The_LiggLzand.1gggggggggézggg=gggomgg=go b3:£§§:§§@§:¥§$§£iglg which normally is the case because the object is immersed in water. Using the results of section 2.1, h1(t) and h2(t) can be written as N h1(t) = a18(t-t1.) (3.1.1) 1 1 31 32 x(t) x(t') _’ «IQ—— y1(t) _ Test object y2(t') ‘— -——' FIGURE 3.1 Bi-directional ultrasonic interrogation. 1 2 3 4 A 2o Z1 Z2 Z3 0(0 “1 012 0(3 T"o 1:1 1‘2 1‘3 1‘1 r2 r3 r4 FIGURE 3.2 i-1 1-1 i-1 i-1 1-1 A layered structure. 1+1 1+2 1+1 “1+1 1:1°+1 1+1 “1+2 N-l N-I N-1 ‘h-1 N-1 33 where, i t1. = Z 2tk (3.1.2) k=0 1"]. 2 ai = exp(-Zflbtb). r. k=1 (1-rk ) exp(-Zfiitk) (3.1.3) Z. - Z. 1 1-1 vs. :2 -— (3.1.4) ‘ Z1 + Z1-1 Similarly, N h2(t) = Z, b.8(t-t'.) (3.1.5) j=1 .1 .1 where, N t'. = Z 211k (3.1.6) J k=j N 2 . = - - N512 . . - - aft bJ exp( 2 J J) rJ kl};1 (1 rk ) exp( 2 k k) (3.1.7) The minus sign accounts for the fact that the reflection coefficient changes its sign when the incident wave is from the opposite side. T23 values a, and b}(i,j4i_1g2ee.JNLuauLJMLJznuLtfixzmdflnojixmLJme imguTge.nesponse_£unctions_hTLtJ_and h2(t).? Using equation (3.1.3), the ratio of ai to a1.+1 is i- exp(-20cot0). r1. T11 (1-rk2) exp(-2&k’Ck) a1 = kfl a. 1 1+1 exp(-Zaoto)- r1+1 "IT (l-rkz) 6Xp(-2°‘ktk) k=1 r1 = _ (3.1.8) ri+1(1-ri2) exp(-Zthi) 34 Similarly, from equation (3.1.7) 2 exp(-2¢{Ci)- ri(1'ri+1 ) N TT 2 .. exp(-Zu-tO). Y'. ‘ (1’1‘ ) exp(-2": ) bi = 1 1 1 k=i+1 k k k bi+1 ( 2“ ‘t ) r INI (1-r 2) ( ZKft ) r1+1 Using equations (3.1.8) and (3.1.9), 2 2 a1 b1 = r1 (1'r1+12) a. 5. 2 1+1 1+1 (1-r1 ) r1+1 Define R, as 2 +6 . ”1 '51 1-r.2 :9 #4.. Then a, bi = Ri a1+1 b1+1 R1+1 or R = R al1+1 b1+1 i+1 i ‘ “SET—EE—T' r. S. RAH 1+1 1+1 Ri+1 + 1 where __ {+1 a1+1 Si '- ‘1 a1+1 (3.1.9) (3.1.10) rln‘TL )2,-R. 5 (3.1.11) 0+RL)V- ;. r34 1,,:—. L \’T v \/z e: Nit: t ( Hei ./// /// /' (3.1.12) / f / lfi L“! (3.1.13) (3.1.14) 35 Dividing equation (3.1.8) by equation (3.1.9), a1 b1+1 = r1 ri+1 a. 5. 2 2 1+1 1 ri+1(1-r1. ) exp(-2°(i‘Ci) - r1.(1-r1.+1 ) exp(-2‘91) which can be written as, a. b «I = 1 1+1 _ 2 _ 2 Hence the attenuation constant is given by, a. b. _ _l_. 1 1+1 2 2 “1 ‘ 41'L"1—1T <1-r.-><1-n-.1> mm The following algorithm can be used to find the reflection coeffi- cient at each interface and the attenuation and impedance in each layer. The value of 3% is assumed to be known. 1. From equations (3.1.3) and (3.1.7) r exp(2u6to) a1 (3.1.16) 1 rN -exp(2¢%1%) bN (3.1.17) El’bN and Tbcan be read from imgglse response functions. 2. Use equations (3.1.11) through (3.1.14) with i=1 to find Ri’ R1 +1 and ri+1 respectively. ”2 < ) R. = ————-——— 3.1.11 1 1 _ r 2 a. . Ri+1 = R, .[TEFFTéFtél (3.1.12) 36 R. *5 rm = 3+1 #T (3.1.13) 1 1+1 where, +1 if a. > 0 51+1 = ”1 (3.1.14) -1 if a. < (3 1+1 3. Use equation (3.1.15) to find the attenuation constant “3, . b. _ 1 , a1 1+1 2 2 “'1 " 747?- . Ln [ET—F.— . (l-Y‘i ) (1-Y‘i+1 4(3.1.15) 1 1+1 1 4. From equation (3.1.4), Z. 1 + r. Z. 1 _ 1 1-1 7. _ 1 _ r. , _7__ (3.1.18) 0 1 o This equation can be used to find the normalized impedance of the i in layer. 5. Repeat steps 2,3 and 4 until all the impedances, attenuation coefficients and reflection coefficients are found. 3.2 RELATIONSHIP OF ATTENUATION TO IMPULSE RESPONSE FUNCTION OF AN INHOMOGENEOUS MEDIUM In this case, the medium is assumed to have a continuous variation of impedance along the path of propagation as shown in Figure 3.3. The following notation is used. x(t) incident waveform y1(t) reflected waveform when the object is interrogated from left x(t) y](t) , Test object FIGURE 3.3 An inhomogeneous medium. t/2 A (t) '(t') Z(t) Z'(t') (iv, T/2 FIGURE 3.4 Time relationships for bi-directional ultrasonic interrogation. y2(t) h1(t) Z(t) Z'(t') °‘(t) «(1') 38 reflected waveform when the object is interrogated from right impulse response function of the medium when interrogated from left, i.e. y1(t) = x(t) * h1(t) impulse response function of the medium when interrogated from right, i.e. y2(t) = x(t) * h2(t) impedance encountered by the acoustic signal, when it has travelled a time t/2 from the left end impedance encountered by the acoustic signal, when it has travelled a time t'/2 from the right end attenuation coefficient (Np/s) encountered by the acoustic signal,when it has travelled a time t/2 from the left end attenuation coefficient (Np/s) encountered by the acoustic signal, when it has travelled a time t'/2 from the right end total round trip time of the ultrasound signal - i.e. time taken for the signal to return after being reflected at the far end. From Figure 3.4 1t 1S seen that h](t) and h2(t ), where t = T-t, corresponds to ref ' fr m he ° ’ m. Hence, Z(t) (x(t) From equation (2.2. 111(1) h2(t') Now, the aim is to z'(1-+)‘/, 1/ o“(T-t) 16), t 39 1:. exp(-I =o((t).dt) . O t 1/2 exp(-0f o<'(t).dt) . DID. t d E1 Ln Z(t) Ln Z'(t') (3.2.1) (3.2.2) (3.2.3) (3.2.4) fiqg:tE§:5itenu5fion>and thqfifipéflgncepof a thin ,_sogment of the body as shown in Figure 3.4. From equation (3.2.3) with t=t and t=t 1 Similarly, h2(‘52 ) h2 t1 ) Also t + t' = and 29 t 13 exp(-f 1 O‘(t).dt) . %‘E L” Z(t) =t1 0 t 15 exp(-d! 2 “(12).dt) . 3? Ln Z(t) =t2 . d t —L Z(t) exp(-f1 x(t).dt) . 3t n t 1 (3.2.5) d t1 —. L Z'(t') 1.! exp(- 20(‘(t).dt) . 3t n It 4:? ti F Ln Z'(t')lt.=ti (3.2.6) 1 (3.2.7) 1 (3.2.8) 40 Using equations (3.2.1), (3.2.2), (3.2.7) and (3.2.8), 3? Ln Z(t) [ t=t1 - 9%?' Ln Z'(t') t|=ti (3.2.9) QE-Ln Z(t) 1:12 = €131) Ln Z'(t') t.=té (3.2.10) 12 ti f «(1) dt -.-. I ot'(t).dt (3.2.11) t 1 Multiplying equations (3.2.5) and (3.2.6), 1111,) 12(1)) h1(t2) h2(ti) t! exp( [111 «(1). dt ).exp(jt_[ 2 .11-(1)31) t2 1 6"L" Z(t)|t= -t dt' L" 2 (t )11' =té ' d dt Ln Z(1")It-"t2-CIT' Ln Z' (t' )|t' 't1 Using equations (3.2.9) and (3.2.10), h 11(t ) h 2(t') t1 = exp(-2 f 0((t).dt ) h12(t ) h 2(t1) t2 i.e. , 1 h t h 1'. 1f 0‘(t).dt = 1 Ln 1(2) 2(1) (3.2.12) 2 h1(t1) h2(t§) From equations (3.2.3) and (3.2.4), . _ , 1 ti . 111(11) 112(11) - 4. exp(- ot(t).dt).exp(-l[ 1x (t).dt) . gE-Ln 2(1)|t=t1 . :7 Ln 2' (1 )‘t. ‘ti 41 Using equations (3.2.1) and (3.2.9), . t1 T h1(t1) 112(11) 1. exp(-6f «(t).dt) . exp(-t] «(t).dt) 1 d 2 . (-1) 3‘1? Ln Z(t) I t=t1 T d 2 -144 exp(-J “(t).dt) . (E L" Z(t) \ t=t 1 This reduces to d 1 exp(% .j «(t).dt) . [-4 h1(t).h2(T-t)] ’2 0 By integrating with respect to time, T -% Ln g 5' [exp(-J ck (t).dt] 1 t /2 .J [—4 h1(t).h2(T-t)] .dt (3.2.13) Let ’X‘ A'3F H = exp(-IIT «(t).dtL (3.2.14) 0 where(§;:ib the total attenuation of the medium, From equations (3.2.13) and (3.2.14), t » Z(t) -P 2 Ln = (A ) 2 -4 h (t).h (T-t) .dt Z(O) T or [ 1 2 (3.2.15) Equation (3.2.12) can be used to find the attenuation variation along the medium. The result can be used to determine AT from equation: (3.2.14). Finally, equation (3.2.15) can be usgg__to find the vari- ation o£_jmpedance along.the.path_o£_propagation. CHAPTER IV EXPERIMENTAL PROCEDURE AND RESULTS The theoretical development in section 3.1 was verified using two models. Section 4.1 outlines the procedure used to obtain the data -the incident and reflected wave forms. The method of analysing the data is given in section 4.2. Next, the results obtained for the two models are compared with the values obtained by direct measure- ments. 4.1 DATA COLLECTION The systems used for sample recording of incident and reflected signals are shown in Figures 4.1 and 4.2. To calculate the attenua- tion and impedance variation in the test object, the incident signal waveform and the reflected signal waveforms, when the ultrasound signal is applied on either side of the object, are required. The waveforms werg_mgngallymsamglgdfig§jng the delayed sweep feature of the Tektronix 465 gathode Bay Qscilloscope. A sampling rate of igflfl; was chosen, as the significant frequency components of the incident signal were below 5 MHz. The experimental procedure is as follows. 1. Record the incident signal waveform. This was done by observing the reflections at the air - water interface as shown in Figure 4.1. As the acoustic impedance of 25 water is about 3700 times that of gin , total reflection with a phase change of.;§g;_was_a§§gmad. The attenuation of ultrasound signal in water was neglected. 42 43 Panametrics 5050 Pulser/ Receiver water Tektronix 465 CRO FIGURE 4.1 Panametrics 5050 Puiser/ Receiver Incident waveform recording Tektronix 465 CRO Figure 4.2 A in Test object S3 Reflected waveform recording 44 2. Record the reflected signal when interrogated from side A of the object. This was done using the system shown in Figure 4.2. 3. Record the reflected signal when interrogated from side B of the object. Same method as in 2 above. 4.2 DATA PROCESSING The following procedure was used to obtain the impedance and the attenuation variation of the object. 1. Estimate the acoustic impulse response function of the object when it is interrogated from either side of the object. This was done using the following two methods. a. Direct Impulse Response Function The reflection function of the object,H(f) is the ratio of the Discrete Fourier Transforms of the reflected and incident signals. H(f) = {(3 (4.2.1) As the spectrum of H(f) is Hfltflliflhle_QutSide_the_signifiggn; f299!§fl§¥_hand_nf_XL£), a Hanning filter H(f) was used to filter out the unreliable frequency components of H(f). m = W (4.2.2) where 2 cos "(I‘m fo-Af< f ' 0 = . “'1 (4.2.9) -1 1f ai+1 < 0 a. b. ‘5 = 31—1—1? - (1 ' riz). (1 " ri+12) (4'2'10) 1+ i I a. b. = 1 1 1+1 2 2 R— Ln[a—1—+-1—b—i . (l-Y‘i )(1-Y‘i+1 )] (4.2.11) _ '| 1-1 ( - . 4.2.12) notation is used (See Figure 4.3) number of interfaces = reflection coefficient at the i th interface = magnitude of the reflected impulse due to the i th interface, when interrogated from side A = magnitude of the reflected impulse due to the i th interface, when interrogated from side B = attenuation coefficient (Np/s) in the layer bounded by i th and i+1 st interfaces = propagation time in the layer between i th and i+1 st interfaces = acoustic impedance of the layer between i th and i+1 st interfaces 4Z 1) hlm a1 /\?3 JN‘N 1\/ \Vi \/’ g: a '. a2 N'Z aN-1 h (t) 4 2 bN-l bN—Z /\"2 b b1 3 b FIGURE 4.3 Bi-directional impulse response functions of a layered object. 48 exp(-20911.) = loss factor - WM propagates forward and backward in the layer bounded by i th and i+1 st interfaces. Equations (4.2.6) through (4.2.12) and either the equation (4.2.4) or equation (4.2.5) are sufficient to find the variation of attenuation and impedance along the path of propagation. Hence it is possible to find the variation of attenuation in two ways - using equations (4.2.4) and (4.2.6) through (4.2.12) or equations (4.2.5) through (4.2.12). The results obtained using both sets of equations are given in section 4.3. 4.3 EXPERIMENTAL RESULTS Two models were investigated using the method outlined in section 4.2. The first object to be tested was a symmetrical structure of three layers. The second was a five layered structure. The results obtained are also compared with the values obtained by direct measure- ment of parameters. 4.3.1 A THREE LAYERED STRUCTURE The first model to be used in the experiment is shown in Figure 4.4. It consisted of three layers - a machine oil layer separated from water by two acrylic cast (plexi glass) layers. The incident signal waveform, its frequency spectrum and the Hanning filter function used in the calculation are shown in Figures 4.5 and 4.6. As_this is a figmmetricallstrugtuge, it was interrogated only from one side. The corresponding reflected signal is shown in Figure 4.7. 49 water a.c. oil a.c. water FIGURE 4.4 Test object 1. The direct impulse response function calculated using the Hanning window of Figure 4.6, before and after amplitude detection are given in Figure 4.8. The impulse response function obtained using the frequency restored method is given in Figure 4.9. For this purpose, frequency components up to 7.27 MHz were estimated. The Hanning filter function shown in Figure 4.6 was used to weight the original and estimated frequency components. The algorithm used for this purpose is given in Appendix B. One iteration was made. The results obtained using each method are given in Tables 4.1 and 4.2. Table 4.3 gives the results obtained by direct measurement of parameters. 50 .ELowm>mz pcmcwocH m.¢ mmszu 51 N12 o~\m .cmurw» mcwccm: mga vcm Escuomqm ucmumucH m.¢ mmszu AmVZm Acvx .Egowm>m3 empompmmm n...» mag: 1P E» 53 .cowuompmu wasprQEm cmpmm vcm mcocmn H pumwno “mop mo :owpucsm wmcoammc mmpzaew pomcmo w.¢ mmzuHu . a mmfi.o-nma m.‘p moo o-" w\/ NN om m m H I m n 4 v 0 ll llllll 4 .4 . n n 0 ammo.oumm NmH.ouHm : g“ mL u \# 4N kmé 3:. ..... :>?>>»>= ><>>>< - L A __r Apvc H.o any; 54 .H pommno pmmw mo cowpucsw mmcoammc mmpzasw cocopmmc xocmzcmgm oaafi.o-nmm q. ) m.¢ mmstm mmmo.onma omo~.ou Hm 3v; 55 TABLE 4.1 Results for test object 1 - using direct i.r.f. a. Reflection coefficients Interface Reflection coefficient Water / Acrylic cast 0.1820 Acrylic cast / Oil - 0.1590 Oil / Acrylic cast 0.1590 Acrylic cast / Water - 0.1820 b. Layer parameters Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 1.45 0.8976 0.0448 Oil 1.05 0.5189 0.0198 Acrylic cast 1.45 0.8976 0.0448 Water 1.00 1.0000 0.0000 56 TABLE 4.2 Results for test object 1 - using restored i.r.f. a. Reflection coefficients Interface Reflection coefficient Water / Acrylic cast 0.1690 Acrylic cast / Oil - 0.1774 Oil / Acrylic cast 0.1774 Acrylic cast / Oil - 0.1690 b. Layer parameters Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 1.41 0.8589 0.0633 Oil 0.98 0.5739 0.0168 Acrylic cast 1.41 0.8589 0.0633 Water 1.00 1.0000 0.0000 TABLE 4.3 measurements. Layer parameters for test object 1 from direct Material Normalized impedance* Attenuation coefficient (Np/s) Acrylic cast Oil 0.13 - 0.18 0.0145 * Normalized with respect to impedance of water. water a. 0 milk oil a.c. water FIGURE 4.10 Test object 2. 58 4.3.2 A FIVE LAYERED ASYMMETRICAL STRUCTURE The model shown in Figure 4.10 was the second object to be used in the experiment. It consisted of five layers - three layers of acrylic cast, a layer of milk and a layer of machine oil. The oil used here was different from the one used in the test object 1. Test object 2 was interrogated using ultrasonic pulses from sides A and B. The corresponding direct and frequency restored impulse response functions are given in Figures 4.11 and 4.12 respectively. The Hanning window used in this experiment is given by. 2 cos 1‘ 1:351:18.” 1.28< f< 4.42 H(f) 0 otherwise where f is the frequency in MHz. In the frequency restoring method of calculating the impulse response function,frequency components up to 7.85 MHz were estimated. The impulse response function obtained by each method was used to calculate the attenuation and impedance variations in the two ways outlined in section 4.2. The corresponding results are tabulated in Tables 4.4 through 4.7. It is seen that when equation (4.2.4) is used, the accuracy of the results become less for the layers closer to the B end. Similarly, the accuracy of the results for layers near the A end decreases if rN =-bN - equation (4.2.5) - is used. This is to be expected since the signal to noise ratio is less for the reflections from the far end of the object. Hence the two sets of results can be combined to give more accurate values by using the parameters of 59 .N pomnno ummp we mcowpocam mmcoammc mmrzasw pumgwu chowuumcwu-vm HH.¢ mmstm . --m . --H ohfio.o-nen omNH o - a oomo o - n mHNo.o-um m.. .< $111--.. mm/ 4? mm ....... NH)? 1 NM nullL. ae~o.on~n OHNH.onQD wamfi.o-umm m; p mama. -uoa may 7 PE? 8 - a... ----.i a . a .8/\ e;(\ _ . . . .4, ----..uul, mhfio.ou . memo.ou a . -m Ammo o- . eHNH.-uHm = AHVN; 1 E: 60 .N pummno pmmp mo mcowuocsm mmcoamw; mmraasw umgoummc hocmzumcm Pacovpomcwu-wm . H . a m;_pemmo o-- n omo.o-mn moeo 0.. a (7..., <41 ..... >>. mNNO.O-Nn ... $851.3 w . .1.“ - a3 . m y... a . a -1. ./\ .,1111l n “ Rad-m. mo¢o.o-:< m< emmo.on m NH.¢ mmzwmm mNNH.o-ma ------- a 1 will... o «HRH.o-cg N 13; 1 memfl.o-Na NH ma ...... . 4 1 -1-- o «HRH.o-Hm H A3; 61 TABLE 4.4 a. Reflection coefficients Interface Reflection coefficient Water / Acrylic cast 1 0.171 Acrylic cast 1 / Milk - 0.185 Milk / Acrylic cast 2 0.101 Acrylic cast 2 / Oil 0.072 Oil / Acrylic cast 3 0.139 Acrylic cast 3 / Water - 0.232 b. Layer parameters Results for test object 2 - using direct i.r.f. and Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 1 1.41 0.9185 0.0708 Milk 0.97 0.6173 0.0201 Acrylic cast 2 1.19 0.9160 0.0730 Oil 1.38 0.2679 0.1090 Acrylic cast 3 1.82 1.0000 0.0000 Water 1.13 1.0000 0.0000 ——r 62 TABLE 4.5 Results for test object 2 - using direct i.r.f. and a. Reflection coefficients Interface Reflection coefficient Water / Acrylic cast 3 0.171 Acrylic cast 3 / Oil - 0.102 Oil / Acrylic cast 2 - 0.052 Acrylic cast 2 / Milk - 0.073 Milk / Acrylic cast 1 0.136 Acrylic cast 1 / Water - 0.125 b. Layer parameters Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 3 1.41 1.0000 0.0000 Oil 1.15 0.2664 0.1100 Acrylic cast 2 1.04 0.9127 0.0760 Milk 0.89 0.6108 0.0206 Acrylic cast 1 1.18 0.9047 0.0830 Water 0.91 1.0000 0.0000 TABLE 4.6 a. Reflection coefficients 63 Interface Reflection coefficient Water / Acrylic cast 1 Acrylic cast 1 / Milk Milk / Acrylic cast 2 Acrylic cast 2 / Oil Oil / Acrylic cast 3 Acrylic cast 3 / Water 0.171 - 0.147 0.111 0.107 0.118 - 0.192 b. Layer paramerers Results for test object 2 - using restored i.r.f. and Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 1 1.41 1.0000 0.0000 Milk 1.05 0.4843 0.0302 Acrylic cast 2 1.31 0.7630 0.2250 Oil 1.63 0.4053 0.0752 Acrylic cast 3 2.06 1.0000 0.0000 Water 1.40 1.0000 0.0000 64 TABLE 4.7 a. Reflection coefficients Interface Reflection coefficient Water / Acrylic cast 3 Acrylic cast 3 / Oil Oil / Acrylic cast 2 Acrylic cast 2 / Milk Milk / Acrylic cast 1 Acrylic cast 1 / Water 0.171 0.106 0.096 0.098 0.131 0.153 b. Layer parameters Results for test object 2 - using restored i.r.f. and Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 3 1.41 1.0000 0.0000 Oil 1.14 0.4042 0.0754 Acrylic cast 2 0.94 0.7611 0.2220 Milk 0.77 0.4826 0.0304 Acrylic cast 1 1.01 1.0000 0.0000 Water 0.74 1.0000 0.0000 65 TABLE 4.8 a. Reflection coefficients Results for test object 2 - using direct i.r.f. Interface Reflection coefficient Water / Acrylic cast 1 0.171 Acrylic cast 1 / Milk - 0.185 Milk / Acrylic cast 2 0.101 Oil / Acrylic cast 2 - 0.052 Acrylic cast 3 / Oil - 0.102 Water / Acrylic cast 3 0.171 b. Layer parameters Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 1 1.41 0.9185 0.0708 Milk 0.97 0.6173 0.0201 Acrylic cast 2 1.19 0.9127 0.0760 Oil 1.15 0.2664 0.1100 Acrylic cast 3 1.41 1.0000 0.0000 Water 1.00 1.0000 0.0000 56 TABLE 4.9 a. Reflection coefficients Results for test object 2 - using restored i.r.f. Interface Reflection coefficient Water / Acrylic cast 1 0.171 Acrylic cast 1 / Milk - 0.147 Milk / Acrylic cast 2 0.111 Oil / Acrylic cast 2 - 0.096 Acrylic cast 3 / Oil - 0.106 ‘ Water / Acrylic cast 3 0.171 b. Layer parameters Layer Normalized Loss factor Attenuation impedance coefficient (Np/s) Water 1.00 1.0000 0.0000 Acrylic cast 1 1.41 1.0000 0.0000 Milk 1.05 0.4843 0.0302 Acrylic cast 2 1.31 0.7630 0.2250 Oil 1.14 0.4042 0.0754 Acrylic cast 3 1.41 1.0000 0.0000 Water 1.00 1.0000 0.0000 67 the layers closer to side A from the set of results obtained by assuming r = a and the parameters of the layers closer to side B 1 1 from the set of results obtained by assuming rN = —bN . These results are given in Tables 4.8 and 4.9. The parameters obtained by direct measurements are given in Table 4.10. TABLE 4.10 Layer parameters for test object 2 from direct measurements. Material Normalized impedance* Attenuation coefficient (Np/s) Acrylic cast 1,2,3 1.50 0.1300 - 0.1800 Oil 1.03 0.0350 - 0.0440 Milk 0.98 0.0147 - 0.0217 * Normalized with respect to impedance of water. CHAPTER v CONCLUSION The conventional ultrasound imaging systems make use of the ampli- tude information of the reflected or transmitted signals. Recent research has resulted in impediography techniques, where the impe- dance of the medium, derived from the reflected signal, rather than the amplitude of the returned echoes is used as the imaging parameter. The impediographic methods make use of the frequency, phase and ampli- tude information available in the reflected signal. However, there are .some_cases, such a§=nggigg=giutge=§ggine_where the attenuation coefficient would be a better imaging parameter than the impedance. Research has begun very recently on the use of the ultrasound attenu- ation coefficient as tbe_imaging_narameten1 The aim of this study was to develop such a method. None of the research, that we came across in this area use reflection techniques. Transmission techniques appear to be an obvious choice in this case, because a signal trans- mitted through an object contains information about the total attenua- tion along the path it has travelled. This thesis has investigated a method which uses a reflection technique for ultrasound attenuation imaging. Only one dimensional images - A mode - has been considered. But this can easily be extended to two or three dimensions. When an ultrasoniic pulse gets reflected from an internal discone- tinuityg of an object, the resulting echo is determined by the reflec- tion coefficient at the interface as well as the attenuation of the subtances in between the transducer and the reflecting point. The 68 69 reflection coefficient is a function of the change in impedance. These two types of information - attenuation and impedance variation along the path of propagation - contained in the reflected signal cannot be separated by the knowledge of reflected signal alone. The additional information necessary to separate the two types of information is derived from another reflected signal. This second reflected signal is obtained by a transmission from the opposite side of the object. This thesis has developed analytical expressions which would allow the separation of information about attenuation and reflection coefficients or the impedance. As a result, the method outlined can extract both the variation of attenuation coefficient and the variation of impedance along the path of propagation. The variation of impedance obtained by this technique should be more accurate than the conventional ultrasonic impediography methods, because those methods neglect the attenuation in the medium, whereas the method outlined in this thesis takes into account the ultrasonic attenuation along the path of propagation. As a matter of fact, the development of an imaging technique utilizing the attenuation coefficient vari- ation is the main concern of this research work. For the purpose of separating the two types of information - ultrasonic attenuation variation and the impedance variation along the path of propagation - the object under investigation was assumed to have one of the two forms.First, it was assumed that the body has a layered structure, which consists of homogeneous layers. Later this was extended to an inhomogeneous medium having a continuous variation of impedance. The expressions developed for a 70 layered structure! were put to test by using models composed of mechine oil, milk and acrylic cast. The experimental results show that this method has not been able to determine the absolute attenuation coefficients of the internal material accurately. But this does not hamper its ability to be used as an imaging techinique, where the relative attenuation coefficient is needed rather than the absolute attenuation coefficient. The main factor which governs the accuracy of the result is the nonlinear beam spreading of the acoustic wave. This could cause the apparent acoustic attenuation coefficient to be varied along the path of propagation even in a homogeneous object. This method of determining the attenuation coefficient has the advantage of suppressing the multiple reflections of the returned signal. The reason is that the impulse response function obtained from either side! of the object is more accurate for reflections closer to the surface. This is because the power and hence the signal to noise ratio is less for reflected echoes from the deep lying parts of the object. But the deep lying parts of the object for one impulse response function are the near parts of the object for the other impulse response function. Thus, by comparing the time intervals between echoes of the two impulse response functions, the echoes due to multiple reflections and noise can be detected and subsequently suppressed. The analytical expressions developed for the inhomogeneous case have not been experimentally ”verified. This has been an extension of the work done on the layered structures. More experiments are needed to establish its validity. 71 This method heavily depends on the ability to determine the impulse response function accurately. The accuracy of the impulse response function in turn depends on the bandwidth of the transducer. Although techniques have been developed to determine the impulse response func- tion using a narrow band source, a wider transducer bandwidth will certainly improve the accuracy. But too wide a bandwidth can give rise to another problem. The attenuation coefficient for biological matter has been found to be dispersive in the frequency domain. Thus a large bandwidth may result in nonuniform attenuation of various frequency components, which in turn would invalidate the assumptions made in this method. Hence, a bandwidth, which makes it possible to determine the impulse response function accurately, to retain the range resolution within desired limits and at the same time does not cause too much variation of attenuation in the frequency band used has to be selected. By using methods such as_pseude_randem_binary_seguenee§, the accuracy of the impulse response function may be improved. As this method allows the use of a higher average power, a considerable improve- ment in signal to noise ratio can be expected. Such methods, which improve the accuracy of the impulse response function, may make this method very attractive in the future developments. APPENDICES APPENDIX A ALGORITHM FOR DIRECT IMPULSE RESPONSE FUNCTION CALCULATION The algorithm given below is used to calculate the direct impulse response function using the incident and reflected waveforms. The following notation is used. x = Xi ; i=1,2,... y = .1. . i=1,2, h = h. , i=1,2, 1 X = Xi , i=1,2, Y = Yi , i=1,2, H = H1 , i=1,2, w = “i ; l=1,2,... A = a1 ; i=1,2,... T = t ; i=1,2, . N Incident signal samples** Reflected signal samples Impulse response samples Discrete Fourier Trans- form (DFT) of x Discrete Fourier Trans- form (DFT) of y Discrete Fourier Trans- form (DFT) of h Hanning filter function Peak values of the impulse response function Epoch times corresponding to peak values of the impulse response function. * N is selected such that there are adequate number of zeroes on either side of the signal samples to prevent aliasing. ** The duration of the incident signal is very small compared to m that of the reflected signal. Hence there are only a small number 01°an 72 73 x and y are the data collected in the experiment. This algorithm calculates the sample values hi of the impulse response function and detepts jts peaks by using the amplitude detection scheme given in Appendix C. was used to determine the Discrete Fourier Transform and the Inverse Discrete Fourier Transform. This as well as the other algorithms were implemented using the Cyber machine at the Michigan State University. The algorithm is as follows. 1. 2. 3. Calculate the DFT X of the incident signal samples x. Calculate the OFT Y of the reflected signal samples y. Calculate the filter function w given by 2 . . Wi COS 134—12310) for io -Ai < i < io +Ai and N+2-io-Ai < i