Solutions Manual for Medical Instrumentation: Application and Design PDF
Preview Solutions Manual for Medical Instrumentation: Application and Design
SOLUTIONS MANUAL MEDICAL INSTRUMENTATION 2 SOLUTIONS MANUAL MEDICAL INSTRUMENTATION: Application and Design Fourth Edition John G. Webster, Editor Contributing Authors John W. Clark Rice University Michael R. Neuman Michigan Technological University Walter H. Olson Medtronic, Inc. Robert E. Peura Worcester Polytechnic Institute Frank P. Primiano, Jr. Consultant Melvin P. Siedband University of Wisconsin–Madison John G. Webster University of Wisconsin–Madison Lawrence A. Wheeler Nutritional Computing Concepts John Wiley & Sons, Hoboken, NJ 3 Copyright © 2009 by John Wiley & Sons, 111 River Road, Hoboken, NJ 07030. All rights reserved. 4 Contents Chapter 1 Basic Concepts of Instrumentation 6 Chapter 2 Basic Sensors and Principles 10 Chapter 3 Amplifiers and Signal Processing 16 Chapter 4 The Origin of Biopotentials 22 Chapter 5 Biopotential Electrodes 32 Chapter 6 Biopotential Amplifiers 42 Chapter 7 Blood Pressure and Sound 56 Chapter 8 Measurement of Flow and Volume of Blood 64 Chapter 9 Measurements of the Respiratory System 72 Chapter 10 Chemical Biosensors 83 Chapter 11 Clinical Laboratory Instrumentation 84 Chapter 12 Medical Imaging Systems 86 Chapter 13 Therapeutic and Prosthetic Devices 91 Chapter 14 Electrical Safety 102 5 Preface Solutions were prepared by the contributing authors and compiled by the editor. We welcome your suggestions for corrections and improvements for subsequent editions and printings. In particular, send any lists of homework problems and examination questions that you have found instructive to the editor John G. Webster or in the care of John Wiley & Sons, 111 River Street, Hoboken, NJ 07030. PowerPoint slides, instructional objectives, lab instructions, and old exams are at http://ecow.engr.wisc.edu/cgi-bin/get/bme/462/webster/ John G. Webster [email protected] February 7, 2009 6 Chapter 1 Basic Concepts of Medical Instrumentation Walter H. Olson 1.1 The following table shows % reading and % full scale for each data point. There is no need to do a least squares fit. Inputs 0.50 1.50 2.00 5.00 10.00 Outputs 0.90 3.05 4.00 9.90 20.50 Ideal Output 1.00 3.00 4.00 10.00 20.00 Difference – 0.10 +0.05 0.0 – 0.10 +0.50 % Reading 11.1 1.6 % 0% – 1.0% +2.4 % = Difference/Output × 100 Full Scale – 0.5% 0.25% 0% – 0.50% + 2.5% = Difference/20 × 100 Inspection of these data reveals that all data points are within the "funnel" (Fig. 1.4b) given by the following independent nonlinearity = ±2.4% reading or ±0.5% of full scale, whichever is greater. Signs are not important because a symmetrical result is required. Note that simple % reading = ±11.1% and simple % full scale = 2.5%. 1.2 The following table shows calculations using equation (1.8). Inputs Xi 0.50 1.50 2.00 5.00 10.00 𝑋 = 3.8 Outputs Yi 0.90 3.05 4.00 9.90 20.50 𝑌 = 7.6 Xi – 𝑋 – 3.3 – 2.3 – 1.8 1.2 6.2 Yi – 𝑌 – 6.7 – 4.55 – 3.6 2.3 12.9 (Xi – 𝑋 )(Yi – 𝑌 ) 22.11 10.465 6.48 2.76 79.98 (Xi – 𝑋 )2 10.89 5.29 3.24 1.44 38.44 (Yi – 𝑌 )2 44.89 20.7 12.96 5.29 166.41 121.795 r = = 0.9997 (7.701)(15.82) 1.3 The simple RC high-pass filter: I C + + Vin R Vout - - The first order differential equation is: d x(t) – y(t) y(t) C = dt R 1 CD + y(t) = (CD) x(t) R y(D) D = Operational transfer function x(D) 1 D + RC 7 Y(j) jRC RC 1 = = = Arctan X(j) jRC + 1 (RC)2 + 1 RC ;where 1/RC is the corner frequency in rad/s. 1.4 For sinusoidal wing motion the low-pass sinusoidal transfer function is Y(j) K = X(j) (j + 1) For 5% error the magnitude must not drop below 0.95 K or K K = = 0.95 K j + 1 2 2 + 1 Solve for with = 2f = 2(100) 2 2 ( + 1) (.95)2 = 1 1 – (0.95)2 1/2 = = 0.52 ms (0.95)2(2100)2 Phase angle = tan–1 (–) at 50z 50 = tan–1 (–2 50 0.0005) = –9.3 at 100 Hz 100 = tan–1 (–2 100 0.0005) = –18.2 1.5 The static sensitivity will be the increase in volume of the mercury per C divided by the cross-sectional area of the thin stem V Hg b K = = 2mm/C A c where 8 3 cm = 1.8210–4 V Hg b 3 cm C V = unknown volume of the bulb b A = cross-sectional area of the column c A = π(0.1 mm)2 = π × 10–4 cm2 c Thus 2 AcK 10–4cm 0.2cm/C 3 V = = = 0.345cm b 3 cm Hg 1.8210–4 3 cm C 1.6 Find the spring scale (Fig. 1.11a) transfer function when the mass is negligible. Equation 1.24 becomes dy(t) B + K y(t) = x(t) s dt When M = 0. This is a first order system with 1 K = static sensitivity = K s B = time constant = K s Thus the operational transfer function is y(D) 1/Ks 1 = = x(D) B K + BD s 1 + D Ks and the sinusoidal transfer function becomes y(j) = 1/Ks = 1/Ks = tan–1 – B B K x(j) 1 +j D 2B2 s Ks 1 + 2 K s 1.7 dy dy a + bx + c + dy = e + fx + g dt dt dy (a – e) + dy = (b + f)x + (g – c) dt This has the same form as equation 1.15 if g = c. (D + 1)y = Kx a – e b + f D + 1 y = x d d a – e Thus = d 1.8 For a first order instrument 9 Y(j) K = X(j) (j + 1) K K = = 0.93 K (j + 1) 2 2 + 1 2 2 + 1 (0.93)2 = 1 1 1 1 – (0.93)2 f = = = 3.15 Hz 2 2 (0.93)2 (.02)2 = tan–1 (–) = –21.6 1.9 – t Ke n 2 –1 2 y(t) = K sin 1 – t + where = sin 1 – n 2 1 – = 0.4; f = 85 Hz n 3 7 – – 2 2 t = = 7.26 ms t = = 20.1 ms n n+1 2 2 1 – 1 – n n 10 10 y(t ) = 10 + e–ntn y(t ) = 10 + e–ntn+1 n n+1 2 2 1 – 1 – = 12.31 = 10.15 3 7 11 1.10. At the maxima y , y , y : sin ( ) = –1 at , , n n+2 n+4 2 2 2 3 – 2 3 2 and 1 – t + = t = n n n 2 2 1 – n 5 9 at the minima y y …: sin ( ) = + 1 at , … n+1, n+3 2 2 5 – 2 5 2 and 1 – t + = t = n n+1 n+1 2 2 1 – n then form the ratio 10 K – t e n n 3 5 2 – – yn 1– 2 2 = = exp = exp yn+1 K e–n tn+1 2 2 1– 1– 2 1– y n = ln = y n+1 2 1– Solve for Chapter 2 Basic Sensors and Principles Robert A. Peura and John G. Webster 2.1 Let the wiper fraction F = x/x i t R ||F R m p v /v = o i Rm||F RP + (1–F)RP R F R m p R + F R m p = R F R m P + (1–F)R P R + F R m p 1 = R F R m p 1 + (1–F)R P R + F R m p 1 = R F + R + F R – F R – FFR m m p m p R F m 1 = 1 + F R /R – FFR /R p m P m F 1 = 1 Rp + (1–F) F R m Let = R /R p m error = F – v /v o i 1 = F – 1/F + (1–F)
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