Solutions Manual for Medical Instrumentation: Application and Design

108 Pages·1978·2.012 MB·English

by  John G. Webster

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.
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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
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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) jRC RC 1
= =   = Arctan
X(j) jRC + 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  = 2f = 2(100)
2 2
(  + 1) (.95)2 = 1

1 – (0.95)2 1/2
 = = 0.52 ms
(0.95)2(2100)2

Phase angle  = tan–1 (–) at 50z
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
 = 1.8210–4
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 1.8210–4
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)