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Signal Processing for Intelligent Sensor Systems DAVID SWANSON The Pennsylvania State University University Park, Pennsylvania
a% M A R C E L
D E K K E R
MARCEL DEKKER,
NEWYORK BASEL
ISBN: 0-8247-9942-9 book
on
Headquarters 10016
270
2
2
Eastern Hemisphere Distribution AG
8 12,
4, 1-8482;
4
filx:
4
World Wide Web
on
book
;i
Copyright
(.
2000 by Marcel Dekker, Inc.
All Rights Reserved.
book
10 9 8 7 6 5 4 3 2
by
1
PRINTED IN T H E LiNITED STATES OF AMERICA
Series Introd uction
50
As
on,
DNA
on
0
0 0 0 0 0
I by
iii
Preface
Sigriul Procvssing.Jiir Iritelligent Sensor Systems
book
by
book
on
on book
book v
vi
Preface
book
on
no
on
by
by
on
do do as:
by
on
Preface
vii
on
on book good
by
book
works
This page intentionally left blank
Acknowledgments
1993
by
on 6,
I
3
3, on
book
PC on
on
you
book.
ix
This page intentionally left blank
Contents
K. J. Rajq Liu
iii v ix
Fundamentals of Digital Signal Processing
1
Series Introduction Preface Acknowledgments Part I
Chapter 1 1.1 1.2 1.3 1.4
Sampled Data Systems
3 5 7 11 14 19
Chapter 2 The Z-Transform 2.1 2.2 2.3 2.4
20 28 32 39
Chapter 3 Digital Filtering 3.1 3.2 3.3 3.4
43 44 47 50 53
Chapter 4 Linear Filter Applications 4.1 4.2 4.3 4.4 4.5
55 56 62 69 80 85 xi
Contents
xii
Part 11
Frequency Domain Processing
Chapter 5 The Fourier Transform 5.1 5.2 5.3 5.4 5.5 5.6
87 93
96 101
106 111 115 123
Chapter 6 Spectral Density 6.1 6.2 6.3 6.4 6.5
127
Chapter 7 Wavenumber Transforms 7.1 7.2 7.3 7.4
183
Part I11
Adaptive System Identification and Filtering
130 144 158 169 179 186 196 203 21 1 215
Chapter 8 Linear Least-Squared Error Modeling 8.1 8.2 8.3 8.4
217
Chapter 9 Recursive Least-Squares Techniques 9.1 9.2 9.3 9.4 9.5
237
Chapter 10 Recursive Adaptive Filtering 10.1 10.2 10.3 10.4
275
Part IV
323
Wavenumber Sensor Systems
Chapter 11 Narrowband Probability of Detection (PD) and False Alarm Rates (FAR) 1 1.1
217 22 1 324 233
238 24 250 257 27 1 27 7 393 31 1 318
327
328
xiii
Contents
11.2 11.3 1 1.4
339 347 356
Chapter 12 Wavenumber and Bearing Estimation 12.1 12.2 12.3 12.4 12.5
361
Chapter 13 Adaptive Beamforming 13.1 MUSIC 13.2 13.3 13.4
407
Part V
Signal Processing Applications
449
Chapter 14 Intelligent Sensor Systems 14.1 14.2 14.3 14.4
451
Chapter 15 Sensors, Electronics, and Noise Reduction Techniques 15.1 15.2 15.3 15.4
521 522 532 563 583
Appendix Answers to Problems Index
589 61 1
362 368 385 396 404
409 414 430 444
454 478 50 1 516
This page intentionally left blank
Part I ~
~
Fundamentals of Digital Signal Processing
1
This page intentionally left blank
Sampled Data Systems
1
control
1
on
O K .
by A/ A /D
A/D 3
4
Chapter 1
Input Sensor System
Input
Convertor
' control i8e
In ell ut ain
Information, Patterns, Etc.
Adaptive Signal Processing System
> Commands, Digital Data Input
I
Intelligent Output Gain Control D/A Convertor
Control Actuator Figure I
A
by
If (D/A)
5
Sampled Data Systems
1.1
A/D CONVERSION
A/D
D/A
2
A/D
A/D
D/A A D/A 1.
A/D up
D/A (LSB) 0
1
D/A
0
Digital Output
1
E Convertor
-
0 bit
Counter
if a>b: count down
if a 0,
sigrwl
r.e,sporzses
19
Chapter 2
20
.w
1
0.80 0.60 0.40
aJ
0.20
K
0.00
(I)
01
-0.20 -0.40
-0.60 -0.80 -1.w
0.00
0.05
0.15
0.10
0.20
0.25
Time (sec)
Figure 1
A
(7
50
.
its
on 0 Hz
all a
.s~~.steni irupirl.sc. r.c'.spom~.
has
2.1
COMPARISON OF LAPLACE AND 2-TRANSFORMS
1) K ( s , I). =
y
K ( . s . r)1r(r)cir
--2
K(s, 1
= (I",
21
The Z-Transform
~ ( s= ) ~
1
{ y ( t ) l = y(t>e-.\‘dl 0 n+/x
y ( t ) = 9-1 ( Y ( s ) ]= -
Y (s)e’“’cls
t
rzT
z = c’‘.
[n],
t1=0
Eq. y[s]
I.
Y(s)
vz]
Eq.
t = 0,
Eq. 11
Eq.
= 0.
no
arid . ~ p c c i f i ~r ~ io~ t .future ~ l l ~ ~iiipirts. on
= a+jw),
by As
(jw
on
irriit
circle. :=
e\ 7’,
analog
on
22
Chapter 2
( f ;= 1 / T
T
1
T. a
Table 1 z
s
1 .v
- .S() 1
(.s - .so)?
The Z-Transform
23
on
on
on no
z-
Linearity
+ h g ( t ) }= ctF(s) + hG(s) Z { u f [ k ]+ hg[k]}= nF[z] + hG[z]
-Ip{c!f(t)
(2.1 A . )
Delay Shift Invariance
, f i r ) =J [ k ] = 0
t,
k < 0,
Chapter 2
24
Convolution
A
Eq.
If‘f’[k]
g[k]
g[k]
by
.f’[k].
g[k]
k
k
Eq.
Initial Value
by ;is ,s
I
/‘(I) =
sF(,s) \”rL
n =
.
F[:]
Final Value a
as
.sF(s)
f ( i )= I - 2
on
(T
j c l ,
i =
(1
-
2
x.
).
on
:-I)F‘[z] 1 1 1
ci
sF(s) F(s)
3 0, on
2 1 . on j(o
on on
F(s)
The Z-Transform
25
s =0 z
flz]
z=1
=e.'T. s
r
on Frequency Translation/Scaling
by Y{e-''Y(t)) = F(s
+N) by
Ix
2 { r k f [ k ] }= E f [ k ]x ( : p F[z/rx] k
Differentiation fit)
by
9
-
= sF(s) - f ( O )
by
..($J
,Y-
= s N F ( s )-
SN-'-kf(k)(0)
k=O
t = 0.
Z { x [ n+ NI) = e Y N ; x [ z ]-
N-l
z.Y-k.Y[k]
on i [ n+
1 T
= -(s[n
+ 11 - x [ n ] )
xk
26
Chapter 2
1
Z ( i [ n+ 11) = -((I T
- zs[O]}
by 1 .\-[II]
11.
1 Z{.i.[IZ]}= - {( 1 - z-l)x[I] - .u[O]} T
1 Z{.;;.[t?]) = -( ( 1 - z - ' ) ' X [ z ] T' -
+ z-'x[I]]]
-
s[iz],
by
z = 1.
z(s[iz]} = __ (( 1 - z - ' ) 3 x [ z ] T-7
-
+ 3Z--').Y[O]
(1 -
- (z-'
-
3z-').u[l] - z - ' s [ 2 ] )
on X[IZ]
z=1
.V
.Y = 0
IV
Mapping Between the s and z Planes
As
1.1 by
t
T = 1 /.f;
11
As f , Hz by
z" =
c"'
nT,
27
The Z-Transform
0 Hz
Hz Hz
c’”
j
0. by
0 CJ
f0,
Hz.
5 tuS/2
on
Hz on
on w,
As
0
1,
0
71
“A”
- 71.
“I”)
on /A
“A”
on
S-Plane
Z-Plane
I 1.
I
Fm E.
Stable Signal Regions Figure 2
on on
“I” 2.
2
Chapter 2
28 s
z
H(s)
h(t),
2,
H(s)
h[rz].
H[=] h[n]
h(t)
11 T
2
jw
on
do
by (a< s
z
As
by s
by
z = 0‘7
2.2
SYSTEM THEORY
up up
M---
t) at’
+R! !! +I Ky( !! t ) = As( t ) at
)
The 2-Transform
29
t >0
As([)
)
A.v(t)
M
y(t)
K on
nT,
1
A.u(t) 1)
on ) ' ( I ) Eq.
+ R { s Y ( s )-
M{s' Y ( s )- s ~ ( O )-
=fo8(t) Y ( s )=
F(s)
+ ( M s+
{Ms' = H(s)G(s)
+ K Y ( S )=,/i) F(s) =.f&
+
+ Rs + K }
H(s) H(s) =
1
(Ms'
+ Rs + K }
G(s)
G ( S ) = F(s)
+ ( M s + R)lfO)+ M j ( 0 )
+ R)j'(O)+ + (MLy {Ms' + Rs + K )
Y ( s )= F ( s ) H ( s )
on F(s).
3
Eq. y(0) F(s)H(s)
y(/)
convolution integral.
F(s)H(s)
1 riT =
Chapter 2
30
H(s) =
Figure 3
1
Ms2+Rs + K
A
=t
-t
Eq. /I( t).
Also
Ms' If(/)
+ Rs + K H(s).
The 2-Transform
31
Eq.
=fo6(t);fo = 1,
jft)
h(r)
h(t).
Eq.
H(s) H(s)=
1
Ms2 i- Rs+ K
H ( s ) = ___ 1 $1
- s,
[-
1
s - SI
1 s - s,
sI
s2
2 =z*J/;-R (5)
2M
Eq.
h(t).
information
A/D
1)
Chapter 2
32
2.3 MAPPING OF S-PLANE SYSTEMS TO THE DIGITAL DOMAIN
Eq. z = e”.
N”’[I]
as
2) z
rz = k
+
1
Eqs
on
7‘. As
0, h‘”[n]
(‘J(/
1
h(nT).
4
on
Hz 57 Hz, 125 Hz,
;= 10
n
Eq.
125 co,,T
57
300
n/2, good
33
The Z-Transform
2.00 1.oo
57 samples/sec 0
.
0.00 -1.oo 0
.
-2.00 0
0.2
0.1
0.3
0.5
0.4
2.00 1.oo
L
125 samples/sec
0.00
-1.oo
-
-2.00
1
1
I
I
I
0 2.00
I
I
I
I
1
1
0.1
I
1
1
1
I
1 I
I 1 1
I
I
0.2
l
l
1
I
I 1
I
0.3
1
1
I
I
I
I 1
I
1 1
0.5
0.4
w
9.
1.oo
300 samples/sec
0.00 -1.oo
-2.00
I l l 1 I
0
I I I I I I
0.1
I
I
I
I
I
I I
I
l
I
1
0.3 Time (seconds)
0.2
Figure 4
CL),,T
I
l
I
I
1
I
1
1
I
I
I
I
l
I
1
1
0.4
1
0.5 on
0
on Eq. (2.3.6).
34
Chapter 2
5
on
2.00
1.oo
0.00 -1 .oo
-2.00
2.00 1.oo
0.00 -1 .oo
b
125 samples/sec
b
-2.00
0.1
0
0.2
0.4
0.3
0.5
2.00 1.oo
300 samples/sec
0.00 -1.oo
t
-2.00
I
0
1
1
I
I
I
1
1
I
I
0.1
1
I
I
1
I
I
l
I
I
I
0.2
I
I
I
I
I
I
I
1
,
1
0.3
Time (seconds) Figure 5
1
1
I
I
l
l
I 1
0.4
0.5
The 2-Transform
35
Eq.
w,I
by T = 1 /fs
3.
2?f;, good
z
H(s)
HI:]
5s5
on H[z]
T, H[z]
H(s)
H(s)
6
< = 10
Hz
H[2]
25
6
300 Wz],
by T2. As
6, At
7 H[z]
Eq.
by
p,
H(s)
57
Chapter 2
36
200 150 100 50 deg 0 -50 -100 -150 -200
t
0
i
k25
50
75
100
125
150
100
125
150
Frequency ( H t ) -60 -70
-80
dB -90 -100
-110
-120 0
25
50
75
Frequency (Hr) Figure 6
(A)
H(.s)(-)
H[:]
properly
300 Hz.
As
(T=
z;
+ 5, j c o / i z = f 130
pi
pi non-r?iinimuni phiist.
0
0
=
= - 10, j w / 2 z = f240
j m / 2 n = f 160 on
The 2-Transform
37
200 150 100 50 deg 0 -50 -100 -150 -200
0
5
10
15
20
25
30
20
25
30
Frequency (Hz) -60 -70 -80
dB -90 -100
-110 -120
0
5
10
15 Frequency (Hz)
Figure 7
H(s)
6,
57
H(s) =
- Zi*) (s - p ; ) ( s- pi*)@ - &)(s (s - Zi)(S
- py)
Eq. (2.3.11). h ( t ) = A,,&';'
+ B,\epi;"+ C,,epi' + D,ePy'
1)
Chapter 2
38
p;.
pi
causes
do
by
H[z]
by
(2 -: ,)(I = T--------
T
A,. ___ [r - p l
c‘, +--] D, +-: - p4,i * +--z-p; -py A ( . ,B,,
.4, =
B,. =
c,.= D, = Eqs.
A - , B,, C-,
D=
D,
Eq.
The Z-Transform
39
8
600
As
8, 9 As
8
9,
up
by As
10 5:1
2.4
11,
3 kHz,
SUMMARY, PROBLEMS, AND BIBLIOGRAPHY
0.005 0.004
0.003 0.002
0.001
I*+++ ++
Linear Scaled
+
++
+
+
O.OO0 -0.001
Modal Scaled
++
-0.002
-0.003 -0.004 -0.005 0.00
0.05
0.10
0.15
0.20
Seconds Figure 8
(-),
600 Hz.
(O),
(+)
40
Chapter 2
-100
-110 -120
dB -130 -140 -150
-160
0
50
100
150
250
200
Figure 9
300
( 0),
( ),
600 Hz.
(+)
0.0015 0.0010
1
,
Linear Scaled
0.0005 0.0000 -0.0005
4.0010 I
0.00
0.01
0.02
0.03
0.04
,
,
,
0.05
Seconds Figure 10
( -),
3000 Hz.
(o),
(+)
41
The Z-Transform
-90 -100
-110 -120
dB -130 -140 -150 -160
0
200
400
600
800
1,000
1,200
Figure 11
1,400
1,600
(-),
3000 Hz.
(+)
on
A/D by
0
fo =
3, fo = 27.65 Hz = 75
78.75 Hz.
Force Unit Impulse Response 0.15
1
1
I
I
1
I
0.1
0.05
0 -0.05
-0.1 -0.15
0
1
1
I
I
2
I
4
3 sec
Figure 3
75 Hz
I
I
5
6
7
Chapter 4
62
/ mcurcrtc’ f;
Eq.
4.2
FIXED-GAIN TRACKING FILTERS
/
on Eq. (4.1.16),
on
no ilk
A
ilk,
on
r
/)
a-/) cx-/)
o,,.
Linear Filter Applications
63
by
a,
x = 0.10, CI
>1 zk
)
xk
H V ~
k.
xklk
k+1
k
k
x;:Ik
A$,,,,
T
r-P-1~
on I
xk+IIk,
x r,
process noise.
0,
y
Chapter 4
64
do
by 2-B
0,.
vA
n
0,.
1 x-p
x-p-;?
x-/I
on
Ix-[I-j*
;.,\[.
o,T’i2.
B, ii
10 on :
s
u
n
ng
; nd non
on ns nt
u
on on
a
ro;. x. /j =
- IX) - 4
6 [j‘ir
7 x
/j (
Linear Filter Applications
65
x,
o,,
on
c,, o,,. a, /?,
7
p.
AM,
a
7 = p2/a cc,
p,
2-B
ZMG x-p-;~
5 15 0.25 up on up 10
3 c,,= 13, o,,= 3,
13 T=0.1
i M= 0.0433,
fl = 0.0374,
-y = 0.0055.
a = 0.2548,
4
U-/?-?
4
good 4 by
on a-p-,,
o,,= 3
13,
(AM = 5. by
66
UJ
Chapter 4
50 40
8 30 *
p 20
10 0
0
1
2
4
3
7
6
5
Seconds
0 0
8
9
10
9
10
0 0
0
0
0 0 0
0
N U
6 4
t%
:
c 8
\
3
1
2
4
3
6
5
6
7
8
2 0
-6 -8 -10 -12 -14
0
Figure 4
1
2
3
4
Seconds
7
8
9
10
3
x p-7
13
on
;i
100
5
Seconds
0
( - - -),
6 (U,, = 100,
(O), =
on
A,,%{, 2, p,
i'
o,,
7,
67
Linear Filter Applications
70 t 60 50 40 0) 3 30 z 20 10 0 -10 0
E
150
1
2
3
4
5
6
Seconds
7
8
9
10
t
I
N i z 0
8
2 0 -2 -4
- 4 f - 8 -10 -12 -14 0
1
2
3
4
5
6
Seconds
7
8
Figure 5
9 CJ,,
10
=3
13
4.
3
7 3
40
8 3
8
o,,,
68
Chapter 4
60 50
40 $ 30 * v)
s ;; 0 -10 0
1
2
3
4
5
8
7
6
Seconds
9
10
c
00
0
0
0
1
2
3
4
6
5
Seconds
Figure 6
7
0
8
9 0,=
4.
on
10 100
13
69
Linear Filter Applications
140 120 100 80 a 60 % 40 20 0 -20
E
=
P
-40
0
1
2
3
0
1
2
3
4
5
6
7
8
9
10
4
5
6
7
8
9
10
Seconds
100
2 Q
s
-50
-100
Seconds
Figure 7
4 n,,. =
7
8
U,,, =
on
4.3
2D FIR FILTERS
70
Chapter 4
160 140
0
120
2100 0)
%
0
80 60
0
I40 20 0 0
1
2
3
0
1
2
3
1
2
3
6
7
8
9
10
4 5 6 Seconds
7
8
9
10
4
7
8
9 10
4
5
Seconds
tgn:
- 0 2 -20 0-40 5-60 -80 -100
6 ' v 4 U
E
2
- 4 Q) e-8 5 -10 -12 -14 0
5
6
Seconds
Figure 8
3 = 40m r , [j. 7
7
(T,,
c,,
8
i.2,.
A
Linear Filter Applications
71
by
by
on 9.
9 on
good by on
Figure 9 640 x
72
Chapter 4
B(.Y.J,),
-v
A
j*
B’(.Y-,J.)
w,, Eq.
on
B(.Y+;.J*)
+s B’(x,j-).
10
c/essinrcrtrd.
10
8x 8
9.
by ‘ 2
Eq.
11
Linear Filter Applications
73
Figure 10
8x8
12 on
by
A A =
N =M
1 Eq.
+
WO. 1
;”’;.‘I
=
[
1 0 2 0 1 0
Eq. by
74
Figure 11
Figure 12
Chapter 4
75
Linear Filter Applications
(4.3.3)
(4.3.4)
45' on estiniute
A x
y
A
by
9
13
(4.3.5). 7
V-B
a'B as-
7
a'B +-
(4.3.5)
(4.3.6).
s
(4.3.6)
by
Chapter 4
76
Figure 13
1.
on
Eq. -V2B=
[
-1
do
8
by Eq.
[:::::I
w3= 14
by 10, 8x 8
15.
As 16
A
77
Linear Filter Applications
Figure 14 A
Figure 15
Wy
Chapter 4
78
16 16
by
on 16
by
on
17
by
Figure 16 A
Figure 17
4
79
Linear Filter Applications
18
on on
7.3. by
on on
8, on
Figure 18
body
body
Chapter 4
80
7.3. body
a
D/A RECONSTRUCTION FILTERS
4.4
to
(A/
(D/A)
upon no
zc~ro-ortlc~r.
A
hold a
a a
A
A
DIA
as
1s a c
Js
a it ii
1000
400 Hz.
f IOOO), f 1400
1980s
by
by
a
a s o \ . c . r . . ~ r i i ? i p l i ~ Io gn
!'
5.
Linear Filter Applications
c 1
0.8 0.6 0.4
0.2
0 -0.2 9.4
-0.6 m0.8 -1 -1.2
~
"
0
"
~
"
~
~
20
40
"
"
'
60
~
"
"
~
80
~
"
"
100
11
1
"~
"
"
"
120
"
" '
' " "
1
140
160
180
200
Sample Number Figure 19
x
x
100
32 x 44,100 Hz do
2 2,822,400
32 x 2,778,300
D/A 2 x, 4 x, 8x 20
3
2x 8x by
4x
7
F I R interpolation filter. by A
by 2x
A
4x 1 x, 2 x , 4 x ,
8x
01,.
21
82
Chapter 4
t
i
0
20
60
40
80
100
140
120
160
180
200
Sample Number Figure 20 D A
21 8x
;I
by
1
ii
Di'A
a
by
22 \+';I\
;I
1
ii
2x
4 x,
23
8x
to
22 atid 23
3x
1x
2x
,
Xx
go
8x
ii
look
4x 24.
;IS
At M
seen
24
IK o\
;i
case. Also
of
leak
03
Linear Filter Applications
- -0 . . 0th Order (1X) - -*- 1st Order (2X) .
(F- . 3rd Order (4X) -+- 7th Order (8X)
-
0.8 0.6
0.4
0.2
0 -0.2
0
8
16
24
32
40
48
56
64
Sample Number Figure 21
0
D/A
1
2
3
Sample Number Figure 22
2x
4
84
Chapter 4
1
0.8 0.6 0.4 0.2
0 -0.2 -0.4
-0.6 -0.8 -1
-1.2
2
1
0
3
4
Sample Number Figure 23
3x Xx
1
1st Order (2X)
-+----
0.6
0.4
.
-
0.2 0 -
-0.2 ;
0
0.5
1
1.5
2
f/fS Figure 24 in
2.5
Linear Filter Applications
85
14 8x
1 8 8x A on
4.5
SUMMARY, PROBLEMS, AND BIBLIOGRAPHY
A
on on on
PROBLEMS
1. (h()+hlZ')l(+ l cI,zl
2.
?
R=
C=
L =2
Chapter 4
86
3. A
by
10%
ct-D-7
cc-a-y?
4.
(2x)
5.
2nd on.
2nd
on.
BIBLIOGRAPHY
X. 1993. A.
J. C.
W. 1989. 1992.
Part II Frequency Domain Processing
"4-E-A") by
to
do
.U( t ) ,
x(t)~~""~~lt X(co).
.I-[/?].
a7
Part II
88
on on
by by by on
2 on by e'"'
(s)
(t)
c)'"
(to)
(k=
t
(o
k
on by
L+
Frequency Domain Processing
89
by do
on by on by
on
12
1
12
121/2
up on
1.059
by
on on
15
do
by
on.
by book.
Part II
90
up (2,
1/3
on by
on
good body,
on
Frequency Domain Processing
by
91
This page intentionally left blank
5 The Fourier Transform
1807. 12. 1817. T/zL;oric’ Arici/j~riqirc~ ck
/U
1822.
Clzaleur
+cc
--oc
(5.0.1)
+oo
Y(w)
~ ( t )
‘y” by
“s”
/
“cu”
+%
y(f) =
-oo
+Y
y(t)e-i2rffdt
(5
-oo
93
94
Chapter 5
(5.0.2)
Hz y( t ) =
(5.0.3)
S
+Tl’
= T+OC -
e+/ W
7-12
2j
/
+ 7-12 p-/2;;+tlt
o -/
dt -
dl
7.-CC
- 7-12
by
Eq. (5.0.5).
(5.0.5)
&fo,Y(f) orthogonulitql
Y(f)
jft).
+fo
-fo,
(5.06)
(5.0.5)
The Fourier Transform
95
k*fo
(5.07)
+
f=fo f=
-.fil
-
-
+
k.fo f=fo
A
+
A
( J ~ I= ) c’2n/of)
j(t),
1 Y(f)
T
by by -
= 1/
T ,Y # 0, Y ( f )= -jS(f;,
6(s), .I‘ = 0.
jft)
=
u)+jn6((o0+ C O ) , Y ( f )= 6(fo + S(fo Y(co)= n6((00 - (U) n6(oo+ 0). Y ( . f )= S ( f o - f), Y(cu)= - CO) Y(co)=
+j6(.fo
+f ) / 2 ,
by 2n
-
+
It([) =
+oc’
=
As 7‘
As
c>’2nfof,
by
96
Chapter 5
5.1 SPECTRAL RESOLUTION ~“121,
T .f, = 1 / T
(5.1.1) y ( i z q , rzT 1/ N
?(n]
1/ N
t,
1/ N
N. hciw t/w
wr prqfrr lzm’ to m?iplititck qf’ tlw .Ji.c.yut.rzc.?’-tiomeiin siizusoichl signtil inck>pencknt qf N ,
6. J*[H]
6 on
0 5 f’ 5 0,5/ T = 0.5/ T,
. ,v-
~~[ii],
I
fLf- I
M j , / M Hz.
Eq.
1.
N
Af=I / ( N O
1/ ( N T ) 0
N
N =M
on 1
Af)
NT, N
97
The Fourier Transform
0.8 0.6 -
wldth 1K
0.4 -
0.2
-
-0.2
-
U
-0.4 -
f)
I
wldth 1K
4.6 -0.8 -
Figure 1
fT / 2
100
Hz 10
on. j*[n], N
Af=fs/ N,
N
Af n T = n l N ,
Y(kAf)= Vk].
(5.1.4)
N
Eq.
p =n.
p # n.
#n
p -n
- (N
-
1)
+(N
1
Eq. p = n,
n do
Eq. p =n
- 1 ), “a” aN e’2n@-t’), ah’=
N j*[n]=j”p]; p = n .
98
Chapter 5
T )=
=
Ajl Aj’=fs
/fs)
.fo
= moAj:
Hz fo=mo = 2 7 c r ~ z ~/r N z ).
by
mo,n, by
nz
fnzO, +uzo
- nzo
-
by N = kn; k =
n(mo -
0, f 1 ,
...,
+moAf Hz
fS
-mo.
Af = j ; /N ,
rziAj;
IT?
Hz up 5,
10,
,r!(f)
ussumes
on
Hz Aj’=
100 16, 6.25 160 2
25
16
10 6.25 Hz
The Fourier Transform
99
“*”
f50
16 m
Y[m]
“m”
160 8,
3. 4. 2, 3,
4
by
‘b*”
by N)
25 Hz
4
160
O3
I
-0.6 I
0 ms
-0.5
Figure 2
1
0.5
100
25 Hz
a (*)
0.4
,
1 -0.5
-0.4
Figure 3
I
0 f/fs
0.5
16
100
Chapter 5
28.125
4.5 100
28.125
5 As
5 16
up spectral Ieukuge.
do
0.6 0.5
-> 0.4 Y
9
0.3 0.2
0.1
flfs
Figure 4
25 (*)
0.6 0.5
-
0.4
9
0.3
t 0.2 0.1
8 Figure 5
28.125 (*)
100
by
The Fourier Transform
101
28.125 112.5
140.612
N.
N
on 4, 32 28.125
3.125 on
9
28.125
on 16 on
2,3,
4.
160
5.3. by
on
on
N2.
5.2
THE FAST FOURIER TRANSFORM
on
do
Chapter 5
102
2
512, 1024,
2
1,048,576 105,240
by by
by by
Hz,
j;
1.3
0
1.7
/
Eq.
W N= e j Z n ' ' ,
by on 2
by
on
odd
+
The Fourier Transform
103
odd
n=O
n=0
N=8,
N
go q
2,
N=
on 3, 4, 5, by
6
Input
Figure 6
8-Point Radix-2 FFT
output
Bit-Reversed Binary Addmss Address
OOO
OOO
100
001
010
010
110
011
001
100
101
101
011
110
111
111
Chapter 5
104
WE-DSP32C, 96002, 6 by
y[3]
y[O]
by
on. h e c m w lit
one node, the two twiddle -fuctors dujc)r on113 in sign! W! = -W$, W,?.= - W { , W i = -Wi. 7
Wi = -W:,
an?l
do by b@
by j*[O]
by
+
0-1 y[4] in-place
W:
-
2 no
-
1
( W:)
2,
unit circle
Figure 7
-
z plane
6,
The Fourier Transform
105
by
Wi,
by
Wd
W: non
N /2 on 2
=
WiT, 4
2Mog2N on
by -
1
0 N,
0.
1
N - 1
on
1.2,
odd
on.
R e [ Y[m]= Re( V
N - t ? ? ] ) , t?? =
0, 1, ... , N - 1. = - lrir [ Y [ N - n ~ ] ; ,
It71 [ in
= 0, 1,
... ,
- 1.
Re[ 1171[ y [ t ? ~ ] = It??[
- 1711
= - Ref y [ N - m])
. by
Eq.
106
Chapter 5
YI[rn]
Y2[nz] Re( Y,[O]}= Re{
Re( Y,[O])= I m [
Inz( YJO])= Inz{ Y2[0])
+
- nz]}}
I!)?(Y,[rzz])= - { I m ( Y[nz]}- I m ( Y" 2
- 1121))
Re( Y?[m]} = - (In2( Y" 2
Y
Im( Y2[tH])= 3 (Rcq Y"
5.3
I I
RP( Y,[r72]}= - {Re(Y [ m ] } Re{ Y" 2
+
1
- 4 ) Inz( Y[nz]}}
1
- nz]) - Re( Y[nz]}}
DATA WINDOWING
by by
also
As
by N A
as
The Fourier Transform
107
2,
8
8
by N 2.0317
2.00
N=64
N)
by N , nurrowhand correction factor is the ratio a rectangular window of the same length. by
the the windoit, integral to the integral of
1
0.8
0.6
0.4
0.2
0
-4
-3
-2
-1
0
1
2
3
4
Relative Bin
Figure 8 N = 64
to
Chapter 5
108
by
on
the brouciband correction .fuctor is determined bqi the squure-root of the integrul of the wincio\t*function squared, then divided by N (the intc>grulof the squured rectangulnr tipinciow jirnction). N = 64 1.6459, 1.6338 N = 1024. N,
Nurrowbund and broadband correction fuctors ure criticullj*importunt to power spectrum unzplitucle culibrution in the frequencqi doniuin. by N N = 64
1.5242
up
-
(
n-i(N{(N-
(5.3.3)
2.0323.
Eq
a by
(5.3.5)
The Fourier Transform
109
1.8768
W E ' ( n )=
8
-c>-;(tl-+-l)~(3.43
10
(5.3.5)
A)
k >_ 4, k = 3,
1.0347.
1.5562.
k
on N , N=64,
2.3229, 9
6 N = 64
1
by on
N = 64,
by by
on do
1
0.8 0.6
0.4 0.2
0 -0.2 -0.2
-+- Hannin
0
0.2
0.6
0.4
n/N Figure 9
0.8
1
Chapter 5
110
Table 1
N=1024
k = 2.6 ~
2.03 17 1.6459
1.5242 1.3801
2.0323 1.7460
1.8768 1.5986
2.3229 1.7890
1 .OOOO 1 .OOOO
2.0020 1.6338
1.5015 1.3700
2.0020 1.7329
1.8534 1.5871
k = 5.2 2.8897 1.6026
N = 1024
1
0.8
0.6
0.4
0.2
0
-4
-3
~~~
1 .OOOO 1 .OOOO
-2
0
-1
1
2
4
3
Relative Bin Figure 10
10
1
N = 64
All
N, 9
The Fourier Transform
111
10. rz = 0
II
=N -
10 on
on
up
7.1, 12.2,
5.4
13.
CIRCULAR CONVOLUTION
by
circwlcrr
(’012\70htk~1?1
Q
by 1
112
Chapter 5
Figure 11 U(.[),
11
N(,/).
12 U(f)
S(J‘)
R ( f ) ,S ( , f ) ,
R(f).
N(f‘)
8. C(j’) no
C ( f ) = S ( f ‘ )= U ( f ) R ( f ) . U(f) C(f) A ( f )= 1 / R ( f ’ ) ,
11
A(f)
by U ( . f )= C ( f ) A ( . f ’ ) .
N(f),
no by
U’(.f)
U’(.f)= C ( f ) A ’ ( f ) / R ( * f ’ ) .
U(J‘) A‘(f)
(5.4.1)
0 2 A’(f’) 2 1
E = [IS[’
(.f)
+ 2 R o { N S )+ ( N / z ] A ’ +z IS]?- 2 R e { S N } * A ’ * -} 2A’(SI‘ + ,4’(N(’ S
N
SN
A’(f)
E(.f‘)
by
E
The Fourier Transform
113
A’ aA’
= ISI’(2A’ - 2 )
aE’ aA‘’
+ 21Nl‘A’
-- 2(ISI’
+
up. on
A’(f)
R(S) B ( . f )= 1 / R ( . f ) . A’(f’)
U(f) H ( f )= A ’ ( f ) / R ( f ) ,
C(f) ’4’(.f’)
no h(t) A’(f) B(f). R”’(z) = E(cr’(t)h(t,- r ) ) A ’ ( f ) B( f ) . B(f)
a’(()
on It is o r i l j i ,fhr tlic~CNSC’ I ~ ~ > I u11 Y J narroii~haizdsignal freyuenclQcomponents are bin-aligncd tlicit the spctral prochrct of tiiv ,functions results in the equivalent linear correlcitio~ior c o ~ i ~ w l i ~ t iino nt/ic tinw-dor~aiii. circular convolution
circular correlation
4),
12, by odd
by
5 12 12
no
(
-
0
-
).
114
Chapter 5
0.4 0.2
0 -0.2
-0.4 -0.6
0
128
64
192
256
Figure 12
(
0 -)
by by
128 256
zc~ro-pudtlc~cl,
12.
128
a
128 a
a
13 (
~
0- )
t
and/ As
13, 128
128
115
The Fourier Transform
-0.6 1
1
1
1
1
1
,
1
1
Figure 13
~
(-
1
1
0-) 128
do
5.5
UNEVEN-SAMPLED FOURIER TRANSFORMS
good
As
Chapter 5
116
by
A good
do
I1
good
by
on
T no
by
t[tz] I I = 0.12, ...,
N N-
, 1 1 [ n ]
11
= 0,1,2,..., N
-
1, j3[11]
117
The Fourier Transform
by (5.5.
N-l
y Lonlh ( ( 0 )= 2a'
Cy[n]- y )
o ( t [ n ]-
tl -I'\
I
1
to(t[n]-
r1=0
(5.5.2)
J
n=o
by
t
(5.5.3)
7
t[n] ulgorithm
verj'
tlio Loriih t r w s f i ~ r uisi N T
by N . A
(5.5.4),
1 YL'"'(tu)I'/N = P"lh((u) (.f/,fi = 0. )
0.1
0.5 14
T 14 15
0.1
on
15.
118
Chapter 5
1.5 1.25 1 0.75 0.5 0.25
YWI
0 -0.25
-0.5
-0.75 -1
o = .5T f / f S = .1
-1.25 L - 1 . 5 - ' 0
~
'
"
32
"
"
'
"
128
96
64
Sample n Figure 14
0
0.5T
0.1.f:~
0.1
0.2
0.4
0.3
0.5
0.6
0.7
0.8
0.9
1
f Jfs Figure 15
14
good
0.9f.S.
-
16
do not SIZO~I, this "nzi,.ror.-ir?ztrKr."
119
The Fourier Transform
0
0.1
0.2
0.3
0.5
0.4
0.6
0.7
0.8
0.9
1
f I fs Figure 16
no
0.9fS
0.5 T
0.1,f.s
As
on
as
19.
17
18 0.45fs
T.
on
0.5T. 19.
14
19
SOMC-
on 20
10T
by
O.7fi
120
Chapter 5
1.25
o=ST f/fs=.45
-1.25
0
32
64
96
128
Sample n Figure 17
0.45fs
0.5T
0.8
Figure 18
14
0.57.
21 do
1
to
0.451s ;is
wave
0.9
The Fourier Transform
121
20 15 10
5
0
dB
-5 -10 -15
-20 -25
-30 0.1
0
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
f Ifs
Figure 19
1.25
on
1
0.5 I 0.25 I
Yml
0 ! -0.25 L -0.5 L
-0.75 I
-1 2
-20
0
20
40
60
80
100
120
140
160
Sample n Figure 20 10
0.7fs
on
122
Chapter 5
20
15
10 5
0
dB
-5 -10
-15 -20 -25
-30
Figure 21 20.
OST? 14 0. I.fs,
16
20
by on up. AI1 thut reullj* niuttc>rsis tliut the correct sample times ure used in the Fourier integrul. As
on by N , 10
N,
21 +I8
The Fourier Transform
123
up
5.6
SUMMARY, PROBLEMS, AND BIBLIOGRAPHY
by by by 10 Hz 100
0.1
10 on
by
no
1024
10,240
1,048,576
by
do
on by N
by
N
Chapter 5
124
by by 1
by /
no
PROBLEMS
1. A
100,000
20,000 20,072 Hz.
2.
25
N N = 128, 1024, by
3.
by
The Fourier Transform
125
1
4. 0.00
11
=1
n=N, 2.00
1.732.
3” BIBLIOGRAPHY
1989. 1977.
H.
B.
S. A. 1986.
J. 1, 1992, pp. 1437. 1974. S.
1975.
REFERENCES 1. J . J. Mclgcizine, 1 , 1992, pp. 1437.
ZEEE Sigttcii Proc*c.s.sitig
This page intentionally left blank
6 Spectral Density
Hz. As
-
T
+T
+
+
t= -
x(t)
on
on -
f
t.
+T/2
(6.0.1) -
-T
X(u)
+ T,
7‘) Sx(w), A good
g(t)
f(t) 127
128
Chapter 6
F(ru),
F(w)G(-co)dtu =
(6.0.2)
(o
Hz (r)
s(t)
- T/2 < t
tiori
no 1
A/D
A
374.5
60.5 1024
240.5 10
(M=
11
4 4 240.5
149
n
% 2
0
64
128
192
256
320
384
448
512
0
64
128
192
256
320
384
448
512
50 40 30 20 10
0
op -10 -20 -30 -40 -50
1.00
cv
*
0.75 0.50 5 0.25 L 0.00
"
' '
"
'''
"
'
'
I '
'
' ' I
' ""' '
I '
I
'
'''' ' '' ''
I
'
''
"
' '
"
'
''
Figure 10 &.(f).
by
I ,*-I
I I
,r-1
12 240
"
''
' I 0
150
Chapter 6
180 135
-135 -180
0
64
128
192
256
320
384
448
512
0
64
128
192
256
320
384
448
512
40 30 20 10 0 -10 rn -20 0 -30 -40 -50
HZ
Figure 11
4
A/D A
151
Spectral Density
180 135
0
64
128
192
256
320
384
448
512
40 30 20
F 'x =m
0
-10 -20 -30 -40 -50
l I , , I / I I
0
64
.
I
,
1
,
1
1
128
1
.
1
1
1
1
1
1
192
,
1
,
1
,
,
,
,
/
,
256
,
,
,
1
,
,
,
.
,
,
320
,
1
,
384
,
,
/
,
,
1
,
,
,
448
,
,
512
Figure 12 by
13.
U(.() + N x ( . f )
X(f)
on V(.f')+
Nj,(.f')
Y(.f')
1) N.I-(,/') Nj*(.f')
C'( I')
152
Chapter 6
Figure 13
X(,f)
H(f’)
Y(,f)
V(.f’).
o,, by ( 1 1.1.14)
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
337
(1
I'
= SO
7
of,
by
So
of.
(1 .Y J'=.Y'
j?'
Rician Densities for Various SNRs
0.7 i Noise Only SNRl SNR2 --SNR 10 +
10 Received Waveform Magnitude
Figure 7
15
2, to
10
Chapter 11
338 4, = A-'
~ 7 " ~ .
Eq.
z =j"' -O.v
c$,=2zdz.
1
(11.1.28)
6
Eq. ( I
z'
4.''
z'
z - So,
z = - z ' - So z
8
z
z
Gtrzrs.sitrrz r~ugnitirdc~ pmhcrhilitjt densitj*fitnctiorz. ( 1 1.1.29)
Gaussian Densities for Various SNRs 0.8
1
0.7
+
SNRl SNR2
0.6
0.5 0.4
0.3 0.2 0.1
n
-0
Figure 8
5 10 Received Waveform Magnitude
15
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
11.2
339
CONSTANT FALSE ALARM RATE DETECTION
to on
by
Pu.stitii(ited
by on
P,,=0.02, 100
A
2 0.1‘%,o n a
15 10 U\.
2
341
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
I I I
0.9 -
Figure 10 T =2
on
ZMG
Pfi1 T, A=T
J ~ o ,
A = To,
12
T
x
(1
(1
342
Chapter 11
1.2
1
i
1
1
0.8
E
0.6 0.4
0.2
0
0
2 3 Relative Threshold T
1
4
5
4
5
Figure 11 T.
1.2 1'
0.8
if
0.6 0.4
0.2 0
0
Figure 12
2 3 Relative Threshold T
1 P,,,
P,,,
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
343
12.
PJ;
1.2.5) (1
by
(1
jfrz]
=A
Rorz + 129[rz]
(1
= 27rfo/.f,
1c1[n]
o-:.
no
1/20:,.
by k , 2.0
5.3
(8/3)12
by
0.8165
50% by
344
Chapter 11
by
by A =T,/mo,, (1
no A. T
P,/
Eq. A,
13
o,,.,
P,, 50'1/;,
P,/,
Eq. by 14,
P,/ by
P,,
0.7
-
0.6
-
--SNR6dB -. --.,SNRlOdB * . * * SNR20dB
0.5 0.4
-
0.3 0.2 -
0' 0
Figure 13
2
4
a s ii
6 0 10 Relative Threshold T
12 T
14
16
345
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
1 0.9 0.8
0.7
-SNR OdB -SNR 6 dB -**SNR 10 dB * - * * Q S N20dB R
-
0.6
z
0.5 0.4
0.3 0.2 0.1 0
-4
Figure 14
-3
-2 by
-1
1
0
T - SNR
2
4
3
SNR
on
0
=
20
=
15
T,
SNR.
on
P,X T , S N R )
T
-
SNR -
1 2 SNR
~
by
346
Chapter 11 1
0.0
0.8 0.7 0.6
z
0.5 0.4
Y
4
1
-3
0 T SNR
-
2
1
3
4
Figure 15 A T, SNR.
book.
by
P,,, P,,
As
11.3,
347
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
11.3
STATISTICAL MODELING OF MULTIPATH
A 6.3.
-
Multi-source multipath 13
on
Chapter 11
340
13.4
by
Coherent multipath on
16
11 by
R
on 16
on
(1
R
A
(k=
=2 n / i ) ,
k
R, 17
(o
Boundary
hl Receiver
Source R 4
Figure 16
I +
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
349
-30 -35
Y
1
1
l i
I
... .
40
%
1
1.
45 I
-50 I
-55
-60
0
100
200
300
400
500
U7
Figure 17
by 100
30
R = 100
l1=30,
345
18
A
(c/f=i).
350
Chapter 11
Image Source \
\ \
\ \
\ \
\ \ \
.
Boundary
Source R 4
Figure 18
7.1. Statistical representation of multipath
on on book. by by by
by
351
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
19
2000 345 by 100
on
dl=0.3
30 2000
16.
0 500 Hz,
19
on 20. on
on
(1
R?,,,
R N = 2000
dl= h / R
h = 30 m. ,
-30
I‘ ‘
-60 0
1
t I
t:
1
I
1
100
200
300 Hz
Figure 19
1
400
500
352
Chapter 11
Boundary Standard Deviation 0.25m -30-
1
1
I
-35-
1 -40
-
-45
-
- 1
-50-
1
’
-55 0
1
100
I
1
I
200
300
400
500
Figure 20 Hz
20 0.25
on
j*= h,
+ > R.
h
h h l R = 0.5.
Random variations in refractive index
At
AM
20
k
354
Chapter 11
by
by
by
good
on
on
by
on on
5
by
21
355
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
c ( z ) = CO
+ z -dc(z) dz z
R 21,
R=-
c ( z ) = 0.
by
dc(z ) Id,x
21,
s
1.3.11 )
s = OR
Refracted Ray s
\
DirectRay x
0
Figure 21 a
Chapter 11
356
/I=R
[
m]
As A-=
1
by
= 345
R=3450
cic'/cl,- =
/1=36.4 3.5
+ 0.1 1003.5
i s - -.y = 3.5
+
+ 350;'
As
1000
by
11.4
SUMMARY, PROBLEMS, AND BIBLIOGRAPHY
11
on
357
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
statistical conjidence
data situatioriul
fzision, awareness.
P,,, Pf,,
A sentient by
by
on on
body
(3) by
good do
body by
1 1.1
Tinge-sjwchronous uveruging is one of the wost siniple and effctive r7ietliod.s for. signal-to-noise iniprovement in signal processing arid should he esploittd ~tiliel.4~ierpossihle.
5.4
11.2.
358
Chapter 11
up
11.3
by
on on
on
PROBLEMS
1
50 Hz
1
400 Hz,
1024
3.2 kHz, 8192
If
8192
no
on
50 1x ‘/U
you
by
359
Narrowband Probability of Detection (PD) and False Alarm Rates (FAR)
5.
by 6.
by by
7.
1%
on no 0.1%)
8.
p(r) p(s).
BIBLIOGRAPHY
26,
I. 1972. 2nd
S. 1991. 1971. 1,
1977.
S. 7,
1986. 3,
K.
1979. 1968. 7, 1996. REFERENCES 1.
K. 103(3),
1998.
This page intentionally left blank
12 Wavenumber and Bearing Estimation
7.1 ).
by
on
361
Chapter 12
362
by
by
12.1
bound
11
8
12.2 by
12.3
12.1
THE CRAMER-RA0 LOWER BOUND
bound
Wavenumber and Bearing Estimation
363
on As
on N by
N+ 1
N
1
N
on N
m
02.
A=[n?02]
N
I;( Y J ) ,
MF
R 8
364
Chapter 1 2
F ( 12.1.4)
F( Y , j , ) .
Eq.
Eq. ( 12.1.4)
]:[
=E -
+ E[F$]
Eq. (
A,
F. F F = 1, E [ 4 ] = O .
F=$
365
Wavenumber and Bearing Estimation
[3
-E - = E[$$T]
E
J
J
Fisher Infomatiori A4atr.i.u.
J
J (1
Eq,
A( Y ) ;I(Y)
R( Y).
$( Y)
8.1
by
/?
p.
E[p]= E[R$7]E[$$‘]-’
$,
Eq.
E[$]
E[e]=E[;I].
A)” = R - E [ i ] Ae = e
-
E[e]
A) A4 = E [ i @ T = ] E[Ah,hT] M
bius of theparmietcv estinicitiorz
(
366
Chapter 12
E[AeAeT]= E[(AR - MJ-'$)(A/ZT - ~ T J - ' M T ) ] = EIAI.AAT] - MJ-'E[$AAT] - EIAI.$']J-'MT = E[AAAAT]- M J - ' M T - M J - ' M T
+ MJ-'$$'J-'MT
+ MJ-'MT
= E[AiAAT]- M J - ' M T ( 12.1.18)
qAeAe'] 2 0
a2(i)=
bound on
E[AAAAT]2 M J - ' M T
( -
on
M = E[A't,bT]= --[A']
-E
= 1 for the unbiased case
Eq.
;I'
no
on
A, A
2.1.2
a2(i) = E[AAAAT]2 J-'
N
Eq.
CRLB efjcient bound.
N
(
Wavenumber and Bearing Estimation
367
i= [m. ' ] a
Y,i)
L.
i= 1
I
-
I)( Y ,
Y , A) =
=
- 172)
N 1 ' -- 4 202 20
CO,- In)'
+
Y
r Y ( Y , E,) =
1
Y , A) =
ai
J = E ( Y (Y , A)} =
1;
0 N 2a4
]
[
E ( A ~ H A ~E{AniAa') ~] a-(i) = '* E(AniAa') E { A a 2 A a 2 )
25
[ ] N
2
9. 1% 1%
1
9 0.25
3 / a . N
144
0.09, N .6
0.09, N
145,800
Chapter 12
368
12.2
BEARING ESTIMATION AND BEAMSTEERING
by by
on by on
by by
1 tI
on
2 3
on
tl
1
0,
= $3
9,. &,
=
-
(
I$~ (o
k = t o / c , c’ 0
Eq. ( (
369
Wavenumber and Bearing Estimation
Sensor 3
d
Sensor 2
Sensor 1
0
Figure 1
by
7.1
(1
(12.2.5)
Chapter 12
370
(r, (I-,,
0, - r/ OA - r,
OJ)
(r,
0,)
(I-,,
8, - r, 0~ - r/
0,) 0,) L
J (
by
1
2
Figure 2
371
Wavenumber and Bearing Estimation 04,
c4 =
{ J=
S
{ -S1N R 1
(1
oNis
x
y
by kd
0 3 by kd 0.01
dli
t
COS
Figure 3 by kd
372
Chapter 12
0.5,
(//A
0.05,
(
a, 0.5
4
5.
on 6. A
25. 100 7. 94 100
1
117 17
0.1 200
217
0.05
”/o
234
0
8.
9.
U,,
T 10.
9.
9
on
519
Intelligent Sensor Systems
BIBLIOGRAPHY
1993. 1983. 1991. 1996.
S. Z. 1988. 1997.
I. 1992. 2nd
1992.
I: 1986. 1992. 1992. 1974. REFERENCES 1. 2.
1992. 1974.
3.
E. 1710,
4. 1992.
1992, pp.
This page intentionally left blank
15 Sensors, EIect ro n ics, and Noise Reduction Techniques
Intelligent Serisor Sjtstcwi
on
sewticvit processing. self-(iH’(ire~i24.s,s,
situcitiorzcil
ci~~ir~vic~.s.s.
no
on
521
522
Chapter 15
by
15.1
ELECTRONIC NOISE
A As ttzerniul popcwrrz
.slzot
c‘orittrct
523
Sensors, Electronics, and Noise ReductionTechniques
by Thermal Noise by
by by
good
no on
on Eq.
v,= J4kTBR
(15.1 .l)
5.1 x
(Q)
524
Chapter 15
up
&E 01
=
J4kTBR
( 1 5.
As
1 10
10
10 (1 g 0.001 g. 100 pg
by 10 0.1
9.81
10
1 10 Hz
1000 jig
30 pg
on
a
Eq. as ‘/z
Shot Noise
A
by
It,(.
B x 10.-
y ii
’’)
by a
525
Sensors, Electronics, and Noise Reduction Techniques
by 1/
1/ 2 N 1
Contact Noise
(1 5 .
K
Thermal or Shot Noise 4
7
1
-
0
1
I
0.2
0.3
I
I
1
1
I
I
0.6
0.7
0.8
0.9
~ ~
~~
0.1
0.4
0.5 sec
1
-20 m U
-40
-601 0
1
50
1
I
1
I
1
1
1
1
I
100
150
200
250
300
350
400
450
500
Hz
Figure 1
526
Chapter 15
by
by
by
R,
o,.(f) = RKI‘/$
/f”
by j ; / N ,
Hz !4
ZMG
-1
0
k
N-k,
k N -k
05k
> j w z ,
17 (T
(
529
Sensors, Electronics, and Noise ReductionTechniques
11,.
0,. =
=p / p o
= 5.82 x 107
11,.
100 Hz,
148 1 MHz,
on good
108
300 x 106
60 Hz
5
6.3
Hz) 10 100 MHz 100
(r-c
by 1
= 271fEOr
(1
r
(3
RE = 321 +
-
dB
-
(r
q), ( q> k ) ,
(15.3.10)
S qsk
S
q.vy
S on
S S
S
2 go by by
k
q
p
p q.vk S E
p
(r,
by p>y
py y
k 2.
f
< 1909.859 Hz
8 rz = 0, 1,2, . . . , 7
Answers to Problems
601
15, 3
14,
on, 16
1
1 9
8
16
by
2
on
on 0.1493 1.5 0.6254
1
35.83
16 15
up
1.4, 1
fmax
k 2
1909.859 828
1500 345 >
8.66 120
5 28.79 - 20 - 10
-6
R R
on X = 103 = 1000 CHAPTER 13
1.
Plane Wave @ 60 deg 2
I
)i
3
3
2,
2
3, 1
d
1
3
0
60
60 60 3
Answers to Problems
602
3
s = [ 1e-jkd
Ue-j2kd
60 X = [X~(~)X~(CU)X~(W)]
3
S = [e+j2kdcosn
0 T
e+jkdcoso
on
XS. up
k = m/lO,OOO 5 kHz,
d=1
2.
1
on 9
< 1 , 0 >, >,
.c
5 on
< 0, 1 >,
0
If 1
16,
3
by 2
on.
15,
0 =0
0
by 16
0
0, S(c0,) = [e
16)
01 -jk[cos( n/ 16)
e
...
e-jk[cos([n-I]n/16)cosn+sin([n-l]n/16)sin 01 e-jk[cos(
-n