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Acquisition time (simple example)
 The first problem is how long it takes to acquire the code. That
is, align the receiver code generator to within a fraction (1/2 or
1/4) of a chip.
 Assume the code has N PN symbols, the probability of detection
is unity ( PD  1 ), the probability of false alarm is zero ( PFA  0 ),
and the dwell time (integration time) is TD ( LTC ) second.
 Then assuming no Doppler shifts and no oscillator instabilities,
the time to search all N chips in half chip increments (2N cells)
is TA  2NTD sec and the mean acquisition time is just
E[T1 ]  NTD sec
When the detection probability is not unity and false alarm
probability is not zero, the calculation is no longer quite so
simple.
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Wireless Communication Technologies 2.4.6
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Example 2.4-7 Mean acquisition time calculation
2  PD
NTD .
 Show that if PD  1 and PFA  0 , then E[Ta ] 
PD
【Sol.】
E[Ta ]  E[T1  PD  (T1  TA )  PD (1  PD )
 (T1  2TA )  PD (1  PD ) 2  (T1  3TA )  PD (1  PD ) 3  ]
 E[T1 ]  PD 
1
1
 TA  PD (1  PD ) 2
1  (1  PD )
PD
1  PD
 NTD  2 NTD
PD

2  PD
NTD
PD
Note that when PD  1 the result agree with E[T1 ]  NTD sec .
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Acquisition time for the single dwell time search
 Consider the simplified filter, square, and integrate detector
acquisition circuit shown in Figure
 Assume that there are q cells to be searched. Now q may be
equal to the length of the PN code to be searched or some
multiple of it.
 For example, if the update size is one-half chip, q will be twice
the code length to be searched.
BPF
(mod
BW)
AMP
(
)2

TD
0
dt
Compare
thres.
Yes
No
VCXO
PN Code
gen.
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Update control
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Acquisition time for the single dwell time search
 Further assume that if a ‘hit’ (output is above threshold) is
detected by the threshold detector, the system goes into a
verification mode that may include both an extended duration
dwell time and an entry into a code loop tracking mode.
 In any event, we model the ‘penalty’ of obtaining a false alarm
as K D sec, and the dwell time itself as  D sec. If a true hit is
observed, the system has acquired the signal, and the search is
completed.
 Assume the false alarm probability PFA and probability of
detection PD are given. Clearly the time to acquire, that is, to
obtain a true hit (not a false alarm) is a random variable.
 Also we assume that PD is constant (time invariant) and as a
consequence, the analysis of the model follows discrete time
invariant Markov processes and flow graphs theory.
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Analysis of acquisition time
 Let each cell be numbered from left to right so that the kth cell
has an a priori probability of having the signal present, given
that it was not present in cells 1 through k-1, of
1
Pk 
q 1 k
 The generating function flow diagram is given in Figure using
the rule that at each node the sum of the probability emanating
from the node equals unity.
 Consider node1. The probability a priori of having the signal
present is P1  1 / q , and the probability of it not being present in
the cell is 1  P1 .
 Suppose the signal were not present. Then we advance to the
next node (node 1a); since it corresponds to a probabilistic
decision and not a unit time delay, no z multiplies the branch
going to it.
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Generating function flow graph for acquisition time
F
F
4
z K 1
PD z
z K 1
q
3
z K 1
(1   ) z
4
z K 1
(1   ) z
(1   ) z
(1   ) z
z K 1
2
P1
3
P2
(1  PD ) z
(1  PD ) z
P3
zK
S
zK
1b
1
2
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z
(1   ) z
1a
...
2b
z
1  P1
(1   ) z
1  P2
(1   ) z
1  P3
...
2a
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Analysis of acquisition time
 At node 1a either a false alarm occurs, with probability PFA   ,
which requires one unit of time to determine (  D sec) and then
K unit of time ( K D sec) are needed to determine that there is
no false alarm.
 Or there is no false alarm with probability (1   ), which takes
one dwell time to determine, which requires the (1   ) z branch
going to node 2.
 Now consider the situation at node 1 when the signal does occur
there. If a hit occurs, then acquisition occurs and the process is
terminated, hence the node F denoting ‘finish’.
 If there was no hit at node 1 (the integrator output was below the
threshold), which occurs with probability 1  PD , one unit of
time would be consumed. This is represented by the
branch (1  PD ) z , leading to node 2.
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Analysis of acquisition time
 At node 2 in the upper left part of the diagram either a false
alarm occurs with probability  and delay (K+1) or a false
alarm does not occur with a delay of 1 unit.
 The remaining portion of the generating function flow graph is a
repetition of the portion just discussed with the appropriate node
changes.
 In order to obtain the transfer function of the flow graph we
shall reduce it by combining sections at a time. Let
H ( z )  z K 1  (1   ) z
Then the flow graph can be drawn as shown in following Figure.
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Reduced flow graph diagram
F
4
PD z
3
q
H (z )
H (z )
H (z )
F
(1  PD ) z[ H ( z )]q 1
F
PD z
(1  PD ) z[ H ( z )]q 1
PD z
H (z )
2
P1
P2
(1  PD ) z
S
Pq  1
2
1
1  P1
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H (z )
1  P2
H(z)
...
q
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Analysis of acquisition time
 Letting  1  PD and
Q( z ) 
z
1   ( H ) q1
we can reduce the flow graph to that of following Figure; since
Q (z ) takes into account the feedback loops.
F
P1 (1   )Q( z )
S
1
P3 (1   )Q( z )
P2 (1   )Q( z )
(1  P1 ) H ( z )
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2
...
(1  P2 ) H ( z )
3
1 (1   )Q( z )
...
q
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Analysis of acquisition time
Pl (1   ) z
and Al ( z )  (1  Pl ) H ( z ) ,
q 1
1  zH
the generating function flow graph is redrawn in a slightly
simpler form.
 Letting Bl ( z ) 
F
S
1
B3 ( z )
B2 ( z )
B1 ( z )
A1 ( z )
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2
...
A2 ( z )
3
Bl (z )
A3 ( z )
...
q
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Analysis of acquisition time
 By inspection the flow graph is given by
U ( z )  B1  A1 B2  A1 A2 B3    A1 A2 A3  Aq1 Bq

(1   ) z
2
[
P

(
1

P
)
P
H

(
1

P
)(
1

P
)
P
H
1
1
2
1
2
3
1  zH q1
   (1  P1 )(1  P2 )  (1  Pq 1 ) H q1
 Writing Pi and 1  Pi in terms of q, we have
(1   ) z 1 q  1 H
q 1 q  2 H 2
H q1
U ( z) 
[ 


]
q 1
1  zH
q
q q 1
q q 1 q  2
q
so
(1   ) z
1 q1 l
U ( z) 
 [  H ( z )]
q 1
1  zH
q l 0
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Analysis of acquisition time

 As a check U (1) should be unity. (Hint: Pij ( z )   z n pij (n) )
(1   ) 1 q 1
U (1) 
 [1]  1
1 
q l 0
n 0
The mean acquisition time is given by (after some algebra)


  (1   ) z
1 q 1 l
E[Ta ]  U ( z )  D 
 [ H ( z )]  D

q 1
z
z 1  zH
q l 0
 z 1
z 1
2  (2  PD )( q  1)(1  KPFA )

 D
2 PD
with  D being included in the formula to translate from our unit
time scale.
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Analysis of acquisition time
 As a partial check on U(z), let PD  1 and PFA  0 . Then we have
1 q
E[Ta ] 
 D
2
 This result can be obtained by direct calculation by noting that
the mean time to acquire is given by (the a priori probability is
1/q )
1
2
q
 D q(q  1) (q  1) D
E[Ta ]   D   D    D 

q
q
q
q
2
2
 For the usual case when q  1 , E[Ta ] is given by
E[Ta ] 
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(2  PD )(1  KPFA )
 q D
2 PD
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Analysis of acquisition time
 The variance of the acquisition time is given by
 d2
d
d
2
Var (Ta )   2 U ( z ) | z 1  U ( z ) | z 1 [ U | z 1 ]    D2
dz
dz
 dz

 It can be shown that (when q  1 and q  K (1  KPFA ) )
1
1
1
Var (Ta )  (1  KPFA ) q ( 
 2 )   D2
12 PD PD
2
2
 As a partial check on the variance result, let PD  1and PFA  0.
Then we have
q 2 D2
Var (Ta ) 
12
which is the variance of a uniformly distributed random variable,
as one would expect for the limiting case.
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