Global Positioning System Reference
In-Depth Information
where
E
[] is the average operation and
represents correlation. For a sampling
rate of 5 MHz, the value of
K
obtained is
K
⊗
10
−
4
. This value is
obtained by the following procedure: Sum the square of the 1-ms cross-correlation
outputs, and divide by 5000 to obtain
N
e
. Next, divide
N
e
by
P
1
to obtain
K
.
Since the autocorrelation peak is 5000,
P
1
=
=
7
.
0654
×
5000
2
.
Since we have the equivalent noise power, we can rewrite the processed signal-
to-noise ratio of the weak signal as
S
N
10 log
P
2
N
e
+
p
=
(
12
.
7
)
N
0
/G
Substituting Equations (12.4) and (12.6) into Equation (12.7) yields
S
N
p
=
S
N
2
+
G
dB
−
10 log
KG
10
(S/N)
1
/
10
+
1
(
12
.
8
)
where
G
is the acquisition processing gain in ratio and
G
dB
is gain in dB. In
Equation (12.7) the thermal noise power
N
0
can be decreased through acquisition,
but the equivalent noise power
N
e
is not affected.
From Section 10.3 we know that the desired weakest signal that can be pro-
cessed is
−
39 dB (reference to 2-MHz bandwidth and
C/N
0
=
24 dB). In order
to detect this signal, a
S
/
N
of 14 dB is required. Thus the required acquisition
gain is 53 dB (14
10
5
.
With the values
G
,
G
dB
,and
K
known,
S/N
p
can be calculated as a function
of
(S/N)
2
. Figure 12.17 shows the relation between
(S/N)
2
and
(S/N)
p
for
different strong signal levels. The dashed line at the left-hand corner of the figure
corresponds to the case where no strong signal is present. This line shows a gain
of 53 dB. For example, when the input
(S/N)
2
is
+
=
×
39) or
G
2
39 dB, the processing output
(S/N)
p
is 14 dB. The solid lines correspond to cases where a strong signal
is present for
(S/N)
1
=−
−
13 dB in one dB step. The required output
threshold of 14 dB is also indicated. The difficulties associated with a weak
signal acquisition in the presence of a strong signal can be illustrated by a simple
example. Suppose that the input is
S/N
2
=−
35 dB (
C/N
0
=
28 dB), and the
corresponding output is
(S/N)
p
=
18 dB without the presence of a strong signal.
When a strong signal of
(S/N)
1
=−
19 dB is present the processed
(S/N)
p
=
13
.
5 dB, which is slightly lower than the desired value of 14 dB. When the strong
signal is
(S/N)
1
=−
13 dB, the processed
(S/N)
p
=
8dB.
From this analysis it is evident that in order to successfully acquire weak
signals, it is necessary to remove the signals with nominal power levels. There
is an obvious procedure to achieve this purpose. It works to remove the strong
signals through direction subtraction. In its operation, the amplitude, the initial
C/A code, the carrier frequency, and the phase of the carrier frequency must be
obtained. To begin, write the strong signal as
19 to
−
S
1
=
A
1
C
1
sin
(
2
πf t
+
θ)
(
12
.
9
)
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