Global Positioning System Reference
In-Depth Information
satellite is at 1000 of the 5456 data points, the pointer is at 1000. If there is no
Doppler shift, after the first ms the mismatch is 0.34 sample. After two ms, the
mismatch is 0.68, which is greater than 0.5 sample. The pointer is subtracted
by one and the value is 999, which indicates a data point shift. After the shift,
the mismatch becomes
0.32. The mismatch accumulator sums the mismatch
of every ms. In this special case the first three values of the mismatch will
be 0.34, 0.32, and 0.02, and accumulator sums these values. In general, the
Doppler frequency must be taken into consideration in obtaining the values of
the mismatch.
The next procedure is to find the fine time. After one second of operation
another C/A code is generated, again, according to the carrier frequency and the
fine time. The C/A code generated after one second of operation must match the
input data C/A code. However, often there is an error in time, and this error is
represented by ε . To see this, in the example above, suppose that at the end of
the first 3 ms, the four mismatches are ε , ε
+ 0 . 02. The
early gate of the first 3 ms is shown in Figure 11.15. Because the prompt gate
is off by ε the early gate will be off by ε
+ 0 . 34, ε
0 . 32, and ε
1. This is because it is one sample
before the prompt gate. Since the three early peak correlations are on a straight
line, as shown in Figure 11.16 the following relation holds:
y e 1
y e 2
y e 3
1 ) =
0 . 34 ) =
P
P
1
+
P
1
0 . 32 )
y e 1 +
y e 2 +
y e 3
=
P
1 )
+
P
1 + 0 . 34 )
+
P
1 0 . 32 )
y e 1 + y e 2 + y e 3
3 P
=
(11.10)
+
( 3 ε
2 . 8 )
where P represents the time when y e approaches zero, as shown in Figure 11.16.
Let us use d n to represent the mismatch and extend this idea to the whole
0.68 sample off
0.02 sample off
0.34 sample off
−0.32 sample off
1st ms
2nd ms
3rd ms
3rd ms
After 1 sample shift
FIGURE 11.15 First 3 ms mismatch of the prompt gate.
 
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