Tracking Weak Signals - Fundamentals of Global Positioning System Receivers: A Software Approach - page 278

Negative time shift

1 ms of C/A code data with zero code phase

offset t of = a initial C/A code phase

b

offset t of = b (positive)

FIGURE 11.5 Offset time of the C/A code.

This equation is identical to Equation (11.1). The actual program for generating
the digitized C/A code is listed at the end of this chapter. The inputs include the
sampling frequency and the offset time.
The actual generation of the C/A code every 20 ms occurs as follows. We will
use a numerical example to illustrate the operation. The important factor is the
determination of the offset time change between the two C/A codes with Doppler
frequency taken into consideration. If the Doppler frequency f d 5kHz, the

t se can be obtained from Equation (11.1). The difference between t s and t se is
−

=

10 − 13 sec. In 20 ms there are 100,000 samples. Therefore the total
difference between two offset times is

6 . 3475 ×

10 − 8

63.475 ns). This
value is less than half of the sample time of 100 ns. It implies that if the first C/A
has zero offset, 20 ms later the C/A code has an offset time of − 63.475 ns. Since
only positive values can be used in the C/A code digitization program, this offset
time must be changed to a positive value as 999.9365 µ s (1000 − 0 . 063475). This
offset time and the correct Doppler frequency will be used to generate the next
1 ms of C/A code. As this process continues, the offset time will accumulate.
For example, after another 20 ms the offset becomes

− 6 . 3475 × sec (or −

− (2 × 63 . 475) ns if the

Initial phase of C/A code at 0

1 ms of C/A code used for the first 20 ms

Initial phase of C/A code at 999.9365 us

1 ms of C/A code used for the second 20 ms

−63.475 ns
FIGURE 11.6 Generation of 1 ms of C/A code twice for 40 ms of operation.