## Ca Code And Data Format125

The C/A code is a bi-phase modulated signal with a chip rate of 1.023 MHz. Therefore, the null-to-null bandwidth of the main lobe of the spectrum is 2.046 MHz. Each chip is about 977.5 ns (1/1.023 MHz) long. The transmitting bandwidth of the GPS satellite in the L1 frequency is approximately 20 MHz to accommodate the P code signal; therefore, the C/A code transmitted contains the main lobe and several sidelobes. The total code period contains 1,023 chips. With a chip rate of 1.023 MHz, 1,023 chips last 1 ms; therefore, the C/A code is 1 ms long. This code repeats itself every millisecond. The spectrum of a C/A code is shown in Figure 5.2.

In order to find the beginning of a C/A code in the received signal only a very limited data record is needed such as 1 ms. If there is no Doppler effect

C/A of sat #12 sampled at 5 MHz for 1 ms

C/A of sat #12 sampled at 5 MHz for 1 ms -2500 -2000 -1500 -1000 -500 0 500 1000 1500 2000 2500

Frequency in KHz

FIGURE 5.2 Spectrum of a C/A code.

-2500 -2000 -1500 -1000 -500 0 500 1000 1500 2000 2500

Frequency in KHz

FIGURE 5.2 Spectrum of a C/A code.

on the received signal, then one millimeter of data contains all the 1,023 chips. Different C/A codes are used for different satellites. The C/A code belongs to the family of Gold codes,(5) which will be discussed in the next section.

Figure 5.3 shows the GPS data format. The first row shows a C/A code with 1,023 chips; the total length is 1 ms. The second row shows a navigation data bit that has a data rate of 50 Hz; thus, a data bit is 20 ms long and contains 20 C/A codes. Thirty data bits make a word that is 600 ms long as shown in the third row. Ten words make a subframe that is 6 seconds long as shown in row four. The fifth row shows a page that is 30 seconds long and contains 5 subframes. Twenty-five pages make a complete data set that is 12.5 minutes long as shown in the sixth row. The 25 pages of data can be referred to as a superframe.

The parameters mentioned in Section 4.10 are contained in the first three subframes of a page. If one can receive the information of these three subframes from four or more satellites, the user location can be found. Theoretically, one can take a minimum of about 18 seconds of data from four satellites and be able to calculate the user position. However, the subframes from each satellite will not reach the receiver at the same time. Besides, one does not know when the beginning of subframe 1 will be received. A guaranteed way to receive the first three subframes is to take 30 seconds (or one page) of data. Thus, one can take a minimum of 30 seconds of data and calculate the user position.