Acquisition of Weak Signals - Fundamentals of Global Positioning System Receivers: A Software Approach - page 234

Incoherent integration loss

10

9

8

P d = 0.9

P fa = 10 − 7

7

6

5

4

3

2

1

0
10 0

10 1
Number of incoherent integration

10 2 10 3

FIGURE 10.3 Noncoherent integration loss.

The discussion above can be summarized as follows: For a desired S / N of
14 dB (or P d = 0 . 9and P fa = 10 − 7 )the D c ( 1 ) = 21. From Equation (10.13)
the integration loss can be found. Then Equation (10.14) can be used to find the
noncoherent integration gain.
Let us use two examples to illustrate the combination of coherent and non-
coherent integrations. Suppose that the input signal strength is at − 150 dBm (or
20 dB below the normal power level.) One approach is to use 1 ms of data for
coherent integration and the rest of the gain is obtained from noncoherent integra-
tion. For 1 ms coherent processing, the corresponding bandwidth is 1 kHz, and
the noise floor is at − 144 dBm; thus the S / N is at − 6dB( − 150 + 144). In order
to achieve S / N = 14 dB, the noncoherent integration should provide 20 dB gain.

The result obtained from Figure 10.4 is that n ≈ 1000. This approach requires

approximately one second of noncoherent integrations. Every 20 ms, however,
there is a possibility of a navigation data transition. If one takes this into consid-
eration, the required noncoherent integration could be slightly longer because a
data transition in a certain ms will decrease its coherent integration.
The other approach uses 10 ms coherent integration. With 10 ms coherent
integration, the equivalent bandwidth is 100 Hz, and the corresponding noise floor
is at 154.) In order to achieve a S / N of

14 dB, an noncoherent gain of 10 dB is required. This requires approximately 19

− 154 dBm. The S / N is 4 dB ( − 150 +