Acquisition of Weak Signals - Fundamentals of Global Positioning System Receivers: A Software Approach - page 239

2500 time domain results

Result from 1st ms
Result from 2nd ms

Result from 10th ms

2500 10-point FFT

2500 time domain results

FIGURE 10.6 Illustration of 10-point FFT operations.

In order to cover more bandwidth than 1 kHz, the process above must be
performed several times. For example, if another kHz should be searched, the
input data must be converted to a baseband using a different frequency, f 0 .To
cover 10 kHz frequency range, the input must be down-converted 11 times, each
separated by 1 kHz. This operation can be achieved by changing the frequency,
f 0 , in Equation (10.15) from f 0 = 1245 kHz to 1255 kHz in 1 kHz steps. The
overall output is a matrix of 100 by 2500 points. Or there are 100 frequency bins
separated by 100 Hz and 2500 time bins separated by 400 ns.
In order to save computation time, the 2500-point operation can, at times, be
reduced to 2048 points. Since 2048 is base 2 number, the operation is slightly
faster. Although the outputs are not evaluated in detail, it appears that this results
in very little noted degradation.

10.11 RECOVER LOSS ON IN-BETWEEN FREQUENCIES ( 13 )

In the above-discussed acquisition method, FFT is used to find the frequency.
Whenever FFT is used with a rectangular time window, the frequency response is
a sinc function (sin x/x ). There are always the possibility that an input frequency
will fall exactly between two frequency bins, where x

= π /2. When this situation

occurs, the amplitude of the signal will drop due to
20 log sin x
x

x = π/ 2 = 20 log ( 0 . 6366 )

=− 3 . 92 dB ( 10 . 16 )

In searching a weak signal, this amount of loss cannot accepted. The goal of this
section is to recover this loss.