Acquisition of GPS C/A Code Signals - Fundamentals of Global Positioning System Receivers: A Software Approach - page 148

Figure 7.7b, the input signal falls in between two frequency bins. Suppose that
the amplitude of X(k) is slightly higher than X(k 1 ) ;then X(k) will be used in

Equations (7.21) and (7.22) to find the fine frequency resolution. The difference
frequency should be close to 500 Hz. The correct result is that the input frequency
is about 500 Hz lower than X(k) . Due to noise the 500 Hz could be assessed
as higher than X(k) ; therefore, a wrong answer can be reached. However, for
this input frequency the amplitudes of X(k) and X(k

−

− 1 ) are close together and

they are much stronger than X(k + 1 ) . Thus, if the highest-frequency bin is X(k)

and the phase calculated is in the ambiguous range, which is close to the centers
between X(k − 1 ) and X(k) or between X(k) and X(k + 1 ) , the amplitude of
X(k − 1 ) and X(k + 1 ) will be compared. If X(k − 1 ) is stronger than X ( k + 1),
the input frequency is lower than X ( k ); otherwise, the input frequency is higher
than X ( k ). Under this condition, the accuracy of fine frequency is determined by
the phase but the sign of the difference frequency is determined by the amplitudes
of two frequency components adjacent to the highest one.
However, the problem is a little more complicated than this, because it is
possible that there is a 180-degree phase shift between two consecutive data sets
due to navigation data. If this condition occurs, the input can no longer be treated
as a cw signal. This possibility limits the ambiguous bandwidth to 250 Hz for
1 ms time delay. If the frequency is off by ± 250 Hz, the corresponding angle is
± π /2. If the frequency is off by + 250 Hz, the angle should be + π /2. However,

a π phase shift due to the navigation will change the angle to − π) ,

which corresponds to a − 250 kHz change. If the phase transition is not taken
account of in finding the fine frequency, the result will be off by 500 kHz.
In order to avoid this problem, the maximum frequency uncertainty must be
less than 250 Hz. If the maximum frequency difference is ± 200 Hz, which is
selected experimentally, the corresponding phase angle difference is ± 2 π /5 as
shown in Figure 7.8. If there is a π phase shift, the magnitude of the phase
difference is 3 π/ 5[

π /2 ( + π/ 2 −

], which is greater than 2 π/ 5. From this
arrangement, the phase difference can be used to determine the fine frequency
without creating erroneous frequency shift. If the phase difference is greater than
2 π /5, π can be subtracted from the result to keep the phase difference less than
2 π /5. In order to keep the frequency within 200 kHz, the maximum separation
between the k values in X ( k ) will be 400 kHz. If the input is in the middle of
two adjacent k values, the input signal is 200 kHz from both of the k values.
In the following discussion, let us keep the maximum phase difference at 2 π /5,
which corresponds to a frequency difference of 200 kHz.
The last point to be discussed is converting the real and imaginary parts of
X ( k ) into an angle. Usually, the phase angle measured is between ± π .The
two angles in Equations (7.20) and (7.21) will be obtained in this manner. The
difference angle between the two angles can be any value between 0 and 2 π
as shown in Figure 7.9. Since the maximum allowable difference angle is 2 π /5
for 200 kHz, the difference angle must be equal or less than 2 π /5. If the result
is greater than 2 π /5, 2 π can be either added or subtracted from the result, and
the absolute value of the angle must be less than 2 π /5. If noise is taken into

|± ( 2 π/ 5 ) ∓ π |