Global Positioning System Reference
In-Depth Information
the lines
A
represent the outputs of the narrow band filters.
The shape of the narrow band filter is not shown. For example, a signal at
A
4,
as shown in the figure, without the narrow band filters, the outputs will appear
at
A
0and
B
0. Since there are narrow band filters, the outputs will appear at
A
4
and
B
4. The signal appears at
B
4 because the negative frequency is represented
by
B
+
,
A
−
,
B
+
,and
B
−
−
. The amplitude difference is caused by the shape of the wideband filters
A
and
B
. The phase difference of these two outputs is close to 180
◦
. When the
input is at filter 5 of
A
(500 Hz from 0 Hz), it has the same amplitude in filter
5of
B
(
−
500 Hz from 1000 Hz.) The difference of these two outputs is equiv-
alent to their summation because of they are 180
◦
out of phase. This operation
doubles the amplitude of the original outputs. Therefore the signal in the center
portion between of filter banks
A
and
B
can be obtained by taking the difference
between outputs
A
and
B
to increase the overall amplitude. For example, the
outputs of filters 4 to 6 of
A
and the outputs of 4 to 6 of
B
can be replaced by
their differences.
In this operation, although the signal amplitude is increased through subtrac-
tion of two corresponding outputs from wideband filters
A
and
B
, the noise also
increases. In order to keep the noise uniform across all channels, the outputs
of the subtraction channels are divided by
√
2 because noise is increased by
√
2. When the signal is 500 Hz away from a channel, its amplitude is 0.6366
(Equation (10.16)). After we double this value and divide by
√
2, the amplitude
becomes 0.9. The corresponding degradation is 0.91 dB, which agrees with the
result of Equation (10.17). Therefore this approach recovers about 3-dB loss.
Figure 10.9 shows the outputs of 4 to 6 obtained from the difference. The ampli-
tudes of points 4 to 6 are divided by
√
2 to compensate for the noise increase.
The best improvement by this method is also shown in this figure.
This same idea applies to the narrow band filters designated as 0's to 9's, as
shown in Figures 10.8 and 10.9. They have a frequency bandwidth of 100 Hz. To
cover a 10 kHz frequency range there are 100 narrow band filters. If a signal falls
at the center between two frequency bins, the amplitude decreases. In order to
improve detection of signals between frequency bins, the outputs from adjacent
bins are subtracted to increase the output amplitude. The amplitude obtained from
an adjacent bin combination is also divided by
√
2 to compensate for the noise
increase. Through this operation there are a total of 199 frequency bins: 100 from
the circular correlation and 99 from the difference between two frequency bins.
The frequency resolution is 50 Hz, and the time resolution is 400 ns. A certain
threshold can be used to determine whether a signal is detected.
A slightly different approach can be used. That is the amplitudes of the adja-
cent channels can be added together rather than subtracting in complex quantity.
Since this operation is equivalent to noncoherent integration, as discussed in
Section 10.7, the improvement is about 2.7 dB rather than 3 dB. To accommo-
date this gain, the summation will be divided by 1.466 rather than
√
2 because
20 log
(
2
/
1
.
466
)
2
.
7dB.
Figure 10.10 shows the filter bank response from the periodic acquisition
through simulation
(
13
)
. In these figures, each individual filter has a bandwidth
=
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