Global Positioning System Reference
In-Depth Information
One approach to remedy this loss is to down-convert the input frequency every
500 Hz, rather than 1 kHz. Under this condition the coarse frequency separation
will be decreased from 1000 to 500 Hz, and the fine frequency resolution will be
decreased from 100 to 50 Hz. A signal falling exactly between two frequency bins
can be calculated from
x
=
π
/4 of the sinc function, which can be expressed as
20 log
sin
x
x
x
=
π/
4
=
20 log
(
0
.
9003
)
=
0
.
91 dB
(
10
.
17
)
This is an acceptable value. Although this operation is very effective, it is usually
computationally intensive.
A more efficient way is to operate on two adjacent frequency bins to generate a
frequency component at the center between the bins. This method can be applied
to any FFT-related operations. The following detailed discussion will continue
the example from the previous section.
Before the frequency domain is shown, the filters in conventional and FFT
outputs are illustrated because there might be a slight difference in appearance.
A conventional filter bank response is shown in Figure 10.7a, whereas a filter
bank produced through FFT is shown in Figure 10.7b.
An input signal at frequency
f
1
is shown in Figure 10.7a. It is easy to under-
stand the outputs from
A
and
B
shown in the figure. The output signal appears
at the same frequency as the input signal.
In a FFT filter bank the outputs are only available on frequency bins. If the
same input signal at
f
1
occurs in Figure 10.7b, the outputs are shown in the two
adjacent bins. Both contain the information of the input signal. For example, in
Figure 10.7b, the amplitudes represent the attenuation of the filter. If the phases of
successive outputs from either bin are also used to find a finer frequency resolution,
the frequency of the input signal can be obtained. Thus one can conclude that the
output from a frequency bin carries the frequency information of the input signal.
The phase difference between the outputs at
A
and
B
is very close to 180
◦
for a
signal in between bins, which can be illustrated from the following equation:
j
2
π(k
+
0
.
5
)n
N
x(n)
=
e
j
2
π(k
+
0
.
5
)n
N
e
−
j
2
πkn
N
jπn
N
N
−
1
N
−
1
1
−
e
jπ
1
−
X(k)
=
e
=
e
=
e
jπ/N
n
=
0
n
=
0
2
2
j
2
N
π
=
≈
jπ/N
=
1
−
e
jπ/N
−
j
2
π(k
+
0
.
5
)n
e
−
j
2
π(k
+
1
)n
e
−
jπn
N
N
−
1
N
−
1
e
−
jπ
1
−
N
N
X(k
+
1
)
=
e
=
=
1
−
e
−
jπ/N
n
=
0
n
=
0
2
jπ/N
=
−
j
2
N
2
=
≈
(10.18)
1
+
e
jπ/N
π
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