Signal Processing - May 2017 - 92
−1
Amplitude
Amplitude
1
0
1,024
2,048
Sample
3,072
100
4,096
0
spk
0
32
64
unf
96
(a)
128 160
Sample
192
224
255
(b)
FIGURE 2. The original and transformed data for the example in the section "Numerical Example." (a) The original signal under analysis (s [·]) and (b) the
first four subbands of the corresponding DWT-packet map, modified and converted based on COR and TEG, respectively (yu [·]) .
Teager's energy operator [6], instead of
the ordinary energy [7], to convert all
the samples of each leaf subband into
their instantaneous potentials, according to the TEG procedure:
For the subband b = 0, 1, ..., 2 j - 1
of the deepest level j, do:
u 0, b ! (y 0, b) 2
■ y
u i, b ! (y i, b) 2 - (y i -1, b) · (y i +1, b), for
■ y
^1 # i # ^ M/2 j h - 2 h, whenever the
length of b is at least four
u (M/2 j) -1, b ! (y (M/2 j) -1, b) 2, whenever
■ y
the length of b is at least two
where the left arrows mean attribution.
Summarizing, the ideal time-frequency analysis of s [·] based on the DWT
requires four steps:
■ Step 1: Select the most appropriate
wavelet family and the support size
of filters, i.e., N. Then, calculate the
M-sample-long transformed signal
y [·] from s [·] based either on the
regular DWT map or the DWTpacket map.
■ Step 2: Apply COR, shifting y [·]
accordingly.
u [·]
■ Step 3: Apply TEG, creating y
from shifted y [·].
u [·] based either
■ Step 4: Interpret y
on STG_A or STG_B, depending on
which map was chosen to carry out
Step 1.
Numerical example
Problem statement
Let s i =1.2 sin ^^2r·44i h /4095)h e -^ 409.6 h
for {0 # i 11, 024}, s i = 0, for {1, 024#
i 1 2, 048}, and s i = sin ((2r·85i/4, 095)
for {2, 048 # i 1 4, 096} to be the 1-s
long signal under analysis, sampled
at 2T = 4, 096 samples/s, as shown in
Figure 2(a). The problem is to find out
the time support of its frequencies with
92
i
2
a balance between temporal and spectral accuracies.
Solution
A balanced time- frequency analysis
of s [·], which contains frequencies
up to T = 2, 048 Hz, requires j =
6^1 + log 2 (M )h /2@ = 6(1+log2(4,096))/2@
= 6. Correspondingly, r = ^T/2 jh =
^2, 048/2 6 h = 32 Hz is the frequency
resolution for this decomposition, which
is based on the DWT-packet map to allow
a uniform analysis, i.e., to keep r intact
for all the subbands. Symmlets of size
N = 16 were chosen to perform the analysis because of their almost linear phase
responses, practically avoiding frequency
components to become misaligned.
Proceeding with the calculations and
applying COR and TEG, we get yu [·], for
which only the first four subbands appear
in Figure 2(b) to allow magnification. Its
inspection reveals two main elements:
spk and unf. The former is a spike decaying from samples 64 to 79, and the latter
consists of a uniform step from samples
160 to 191. Observing that each one of
the 2 j = 2 6 = 64 subbands contains
^ M/2 j h = ^4096/2 6 h = 64 samples, indexed from 0 to 63, we conclude that
spk occurs within the interval {i = 0, i =
15} of the second subband, i.e., b = 1,
and unf occurs within the interval
{i = 32, i = 63} of the third subband,
i.e., b = 2. Thus, according to STG_B:
■ For spk, the energy of the sample i = 0
of the subband b = 1 at the level
j = 6 contains the amplitude of
frequencies between ^1·2, 048/2 6 h =
32 and ^(1 + 1) ·2, 048h /2 6 = 64 Hz,
located within the range {s 0·2 6,
s (0 + 1) 2 6 - 1}, i.e., {s 0, s 63}. Accordingly, the energy of the sample i = 15
of that subband contains the ampliIEEE Signal Processing Magazine
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May 2017
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■
tude of those frequencies, which
a r e located within the range
{s 15·2 6, s (15 +1) 2 6 -1}, i.e., {s 960, s 1023}.
It means that a frequency between 32
and 64 Hz exists in s [·], starting and
ending somewhere between {s 0, s 63}
and {s 960, s 1023}, respectively. In fact,
the frequency of 44 Hz was defined in
s [·] within the interval {s 0, s 1023} .
Furthermore, the decay starting from
sample 64 of Figure 2(b) shows the
corresponding power reduction clearly
seen within the corresponding interval
of Figure 2(a).
For unf, the energy of the sample
i = 32 of the subband b = 2 at the
level j = 6 contains the amplitude of
frequencies between 2 $ 2048/2 6 = 64
a n d ^(2 + 1) ·2048 h /2 6 = 96 H z ,
l o cated within the range {s 32·2 6,
s (32 + 1) 2 6 - 1}, i.e., {s 2048, s 2111}. Complementarily, the energy of the sample
i = 63 of that subband contains the
amplitude of those frequencies,
which were placed within the range
{s 63·2 6, s (63 +1) 2 6 -1}, i.e., {s 4032, s 4095}.
It means that a frequency between
64 and 96 Hz exists in s [·], starting
and ending somewhere between
{s 2048, s 2111} and {s 4032, s 4095}, respectively. In fact, the frequency of 85 Hz
was uniformly defined in s [·] within
the interval {s 2048, s 4095}.
What we have learned
Based on this article, readers could
have learned how to efficiently interpret discrete wavelet-transformed signals, while extracting from them the
time support of frequencies contained
in the original signal under analysis.
Supplementary to this, the criteria used
(continued on page 100)
Table of Contents for the Digital Edition of Signal Processing - May 2017
Signal Processing - May 2017 - Cover1
Signal Processing - May 2017 - Cover2
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Signal Processing - May 2017 - Cover3
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