Biomedical Engineering Reference
In-Depth Information
Taking into account the linearity of
Z
we can compute the transform of a linear
combination of the
p
signal samples:
a
1
z
−
1
a
p
z
−
p
Z
{
x
[
n
]+
a
1
x
[
n
−
1
]+
···
+
a
p
x
[
n
−
p
]
}
=(
1
+
+
···
+
)
X
(
z
)
(1.36)
=
A
(
z
)
X
(
z
)
This result will be very useful for discussion of AR model and filtering.
1.4.7 Uncertainty principle
In the previous sections we considered signals represented in either time or fre-
quency domain. For stationary signals it is enough to know one of the representa-
tions. Real-life signals are often a mixture of stationary and non-stationary elements,
e.g., alpha spindles, sleep spindles, K-complexes, or other graphoelements together
with ongoing background EEG activity. We are often interested in characterizing the
non-stationary transients. It is natural to think about their frequency
f
0
and localiza-
tion in time
t
0
. It is also obvious that such a transient has certain duration in time—
that is, it is not localized in a single time point but it has a span in time. The time span
can be characterized by σ
t
. The localization of the transient in the frequency domain
also has a finite resolution characterized by frequency span σ
f
. Those two spans are
bounded by the uncertainty principle. Before we continue with the formal notation,
let's try to understand the principle heuristically. Let's think about a fragment of a
sinusoid observed over time
T
, as shown in the
Figure1.5.
We can
calculate its frequency dividing the number of periods observed during time
T
by the
length of
T
. As we shrink the time
T
the localization in time of the fragment of the
sinusoid becomes more precise, but less and less precise is estimation of the number
of cycles, and hence the frequency.
Now let's put it formally. We treat the signal energy representation in time
=(
t
−
σ
t
,
t
+
σ
t
)
2
|
x
(
t
)
|
2
as probability distributions with normalizing
and representaion in frequency
|
X
(
f
)
|
=
R
∞
−
2
dt
. Then we can define mean time:
factor
E
x
|
x
(
t
)
|
∞
Z
∞
1
E
x
2
dt
t
0
=
t
|
x
(
t
)
|
(1.37)
−
∞
Mean frequency:
Z
∞
1
E
x
2
df
=
|
(
)
|
f
0
f
X
f
(1.38)
−
∞
Time span:
Z
∞
1
E
x
σ
t
=
2
2
dt
∞
(
t
−
t
0
)
|
x
(
t
)
|
(1.39)
−
Frequency span:
Z
∞
1
E
x
σ
f
=
2
2
df
∞
(
f
−
f
0
)
|
X
(
f
)
|
(1.40)
−
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