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Although the Ztransform method
is a great aid in studying cases
where divergence (as 1/t) plays a role,
it has the disadvantage that it destroys the formal interchangeability
between the time domain and the frequency domain.
To take advantage of the analytic simplicity
of the Ztransform,
we consider instead the dual to the risetime problem.
Instead of a signal
whose square vanishes at negative time, we have a spectrum
that vanishes at negative frequencies.
We measure how fast this spectrum can rise after .We will find this time interval to
be related to the time duration of the complexvalued signal b_{t}.
More precisely, we now define the lowest significant frequency
component in the spectrum, analogously to
(3), as
 
(4) 
where we have assumed the spectrum is normalized,
i.e., the zero lag of the autocorrelation of b_{t} is unity.
Now recall the bilinear transform,
equation (), which represents as
the coefficients of , namely,
.
The pole right on the unit circle at Z = 1 causes some nonuniqueness.
Because is an imaginary, odd, frequency function,
we will want an odd expression
(such as on page )
to insert into (4):
 
(5) 
Using limits on the integrals for timesampled functions and
inserting (5) into (4) gives
 
(6) 
Let s_{t} be the autocorrelation of b_{t}.
Since any integral around the unit circle
of a Ztransform polynomial selects
the coefficient of Z^{0} of its integrand, we have
 
(7) 
 (8) 
The height of the autocorrelation has been normalized to s_{0}=1.
The sum in (8) is an integral representing area under the
s_{t} function.
So the area is a measure of the autocorrelation width
.Thus,
 
(9) 
Finally, we must relate the duration of a signal to the
duration of its autocorrelation .Generally speaking,
it is easy to find a long signal that has short autocorrelation.
Just take an arbitrary short signal
and convolve it using a lengthy allpass filter.
Conversely, we cannot get a long autocorrelation function
from a short signal.
A good example is the autocorrelation of a rectangle function,
which is a triangle.
The triangle appears to be twice as long,
but considering that the triangle tapers down,
it is reasonable to assert that the 's are the same.
Thus, we conclude that
 
(10) 
Inserting this inequality into (9),
we have the uncertainty relation
 
(11) 
Looking back over the proof, I feel that the basic
timebandwidth idea is in the equality (7).
I regret that the verbalization of this idea, boxed following,
is not especially enlightening.
The inequality arises from
,which is a simple idea.
The inverse moment of the normalized spectrum of an analytic signal
equals the imaginary part of the mean of its autocorrelation.

EXERCISES:

Consider B(Z) = [1  (Z/Z_{0})^{n}]/(1 Z/Z_{0}) as Z_{0} goes to
the unit circle.
Sketch the signal and its squared amplitude.
Sketch the frequency function and its squared amplitude.
Choose and .

A time series made up of two frequencies can be written as
Given , , b_{0}, b_{1}, b_{2}, and b_{3},
show how to calculate
the amplitude and phase angles of the two sinusoidal components.
Next: FT OF RANDOM NUMBERS
Up: My risetime proof of
Previous: My risetime proof of
Stanford Exploration Project
10/21/1998