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Proof by way of the dual problem

Although the Z-transform 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 Z-transform, we consider instead the dual to the rise-time problem. Instead of a signal whose square vanishes at negative time, we have a spectrum $\overline{B} (1/Z) B(Z)$ that vanishes at negative frequencies. We measure how fast this spectrum can rise after $\omega = 0$.We will find this time interval to be related to the time duration $\Lambda T$ of the complex-valued signal bt. More precisely, we now define the lowest significant frequency component $\Lambda F$ in the spectrum, analogously to (3), as  
 \begin{displaymath}
{1 \over \Lambda F}
\eq
\int^{\infty}_{-\infty} 
{1 \over f}...
 ...nt^{\infty}_{-\infty} 
\overline{B} B\,
{d \omega \over \omega}\end{displaymath} (4)
where we have assumed the spectrum is normalized, i.e., the zero lag of the auto-correlation of bt is unity. Now recall the bilinear transform, equation ([*]), which represents $1/(-i\omega)$ as the coefficients of ${1\over 2}(1+Z)/(1-Z)$, namely, $(\ldots 0, 0, 0, {1\over 2}, 1, 1, 1 \ldots )$.

The pole right on the unit circle at Z = 1 causes some nonuniqueness. Because $1/i \omega$ is an imaginary, odd, frequency function, we will want an odd expression (such as on page [*]) to insert into (4):  
 \begin{displaymath}
{1 \over -i \omega }
\eq
{(\cdots - Z^{-2} - Z^{-1} + 0 + Z + Z^2 + \cdots) \over
2}\end{displaymath} (5)
Using limits on the integrals for time-sampled functions and inserting (5) into (4) gives  
 \begin{displaymath}
{1 \over \Lambda F}
\eq
{-i \over 2\pi} \int^{+\pi}_{-\pi} {...
 ... )\ \overline{B} \left( {1 \over Z}
 \right) \, B(Z)\, d \omega\end{displaymath} (6)
Let st be the autocorrelation of bt. Since any integral around the unit circle of a Z-transform polynomial selects the coefficient of Z0 of its integrand, we have
      \begin{eqnarray}
{1 \over \Lambda F}
&=&
{-i \over 2} \left[
 (s_{-1} - s_1) +
 ...
 ...} - \Im s_t
\quad \leq \quad
 \sum^{\infty}_{t = 1} \vert s_t\vert\end{eqnarray} (7)
(8)
The height of the autocorrelation has been normalized to s0=1. The sum in (8) is an integral representing area under the |st| function. So the area is a measure of the autocorrelation width $\Lambda T_{\rm auto}$.Thus,  
 \begin{displaymath}
{1 \over \Lambda F}
\quad \leq \quad
\sum^{\infty}_{t = 1} \vert s_t\vert
\eq
\Lambda T_{\rm auto}\end{displaymath} (9)

Finally, we must relate the duration of a signal $\Lambda T$ to the duration of its autocorrelation $\Lambda T_{\rm auto}$.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 all-pass 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 $\Lambda T$'s are the same. Thus, we conclude that  
 \begin{displaymath}
\Lambda T_{\rm auto}
\quad \leq \quad
\Lambda T\end{displaymath} (10)
Inserting this inequality into (9), we have the uncertainty relation  
 \begin{displaymath}
\Lambda T \ \Lambda F \quad \geq \quad 1\end{displaymath} (11)

Looking back over the proof, I feel that the basic time-bandwidth idea is in the equality (7). I regret that the verbalization of this idea, boxed following, is not especially enlightening. The inequality arises from $\Lambda T_{\rm auto} < \Lambda T$,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:

  1. Consider B(Z) = [1 - (Z/Z0)n]/(1 -Z/Z0) as Z0 goes to the unit circle. Sketch the signal and its squared amplitude. Sketch the frequency function and its squared amplitude. Choose $\Lambda F$ and $\Lambda T$.
  2. A time series made up of two frequencies can be written as

    \begin{displaymath}
b_t = A \cos \omega_1 t + B\sin \omega_1 t + C \cos \omega_2 t + D\sin
\omega_2 t\end{displaymath}

    Given $\omega_1$, $\omega_2$, b0, b1, b2, and b3, show how to calculate the amplitude and phase angles of the two sinusoidal components.

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Next: FT OF RANDOM NUMBERS Up: My rise-time proof of Previous: My rise-time proof of
Stanford Exploration Project
10/21/1998