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 b_t. More precisely, we now define the lowest significant frequency component $\Lambda F$ in the spectrum, analogously to (3), as  

$${1 \over \Lambda F} \eq \int^{\infty}_{-\infty} {1 \over f}... ...nt^{\infty}_{-\infty} \overline{B} B\, {d \omega \over \omega}$$ (4)

where we have assumed the spectrum is normalized, i.e., the zero lag of the auto-correlation of b_t 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):  

$${1 \over -i \omega } \eq {(\cdots - Z^{-2} - Z^{-1} + 0 + Z + Z^2 + \cdots) \over 2}$$ (5)

Using limits on the integrals for time-sampled functions and inserting (5) into (4) gives  

$${1 \over \Lambda F} \eq {-i \over 2\pi} \int^{+\pi}_{-\pi} {... ... )\ \overline{B} \left( {1 \over Z} \right) \, B(Z)\, d \omega$$ (6)

Let s_t be the autocorrelation of b_t. 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 |s_t| function. So the area is a measure of the autocorrelation width $\Lambda T_{\rm auto}$.Thus,  

$${1 \over \Lambda F} \quad \leq \quad \sum^{\infty}_{t = 1} \vert s_t\vert \eq \Lambda T_{\rm auto}$$ (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  

$$\Lambda T_{\rm auto} \quad \leq \quad \Lambda T$$ (10)

Inserting this inequality into (9), we have the uncertainty relation  

$$\Lambda T \ \Lambda F \quad \geq \quad 1$$ (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

   $$b_t = A \cos \omega_1 t + B\sin \omega_1 t + C \cos \omega_2 t + D\sin \omega_2 t$$

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