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Spectra in terms of Z-transforms

Let us look at spectra in terms of Z-transforms. Let a spectrum be denoted $S(\omega)$, where  
 \begin{displaymath}
S(\omega) \eq \vert B(\omega)\vert^2 \eq \overline{B(\omega)}B(\omega)\end{displaymath} (32)
Expressing this in terms of a three-point Z-transform, we have
\begin{eqnarray}
S(\omega) & = & (\bar{b}_0+\bar{b}_1 e^{-i\omega} + 
 \bar{b}_2...
 ... b_2Z^2 ) \ S(Z) & = & \overline{B} \left(\frac{1}{Z}\right) B(Z)\end{eqnarray} (33)
(34)
(35)
It is interesting to multiply out the polynomial $\bar{B}(1/Z)$ with B(Z) in order to examine the coefficients of S(Z):
   \begin{eqnarray}
S(Z) &=& \frac{\bar{b}_2b_0}{Z^2} + 
 \frac{(\bar{b}_1b_0 + \ba...
 ...) &=& \frac{s_{-2}}{Z^2} + \frac{s_{-1}}{Z} + s_0 + s_1Z + s_2 Z^2\end{eqnarray}
(36)
The coefficient sk of Zk is given by  
 \begin{displaymath}
s_k \eq \sum_{i} \bar{b}_i b_{i+k}\end{displaymath} (37)
Equation (37) is the autocorrelation formula. The autocorrelation value sk at lag 10 is s10. It is a measure of the similarity of bi with itself shifted 10 units in time. In the most frequently occurring case, bi is real; then, by inspection of (37), we see that the autocorrelation coefficients are real, and sk=s-k.

Specializing to a real time series gives
      \begin{eqnarray}
S(Z) & = & s_0 + s_1\left(Z+\frac{1}{Z} \right) +
 s_2\left(Z^2...
 ...os k\omega
\ S(\omega ) & = & \mbox{cosine transform of }\;\; s_k\end{eqnarray} (38)
(39)
(40)
(41)
(42)
This proves a classic theorem that for real-valued signals can be simply stated as follows:

For any real signal, the cosine transform of the autocorrelation equals the magnitude squared of the Fourier transform.


next up previous print clean
Next: Two ways to compute Up: CORRELATION AND SPECTRA Previous: CORRELATION AND SPECTRA
Stanford Exploration Project
10/21/1998