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A Z-transform view of Hilbert transformation

Let xt be an even function of t. The definition $Z=e^{i\omega}$ gives $Z^{-n}+Z^n=2 \cos\omega n$; so
   \begin{eqnarray}
X(Z) &= & \cdots + x_1 Z^{-1} + x_0 + x_1 Z + x_2 Z^2 + \cdots \ X(Z) &= & x_0 + 2x_1 \cos \omega + 2x_2 \cos 2\omega + \cdots\end{eqnarray} (1)
(2)
Now make up a new function Y(Z) by replacing cosine by sine in (2):  
 \begin{displaymath}
Y(Z)\eq 2x_1 \sin\omega + 2x_2 \sin 2\omega + \cdots\end{displaymath} (3)
Recalling that $Z=\cos\omega +i\sin\omega$, we see that all the negative powers of Z cancel from X(Z)+iY(Z), giving a causal C(Z):
\begin{displaymath}
C(Z) \eq
{1\over 2} [X(Z)+iY(Z)] \eq
{1\over 2} x_0 + x_1 Z + x_2 Z^2 + \cdots \ \end{displaymath} (4)
Thus, for plot pairs, the causal response is ct, the real part of the FT is equation (2), and the imaginary part not usually shown is given by equation (3).


next up previous print clean
Next: The quadrature filter Up: HILBERT TRANSFORM Previous: HILBERT TRANSFORM
Stanford Exploration Project
10/21/1998