A Z-transform view of Hilbert transformation

Let x_t 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):  

$$Y(Z)\eq 2x_1 \sin\omega + 2x_2 \sin 2\omega + \cdots$$ (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):

$$C(Z) \eq {1\over 2} [X(Z)+iY(Z)] \eq {1\over 2} x_0 + x_1 Z + x_2 Z^2 + \cdots \$$ (4)

Thus, for plot pairs, the causal response is c_t, the real part of the FT is equation (2), and the imaginary part not usually shown is given by equation (3).