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Causal integration

Begin with a function in discretized time pt. The Fourier transform with the substitution $Z = \exp (i \omega \Delta t)$is the Z-transform  
 \begin{displaymath}
P(Z) \eq
\cdots p_{{-2}} \, Z^{{-2}}\ +\ p_{{-1}} \, Z^{{-1}}\ +\ 
p_0\ +\ p_1 \, Z\ +\ p_2 \, Z^2\ +\ \cdots\end{displaymath} (29)
Define $ -i \hat \omega $ (which will turn out to be an approximation to $ -i \omega $) by  
 \begin{displaymath}
{1 \over -i \hat \omega \,\Delta t } \eq {1 \over 2 }\ 
{ 1\ +\ Z \over 1\ -\ Z }\end{displaymath} (30)
Define another time function qt with Z-transform Q(Z) by applying the operator to P(Z):  
 \begin{displaymath}
Q(Z) \eq {1 \over 2 }\ { 1\ +\ Z \over 1\ -\ Z} \ P(Z)\end{displaymath} (31)
Multiply both sides by (1-Z):  
 \begin{displaymath}
(1\ -\ Z)\ Q(Z) \eq {1 \over 2 }\ (1\ +\ Z)\ P(Z)\end{displaymath} (32)
Equate the coefficient of Zt on each side:  
 \begin{displaymath}
q_t\ -\ q_{{t-1}} \eq {p_t\ +\ p_{{t-1}} \over 2 }\end{displaymath} (33)
Taking pt to be an impulse function, we see that qt turns out to be a step function, that is,
      \begin{eqnarray}
p \ \ \ &=&\ \ \ \cdots \ 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, \...
 ...\ \ \ \cdots 0, 0, 0, 0, 0, {1 \over 2 }, 1, 1, 1, 1, 1, 1, \cdots\end{eqnarray} (34)
(35)
So qt is the discrete domain representation of the integral of pt from minus infinity to time t. It is the same as a Crank-Nicolson-style numerical integration of the differential equation dQ/dt = P. The operator (1+Z)/(1-Z) is called the bilinear transform. The accuracy of the approximation to differentiation can be seen by multiplication on top and bottom by Z-.5 and substitution of $Z = e^{{i} \omega \Delta t}$.
      \begin{eqnarray}
-\,i \ { \hat \omega \, \Delta t \over 2} \ \ &=&\ \ 
{1\,-\,Z ...
 ...elta t /2 }
\ \ =\ \ -\,i \, \tan \ { \omega \, \Delta t \over 2 }\end{eqnarray} (36)
(37)

The integration operator has a pole at Z = 1, which is exactly on the unit circle. This raises the possibility of the paradox of infinity. In other words, there are other noncausal expansions too. For example, taking $1/(-i \omega )$ to be an imaginary, antisymmetric function of $\omega$ implies a real, antisymmetric time function, namely, $\sgn(t) = t / \vert t\vert$, which is not usually regarded as the integration operator. To avoid any ambiguity, we introduce here a small positive number $\epsilon$ and define $\rho = 1- \epsilon$.The integration operator becomes
         \begin{eqnarray}
I \ \ \ &=&\ \ \ {1 \over 2 }\ {1\ +\ \rho Z \over 1\ -\ \rho Z...
 ...1 \over 2 }\ +\ \rho Z\ +\ (\rho Z)^2\ +\ ( \rho Z )^3 \ +\ \cdots\end{eqnarray} (38)
(39)
(40)
Because $\rho$ is slightly less than one, this series converges for any value of Z on the unit circle. If $\epsilon$ had been slightly negative instead of positive the expansion would have come out in negative instead of positive powers of Z.

Now the big news is that the causal integration operator is an example of an impedance function. The operator is clearly causal with a causal inverse. Let us check in the frequency domain that the real part is positive. Rationalizing the denominator gives
      \begin{eqnarray}
I \ \ =\ \ 
{1 \over 2 }\ {(1\ +\ \rho Z) \over ( 1\ -\ \rho Z ...
 ... \ +\ 2\,i\,\rho \, \sin \, \omega \Delta t 
\over \rm{positive} }\end{eqnarray} (41)
(42)
Again, it is the choice of a positive $\epsilon$ that has caused $1\ -\ \rho^2 $, and hence the real part to be positive for all $\omega$, as shown in Figure 5.

 
cintegral
cintegral
Figure 5
The causal integration operator I. The frequency axis is represented by a discrete Fourier transform over 256 points. Zero time and zero frequency are on the left end of their respective axes.


view

As multiplication by $ -i \omega $ in the frequency domain is associated with differentiation d/dt in the time domain, so is division by $ -i \omega $ associated with integration. People usually associate the asymmetric operator (1,-1) with differentiation. But notice that the inverse to the causal integration operator, namely,
      \begin{eqnarray}
I^{{-1}} \ \ \ &=&\ \ \ 2\ { 1\ -\ \rho Z \over 1\ +\ \rho Z }
...
 ... 2\ -\ 4 \rho Z\ +\ 
4 ( \rho Z )^2\ -\ 4 ( \rho Z )^3
\ +\ \cdots\end{eqnarray} (43)
(44)
also represents differentiation, although it is completely causal and not at all asymmetric. In linear systems analysis this representation of discrete differentiation is often the preferred one. The construction of higher-order, stable differential equations is subject to certain rules, to be covered, for combining impedances.

Occasionally it is necessary to have a negative real part for the differentiation operator. This can be achieved by taking $\epsilon$ to be negative, which means taking $ \rho \gt 1 $, and doing the infinite series expansion in powers of Z-1, that is, anticausally instead of causally. In either the anticausal or the causal case the imaginary part will still be $ -i \omega $, but the real part will have the opposite sign.


previous up next print clean
Next: Muir's rules for combining Up: IMPEDANCE Previous: Review of impedance filters
Stanford Exploration Project
10/31/1997