CAUSAL INTEGRATION FILTER

Begin with a function in discretized time x_t. The Fourier transform with the substitution $Z=e^{i \omega\,\Delta t}$ is the Z-transform

$$X(Z) \eq \cdots\ + x_{-2} \,Z^{-2} \ +\ x_{-1} \,Z^{-1} \ +\ x_0 \ +\ x_1 \,Z \ +\ x_2 \,Z^2 \ +\ \cdots$$ (12)

Define $-i \hat \omega$ (which will turn out to be an approximation to $-i \omega$) by  

$${1 \over -i \hat \omega \,\Delta t } \eq {1 \over 2 }\ { 1\ +\ Z \over 1\ -\ Z }$$ (13)

Define another signal y_t with Z-transform Y(Z) by applying the operator to X(Z):

$$Y(Z)\ \ \ =\ \ \ {1 \over 2 }\ { 1\ +\ Z \over 1\ -\ Z} \ X(Z)$$ (14)

Multiply both sides by (1-Z):

$$(1\ -\ Z)\ Y(Z)\ \ \ =\ \ \ {1 \over 2 }\ (1\ +\ Z)\ X(Z)$$ (15)

Equate the coefficient of Z^t on each side:

$$y_t\ -\ y_{{t-1}}\ \ \ =\ \ \ {x_t\ +\ x_{{t-1}} \over 2 }$$ (16)

Taking x_t to be an impulse function, we see that y_t turns out to be a step function, that is,

$$x_t \eq \cdots 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, \cdots$$ (17)

$$y_t \eq \cdots 0, 0, 0, 0, 0, {1 \over 2 }, 1, 1, 1, 1, 1, 1, \cdots$$ (18)

So y_t is the discrete-domain representation of the integral of x_t from minus infinity to time t. The operator (1+Z)/(1-Z) is called the ``bilinear transform."