of Microprocessors

DIGITAL FILTERING 485
the input wave by 90°. This behavior is just what would be expected of a real
analog integrator. Even the de offset at the output would be expected, since
the input was suddenly applied at zero-crossing time. Although not apparent
in the figure, the integrator does not contribute any noise or distortion to the
signals passing through but may alter the signal-to-noise ratio due to its
filtering effect.
Digital R-C Low-Pass Filter
Returning to the leaky integrator equivalent of an R-C low-pass filter,
let us see how the same effect can be accomplished digitally. It should be
obvious that the magnitude of the leakage current is proportional to the
voltage across the capacitor and the direction is such that the capacitor is
discharged. Since the capacitor voltage is the same as the output voltage (the
capacitor lead connected to the op-amp's inverting input is at virtual
ground), the leakage current is proportional to the instantaneous output
voltage. This same discharging effect can be simulated by subtracting a constant
proportion of the accumulator's content from the accumulator every sample
period. In fact, the percentage of accumulator "charge" that should be
subtracted is exactly equal to the percentage that would have leaked through
the leakage resistor during one sample period in the analog filter.
It is well known that the capacitor voltage during discharge follows an
inverse exponential curve: E=Eoexp(-T/RC), where E is the voltage after
time T, Eo is the initial voltage, and Rand C are the component values in
ohms and farads. Since the cutoff frequency is l/2nRC, a substitution can be
made in the equations, and it is found that E =Eoexp(21TFcT), where Fe is the
cutofffrequency of the filter. Converting the sample period, T, to the sample
rate, Fs, makes the final result: E=Eoexp(-21TFc!Fs). Thus, the proper
percentage to subtract from the accumulator each sample period is:
l-exp(- 21TFc!Fs) which can be designated K. A BASIC statement for implementing
the digital filter then would be:
1000 LET A=A-K*A+I
where A is the accumulator content, K is the leakage constant, and I is the
input sample. This statement is executed for each input sample and successive
values of A are the output samples.
One could also simply multiply the accumulator contents by l-K,
which can be called L, and put the product back into the accumulator with
the input added in. In BASIC, the result would be:
1000 LET A=A*L+l
which is about as simple as one can get. Note that L can never be greater than
1.0. If it were, the filter would be unstable and eventually overflow evenfloating-point
arithmetic. (Imagine, a universe full of energy circulating in
this little filter!)