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67

The combination of kxiowmg the spectral behaviour of a given design of moving average and knowing in advance the general nature of the signal against which it is to be applied (per the price-motion model), generates a terrific advantage in the apphcations. Moving averages can now be designed to clarify and allow inference of spectral status of stock prices at any given time, thus providing insight into the future of the price time-series. This is the basis of the several applications of the chapters.

In using moving averages to help define the turning points of price-motion modei components, one is now aware of the potential for high-frequency "creep-through"-either in phase or 180 degrees out of phase. Such knowledge permits identification of these unwanted residuals, so that they do not mislead conclusions.

RESPONSE OF THE INVERSE CENTERED MOVING AVERAGE

The chapters have pomted up the utility of this version of centered moving averages. Knowledge of the frequency response associated with them can be used to sharpen your ability to interpret results obtained through their usage.

The phase response and lag characteristics of such an operation are similar to that of the conventional moving average. The principal difference lies in the fact that no phase inversion is experienced at frequencies output via error lobes. The output is, therefore, perfectly in phase with input across the entu-e frequency spectrum, and time lag remains precisely one-half the span of the filter.

Amplitude is simply derived by taking 1.0 minus that for the conventional moving average, with the result shown in Figure A IV-2.

ERROR LOSES

FREQUENCY RESPONSE OF INVERSE MOVING AVERAGE OFCRATION

FK30¹ AK-Z

FREQUENCr

How An Inverse Moving Average Works



From this figure it is noted that the inverse moving average behaves like a high-pass filter-again with error lobes to 23%. The firequency for which amplitude

ratio first become 1.0 is -j-, or the frequency corresponding to a period equal to

the span of the average, „. The actual maximum response point occurs at a frequency somewhat higher. You will sometimes want to conect output for the nearly one-fourth excess amplitude of the output at these frequencies in order to arrive at a true estimate of the magnitude of a spectral model component.



appendix five

Parabolic Interpolation

Three-Point Interpolation Equation Derivation

As indicated in Chapter Eleven, digital data spacing of the time series of numbers on which a numerical filter operates is a design parameter of the filter. As such, it must be varied according to need, along with other factors, in order to achieve the desired fBter characteristics.

This is sometimes inconvenient, as when the outputs of two filters must be combined or compared numerically, but the output of one is spaced at five-week intervals, while the other is spaced at three (for example).

In such cases, paraboUc interpolation is useful. This process defines the equation of a least-square-error parabola fitted through any specified number of filter output data points. The equation is then solved for best-estimate intervening values that are a common interval apart. The results of the two or more incompatible filter outputs may then be compared directly, or operated upon together numerically.

The following material is intended for use with the output of band-pass filters where the high frequency content is small. I te olation of high-pass (or otherwise non-smooth) data may require adaptation of the methods to use more data points, or even the use of other forms of curve fitting. If the need arises, a representative sample of the literature on such additional methods is to be found in the bibliography.

THREE-POINT INTERPOLATION The process operates as follows:

Consider any three consecutive data points in the output of a filter spaced "f" tirae units apart. Assign f = 0 at the central point, and symbolize this point as Sq. Call the one just preceding it in time 5-,, and the one following it S*i. Form the least-square-error parabola fitted through 5-i, 5o, and 5+j. Solve the resulting equation from f = 0 to f = to find the interpolated values of "s" at any required time intervals.



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