Calculating crosscorrelation and autocorrelation

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Four causal wavelets are given by , , , . Calculate the crosscorrelations and autocorrelations , , , , and in both the time and frequency domains.


A causal wavelet as defined in problem 5.21 has zero values when .

The crosscorrelation of and tells us how similar the two functions are when is shifted the amount relative to . The crosscorrelation is given by the equation


This equation means that is displaced units to the left relative to , corresponding values multiplied, and the products summed to give . The result is the same if we move to the right units, that is,


The functions and are closely related. Using two simple curves, it is easily shown that we can crosscorrelate by reversing and convolving the reversed function with , that is,


(since convolution is commutative). Reversing a function in time changes the sign of , so the exponent of changes sign also, becoming and becoming the conjugate complex . The convolution theorem (equation (9.3f)) now becomes the crosscorrelation theorem:


If is the same as , we get the autocorrelation of and equations (9.8a,d) become


the transform relation being the autocorrelation theorem. Since the two functions are the same, does not depend upon the direction of displacement. The autocorrelation for equals the sum of the data elements squared, hence is called the energy of the trace;


Both the autocorrelation and the crosscorrelation are often normalized; in the case of the autocorrelation, equation (9.8e) becomes


The normalized crosscorrelation is



Time-domain calculations:

Using equation (9.8a) to calculate , i.e., is first shifted to the left; we have:


where marks the value at . Proceeding in the same way, we find that

To find we displace instead of and obtain , which equals .

Autocorrelations are found in the same way:

Frequency-domain calculations

The -transforms of , , and are , , . The conjugate complexes of these transforms are are , , . Using these transforms, we get

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