Why does the XCORR function from the Signal Processing Toolbox 6.11 (R2009a) not output a value of 1 for perfect correlation?

Why does the XCORR function from the Signal Processing Toolbox not output a value of 1 for perfect correlation?

The following code exemplifies the problem:

x=0:0.01:10; 
X = sin(x);
Y = sin(x+pi/2);
[r,lags]=xcorr(X,Y,'coeff');  
max(r)

ans =

    0.8090

Answer 1

[MathWorks Support Team]

This change has been incorporated into the documentation in Release 2009b (R2009b). For previous releases, read below for any additional information:

XCORR will produce correlations identically equal to 1.0 at zero lag only when performing an auto-correlation and only when the 'coeff' flag is passed to XCORR. For example:

x=0:0.01:10; \nX = sin(x);\n[r,lags]=xcorr(X,'coeff');  \nmax(r)

ans =\n\n     1

When performing a cross-correlation, XCORR will behave as outlined in the documentation located at the following URL:

https://www.mathworks.com/help/matlab/ref/xcorr.html

For example:

x=0:0.01:10; \nX = sin(x);\nY = sin(x+pi/2);\n[r,lags]=xcorr(X,Y);\nmax(r)

ans =\n\n  404.4763

% biased   - scales the raw cross-correlation by 1/M.\n[rb,lagsb]=xcorr(X,Y,'biased');  \nmax(rb)

ans =\n\n    0.4041

% Now, if you take the raw correlation and scale it by 1/M, where M is the length of the signal, \n% you get the biased cross-correlation.\nM = length(X);\nmax(r./M)

ans =\n\n    0.4041

%unbiased - scales the raw correlation by 1/(M-abs(k)), where k \n%is the index into the result.\n[ru,lagsu]=xcorr(X,Y,'unbiased');\nmax(ru) 

ans =\n\n    0.5441

% Now, if you take the raw correlation and scale it by 1/(M - abs(lags))\n% you get the unbiased cross-correlation. \nk = lags;\nmax(r./(M-abs(k)))

ans =\n\n    0.5441

For more information, see the following references:

[1] J.S. Bendat and A.G. Piersol, "Random Data: Analysis and Measurement Procedures", John Wiley and Sons, 1971, p.332.

[2] A.V. Oppenheim and R.W. Schafer, Digital Signal Processing, Prentice-Hall, 1975, pg 539.