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98                                                5 Time-Series Analysis

            the complex Fourier transform X(f) also contains information on the phase
            relationship W(f) of the two signals:





            The phase difference is important in calculating leads and lags between two
            signals, a parameter often used to propose causalities between the two pro-
            cesses documented by the signals. The correlation between the two spectra
            can be calculated by means of the coherence:







            The coherence is a real number between 0 and 1, where 0 indicates no cor-
            relation and 1 indicates maximum correlation between x(t) and y(t) at the

            frequency f. Significant degree of coherence is an important precondition for
            computing phase shifts between the two signals.
               We use two sine waves with identical periodicities τ=5 (equivalent to
            f=0.2) and amplitudes equal to two. The sine waves show a relative phase





                      Cross PSD Estimate                Phase spectrum
                 20                              4
                       f 1 =0.02                 3    f 1 =0.02
                 15                              2
               Power  10                       Phase angle  1

                                                        Corresponding phase
                                                 0
                                                        angle of 1.2568, equals
                  5                                     (1.2568*5)/(2*π)=1.001
                                                 −1
                  0                              −2
                   0    1    2   3    4    5       0    1   2    3    4    5
                            Frequency                      Frequency
               a                               b

            Fig. 5.8 Crossspectrum of two sine waves with identical periodicities τ=5 (equivalent to
            f=0.2) and amplitudes two. The sine waves show a relative phase shift of t=1. In the argument

            of the second sine wave this corresponds to 2›/5, which is one fifth of the full wavelength
            of τ=5. a The magnitude shows the expected peak at  f=0.2. b The corresponding phase
            difference in radians at this frequency is 1.2568, which equals (1.2568*5)/(2*›) = 1.0001,
            which is the phase shift of one we introduced at the very beginning.
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