Appendix A

Cross-correlation integral solution

Integral solution for cross-correlation of a sine function \( Asin(ωt) \) with its leading copy \( Asin(ωt+φ) \):

\[ R_{xy}(\tau) = \frac{1}{T} \int\limits_{-T/2}^{T/2} A\sin(\omega t) \cdot A\sin\left(\omega(t - \tau) + \phi\right) dt \] \[ = \frac{A^2}{T} \int\limits_{-T/2}^{T/2} \sin(\omega t)\sin\left(\omega t - \omega \tau + \phi\right) dt \]

Let us use the trigonometric identity:

\[ \sin(\alpha)\sin(\beta) = \frac{1}{2} \left[ \cos(\alpha - \beta) - \cos(\alpha + \beta) \right] \]

The cross-correlation integral becomes:

\[ R_{xy}(\tau) = \frac{A^2}{2T} \int\limits_{-T/2}^{T/2} \left[ \cos(\omega t - \omega t + \omega \tau - \phi) - \cos(\omega t + \omega t - \omega \tau + \phi) \right] dt \] \[ = \frac{A^2}{2T} \int\limits_{-T/2}^{T/2} \left[ \cos(\omega \tau - \phi) - \cos\left(2\omega t - \omega \tau + \phi\right) \right] dt \] \[ = \color{red}{\frac{A^2}{2T} \int\limits_{-T/2}^{T/2} \cos(\omega \tau - \phi) dt} \color{#222832} {\ -\ } \color{green}{\frac{A^2}{2T} \int\limits_{-T/2}^{T/2} \cos\left(2\omega t - \omega \tau + \phi\right) dt} \]

First integral solution:

Factor out the part of the integral that doesn't depend on \( t \).

\[ \color{red}{\frac{A^2}{2T} \cos(\omega \tau - \phi) \int\limits_{-T/2}^{T/2} dt = \frac{A^2}{2T} \cos(\omega \tau - \phi) \left[ t \right]_{-T/2}^{T/2}} \] \[ = \color{red}{\frac{A^2}{2T} \cos(\omega \tau - \phi) \left[ \frac{T}{2} + \frac{T}{2} \right] = \frac{A^2}{2} \cos(\omega \tau - \phi)} \]

Second integral solution:

Since cosine is an even function \( \cos(-\alpha) = \cos(\alpha) \), the integral of a cosine function over its full period averages to zero.

\[ \color{green}{\frac{A^2}{2T} \int\limits_{-T/2}^{T/2} \cos\left(2\omega t - \omega \tau + \phi\right) dt = 0} \]

The cross-correlation function is:

\[ R_{xy}(\tau) = \frac{A^2}{2} \cos(\omega \tau - \phi) \]