I am currently reading chapter 13 of 'Adaptive Signal Processing' by Widrow and Stearns, so if anyone happens to have a copy of the book to hand it could be helpful.
I am reading the chapter on "Introduction to Adaptive Arrays and Adaptive Beamforming" and am struggling understanding the auto-correlation/cross correlation between the primary and reference signals at a delay of one (as shown in the diagram below) and in the more general case (where the signal is arriving at an angle), where the general cross correlation vector is described by:
$$ \mathbf{P} = \begin{bmatrix} \phi_{dx}(0) \\ \phi_{dx}(1) \end{bmatrix} = \begin{bmatrix} E [C \cos k\omega_0 \cdot C \cos (k + \delta_0)\omega_0] \\ E [C \cos k\omega_0 \cdot C \sin (k + \delta_0)\omega_0] \end{bmatrix} $$
where $k$ is the sample number, $C$ is a constant amplitude, $\delta_0$ is the time delay of arrival between the primary and reference sensor and $\omega_0$ is the angular frequency.
The book assumes that the refence signal to the array is a cosine of the form: $$ \text{reference signal} = C \cos [(k + \delta_0) \omega_0]. $$ (The above equation is 13.3 in the book).
My question is, I don't understand why, for $\phi_{dx} (1) $ the 'lag' part of the equation becomes a sine rather than a cosine with a shift 1 (sample)? It's as if there is a shift of $-\pi/2$, hence the sine, but I can't see where this is from? Why for a lag of 1 does this cosine seemingly change to a sine?
My understanding of cross (or auto) correlation is that for $\phi_{dx} (1)$ you compare one signal with another with a shift of 1 sample (hence a lag of 1).
I understand that a sine is the quadrature of a cosine and understand why these equal zero (as the book describes; "$\phi_{dx} (1) $, is zero because it represents the correlation of (13.3) with its quadrature (sine) component"). My question is why is a sine introduced for a lag of 1?
Diagram for completeness:
Many thanks
