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:

Adaptive Beamformer from "Adaptive Signal Processing, page 371"

Many thanks

  • $\begingroup$ I've read these pages in Widrow, there is no direct explanations why 1 sample shift is equal to $-\pi/2$.. It is so if sample rate is $2\omega_0$ but it isn't told to us. But I think it isn't too important for this case. In the scheme above you have only one tap and it exactly shifts the signal for $\pi/2$, so maybe author means "lag of 1" is this tap. It's only my suggestion. $\endgroup$
    – Serj
    Commented Jan 12, 2015 at 20:35

1 Answer 1


The $90$ degree shift introduces the sine component to the cross-correlation vector $P$ at lag 1 ($\phi_{dx}(1)$).

The block should be considered to be a 'high-level' description, and does not mean that a shift of one sample is equivalent to a $90$ degree phase shift, or anything like that. Rather it should be thought of as "shift by as many samples as necessary for a phase shift of $90$ degrees".


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