Hello, Signal Processing Experts, There is something that I have been so used to and does not look so obvious today. On page 98 of Communication Systems Engineering by Proakis (1st edition), it defines the autocorrelation function as the convolution between the signal itself and time reversed version of its complex conjugate: Rx(tau) = x(tau) convolve x*(-tau) --- (1.1) = integral from -inf to +inf { x(t) x*(t-tau) } dt --- (1) However, when we talk about convolution of two signals we always assume that the two signals have the same independent variable, such as what follows that the independent variable is tau: x1(tau) convolve x2(tau) = integral from -inf to +inf { x1(t) x2(tau- t) } dt --- (2) How do we get (1) from (1.1) and (2)? I mean that I just cannot apply the definition of convolution in (2) to (1.1) since the second function in the convolution in (1.1) does not have the same independent variable as the first function -- the first one is tau, the second one is -tau. Could anyone help me with this? Thanks, Carl

# Definition of autocorrelation -- it was obvious but not now

Started by ●April 22, 2009

Reply by ●April 23, 20092009-04-23

carl.horton08@gmail.com wrote:> Rx(tau) = x(tau) convolve x*(-tau) --- (1.1) > = integral from -inf to +inf { x(t) x*(t-tau) } dt --- (1) > > However, when we talk about convolution of two signals we always > assume that the two signals > have the same independent variable, such as what follows that > the independent variable is tau: > > x1(tau) convolve x2(tau) > = integral from -inf to +inf { x1(t) x2(tau- t) } dt --- (2) > > How do we get (1) from (1.1) and (2)? I mean that I just cannot > apply the definition of convolution in (2) to (1.1) since the > second function in the convolution in (1.1) does not have the > same independent variable as the first function -- the first > one is tau, the second one is -tau.No, in fact the tau's in (1) and (2) are separate entities. What's confusing you is the engineers' (mostly expedient) habit to write a *function* as what a mathematician would regard as a function *application*. That is, the content of (1.1) is that x1 = x and x2 = x* compose (-) in (2). Therefore, x2(tau-t) = x*(-(tau-t)) = x*(t- tau) as (1) claims. Martin -- Quidquid latine scriptum est, altum videtur.

Reply by ●April 23, 20092009-04-23

On Apr 22, 10:33�pm, Martin Eisenberg <martin.eisenb...@udo.edu> wrote:> No, in fact the tau's in (1) and (2) are separate entities. What's > confusing you is the engineers' (mostly expedient) habit to write a > *function* as what a mathematician would regard as a function > *application*. That is, the content of (1.1) is that x1 = x and x2 = > x* compose (-) in (2). Therefore, x2(tau-t) = x*(-(tau-t)) = x*(t- > tau) as (1) claims. > > Martin > > -- > Quidquid latine scriptum est, altum videtur.- Hide quoted text - > > - Show quoted text -Thanks Martin, Yes, I get it now. Just a note to myself, 1. The x*(-tau) in (1.1) is a function of -tau and it is a function of tau as well. 2. So, let x1(tau) =x(tau) and x2(tau) = x*(-tau), we have x2(tau-t) =x*(t-tau) 3. Substitue x1 and x2 in (2) with x and x* to get (1). Thanks again for your help! Carl