I have read quite a few papers on teh subject and have a question. Suppose we have two sensors x1 and x2 x1(t)=s1(t)+n1(t) and x2(t)=a.s2(t-D)+n2(t) where n1(t) and n2(t) are uncorrelated white noise sources also uncorrelated with the two signals s1 and s2. I need to find the delay D.The constant a is an attenuation factor. The Knapp-Carter give an expression for cross-correlation as Cx1x2(tau) = a.Css(tau) * delta (tau - D) where * is convolution and delta is the delta function and in other papers it is given as Cx1x2(tau) = a.Css(tau-D) are these the same? I Think they are but the Knapp-Carter one makes more sense to me than the other one. If I take the Fourier Transform of both I should get the cross-spectral density of x1 with x2. Thanks Tom

# Time delay estimation question

Started by ●July 22, 2003

Reply by ●July 24, 20032003-07-24

aberdonian_2000@yahoo.com (Tom) wrote in message news:<e1b1658f.0307221459.5a2d5e56@posting.google.com>...> I have read quite a few papers on teh subject and have a question. > > Suppose we have two sensors x1 and x2 > > x1(t)=s1(t)+n1(t) > > and > > x2(t)=a.s2(t-D)+n2(t) > > where n1(t) and n2(t) are uncorrelated white noise sources also > uncorrelated with the two signals s1 and s2. I need to find the delay > D.The constant a is an attenuation factor. > > The Knapp-Carter give an expression for cross-correlation as > > Cx1x2(tau) = a.Css(tau) * delta (tau - D) where * is convolution and > delta is the delta function > > and in other papers it is given as > > Cx1x2(tau) = a.Css(tau-D) > > are these the same? I Think they are but the Knapp-Carter one makes > more sense to me than the other one. If I take the Fourier Transform > of both I should get > the cross-spectral density of x1 with x2. > > Thanks > > TomHi Tom, I don't know why it makes you to think so deep. From fundamental convolution theorem we get, css(tau-D) = integral [css(tau) deltafunction(tau-D)d(tau) where, d(tau) is derivative of tau. or in other words, css(tau-D) = css * deltafunction(tau-D) where, * is convolution. Frequency domain also does not show any differnce. Conclusion: I don't see any difference ; also no special sense which you try to hint. I guess "a" is scalar here. Regards, Santosh

Reply by ●July 24, 20032003-07-24

aberdonian_2000@yahoo.com (Tom) wrote in message news:<e1b1658f.0307221459.5a2d5e56@posting.google.com>...> I have read quite a few papers on teh subject and have a question. > > Suppose we have two sensors x1 and x2 > > x1(t)=s1(t)+n1(t) > > and > > x2(t)=a.s2(t-D)+n2(t) > > where n1(t) and n2(t) are uncorrelated white noise sources also > uncorrelated with the two signals s1 and s2. I need to find the delay > D.The constant a is an attenuation factor. > > The Knapp-Carter give an expression for cross-correlation as > > Cx1x2(tau) = a.Css(tau) * delta (tau - D) where * is convolution and > delta is the delta function > > and in other papers it is given as > > Cx1x2(tau) = a.Css(tau-D) > > are these the same? I Think they are but the Knapp-Carter one makes > more sense to me than the other one. If I take the Fourier Transform > of both I should get > the cross-spectral density of x1 with x2. > > Thanks > > TomYes they are the same, convolution with a delta signal will result in the input signal.

Reply by ●July 24, 20032003-07-24

aberdonian_2000@yahoo.com (Tom) wrote in message news:<e1b1658f.0307221459.5a2d5e56@posting.google.com>...> I have read quite a few papers on teh subject and have a question. > > Suppose we have two sensors x1 and x2 > > x1(t)=s1(t)+n1(t) > > and > > x2(t)=a.s2(t-D)+n2(t) > > where n1(t) and n2(t) are uncorrelated white noise sources also > uncorrelated with the two signals s1 and s2. I need to find the delay > D.The constant a is an attenuation factor. > > The Knapp-Carter give an expression for cross-correlation as > > Cx1x2(tau) = a.Css(tau) * delta (tau - D) where * is convolution and > delta is the delta function > > and in other papers it is given as > > Cx1x2(tau) = a.Css(tau-D) > > are these the same? I Think they are but the Knapp-Carter one makes > more sense to me than the other one. If I take the Fourier Transform > of both I should get > the cross-spectral density of x1 with x2. > > Thanks > > TomAsk yourself, wha is the value of delta(tau-D) at any time other than tau-D

Reply by ●July 24, 20032003-07-24

aberdonian_2000@yahoo.com (Tom) wrote in message news:<e1b1658f.0307221459.5a2d5e56@posting.google.com>...

Reply by ●July 24, 20032003-07-24

santosh nath wrote:> aberdonian_2000@yahoo.com (Tom) wrote in message news:<e1b1658f.0307221459.5a2d5e56@posting.google.com>... > > I have read quite a few papers on teh subject and have a question. > > > > Suppose we have two sensors x1 and x2 > > > > x1(t)=s1(t)+n1(t) > > > > and > > > > x2(t)=a.s2(t-D)+n2(t) > > > > where n1(t) and n2(t) are uncorrelated white noise sources also > > uncorrelated with the two signals s1 and s2. I need to find the delay > > D.The constant a is an attenuation factor. > > > > The Knapp-Carter give an expression for cross-correlation as > > > > Cx1x2(tau) = a.Css(tau) * delta (tau - D) where * is convolution and > > delta is the delta function > > > > and in other papers it is given as > > > > Cx1x2(tau) = a.Css(tau-D) > > > > are these the same? I Think they are but the Knapp-Carter one makes > > more sense to me than the other one. If I take the Fourier Transform > > of both I should get > > the cross-spectral density of x1 with x2. > > > > Thanks > > > > Tom > > Hi Tom, > > I don't know why it makes you to think so deep. > > From fundamental convolution theorem we get, > > css(tau-D) = integral [css(tau) deltafunction(tau-D)d(tau) > where, d(tau) is derivative of tau. > > or in other words, > css(tau-D) = css * deltafunction(tau-D) > where, * is convolution. > > Frequency domain also does not show any differnce. > Conclusion: I don't see any difference ; also no special sense which you > try to hint. > > I guess "a" is scalar here. > > Regards, > SantoshYes of course it is simple. I only needed to substiitute h =tau-D and take the FT and they are identical. Now can somebody anwswer a (vaguely) related question. I have seen books which define sampling as multiplication by a train of impulses and somebody on this group said it was convolution by impulses - do they give the same end result? ie suppose I have an analugue signal f(t) then is the FT of f(t)*sum delta(t-tau) * = convolution the same as f(t).sum delta (t-tau) where . is multiplication. We should get the standard sampled spectrum . Thanks Tom

Reply by ●July 25, 20032003-07-25

aberdonian_2000@yahoo.com (Tom) wrote in message news:<e1b1658f.0307221459.5a2d5e56@posting.google.com>...

Reply by ●July 25, 20032003-07-25

Tom wrote: (snip)> Now can somebody anwswer a (vaguely) related question. I have seen books which define sampling as > multiplication by a train of impulses and somebody on this group said it was convolution by impulses - do they > give the same end result? > ie suppose I have an analugue signal f(t) then is the FT of > > f(t)*sum delta(t-tau) * = convolution > > the same as > > f(t).sum delta (t-tau) where . is multiplication. > > We should get the standard sampled spectrum . > > Thanks > > Tom > >Convolution of a continuous function by the delta function would result in a continuous function, wouldn't it? There would have to be something about an infinite sum of shifted continuous functions for the result to be discrete sequence which doesn't seem obvious to me.