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I have 1 transmitted signal transmits to the receiver. Every time transmit, I collect data from the receiver, I use the transmitted which I just transmitted and received data, correlate them, then I can get the delay using xcorr tool in MATLAB.

I repeat 10 times periodically, they are different because every time transmit, noise which is added, has to be exist. I have this delay vector :

delay =   

  Columns 1 through 2

   0.010346125000000   0.010349000000000

  Columns 3 through 4

   0.010349875000000   0.010349000000000

  Columns 5 through 6

   0.010349750000000   0.010349750000000

  Columns 7 through 8

   0.010349750000000   0.010349750000000

  Columns 9 through 10

   0.010348875000000   0.010348875000000

from this vector, how much variation in delay do I know?

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  • $\begingroup$ If that is a vector of delays, then don't you just calculate the standard deviation? Note if the noise is white (meaning uncorrelated and also should add identically distributed), your processing gain in dB is 10Log(N) where N is the number of samples- that should give you an idea on how to predict the delay variation (standard deviation) you are seeing relative to your input SNR $\endgroup$ – Dan Boschen Apr 12 '17 at 21:49
  • $\begingroup$ @DanBoschen : Average tells you the central part distribution where that is? and Standard-deviation tells you how wide of distribution. As you mentioned, "N is the number of samples", but I've never known the value of N, but I know the gain between 0 to 60dB. Since I know the gain, How to predict delay variation? $\endgroup$ – Nate Duong Apr 13 '17 at 0:45
  • $\begingroup$ @DanBoschen: I am computing the time delay by interpolating between sample, and compute xcorr. So, I should not see when I am doing that identical value, just slight changed. I can take standard-deviation of those values, I have number in nanosecond, and each nanosecond that I have correspond to this much variation, so that should be possible for me to determine what the accuracy of my measurements in meters? Therefore, I need to compute the variation of delay. $\endgroup$ – Nate Duong Apr 13 '17 at 0:55
  • $\begingroup$ @DanBoschen: if I take 10 measurements (as I mentioned in previous topic: I have 10 seconds of data and cut off 10 windows, and do 10 time xcorr to get vector of delay 1x10), I compute standard deviation, and I have number in seconds. they are very small number seconds (like a few nanoseconds). I can multiply that with speed of light, and I have the accuracy of measurement in meter. That is my goal. $\endgroup$ – Nate Duong Apr 13 '17 at 1:02
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    $\begingroup$ I would I just want to make sure I really understand what you were asking as a question. So was the issue that you did not know that the standard deviation is how you measure the "noise" for an arbitrary signal or set of data? $\endgroup$ – Dan Boschen Apr 13 '17 at 14:50
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For your dataset, the mean is the estimate of the delay and the standard deviation is the estimate of the noise, The SNR of your delay measurement is 20Log(mean/ standard deviation).

The number of samples is very small so the result will not be an accurate measurement of the what the "true" mean and standard deviation of your random process is. To understand that better, look into "confidence interval". As a start here is a simple calculator demonstrating the concept:

https://www.mathsisfun.com/data/confidence-interval-calculator.html

Importantly we are assuming that each of your samples are independent and identically distributed random variables.

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  • $\begingroup$ almost closed what I am looking for? there are some key words you mentioned: interval, SNR, ** standard deviation**, mean ... during the comments from you. They are also matched with my professor. Now, I need to look at couple more times what you suggested and advised me from the beginning till now. I hope I will catch up with this knowledge. Thank you for your patient, I wish I can vote you up million times in this forum, It's very kind of you, I hope I can see you in the next topics. Thank you very much, Mr. Boschen! $\endgroup$ – Nate Duong Apr 13 '17 at 15:27

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