Hi everyone, I need help in autocorrelation. I have generated a particular sequence code, C of period 15 and need to find its autocorrelation. I need to get the values at given lags from i = 0 to i = 15. For example when i = 0, it's a perfect autocorrelation (no lag) so I should get the peak value of the autocorrelation here. Theoretically, this value should be equal to 1. But what I got was more that. Subsequently, I got a constant value (which was lower than the peak) when there is lag. Theoretically, I should be getting a constant 0 value... What I did was multiply C by a cyclic shift of it for each lag. What did I do wrong or forgot? Really need suggestions.... Thanks in advance! Kat~ 

autocorrelation problem
Started by ●February 22, 2003
Reply by ●February 24, 200320030224
Hi katrina, I do not really understand what you mean (my English is no that good). Could you send the sequence and describe a little more what you intend to do? What kind of process is it and what are you using to calculate the autocorrelation? I guess you are using xcorr, or have you done your own code? Sergio A las 07:24 del 22 de Feb de 2003, t_katrina <> <> dijo: > Hi everyone, > > I need help in autocorrelation. I have generated a particular > sequence code, C of period 15 and need to find its autocorrelation. > I need to get the values at given lags from i = 0 to i = 15. For > example when i = 0, it's a perfect autocorrelation (no lag) so I > should get the peak value of the autocorrelation here. > Theoretically, this value should be equal to 1. But what I got was > more that. Subsequently, I got a constant value (which was lower > than the peak) when there is lag. Theoretically, I should be > getting a constant 0 value... > > What I did was multiply C by a cyclic shift of it for each lag. > What did I do wrong or forgot? Really need suggestions.... > > Thanks in advance! > Kat~ > > _____________________________________ > Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group. > > _____________________________________ > About this discussion group: > > To Join: > > To Post: > > To Leave: > > Archives: http://www.yahoogroups.com/group/matlab > > More DSPRelated Groups: http://www.dsprelated.com/groups.php3 > > ">http://docs.yahoo.com/info/terms/ >  Did you ever notice that everyone in favour of birth control has already been born?  Benny Hill 
Reply by ●February 24, 200320030224
I'm just guessing here, but it sounds as if you have a nonzero average value in C when you expected C to be zeromean. Is the constant value at lags not equal to zero positive or negative? Sincerely, Glen Ragan From: "t_katrina <>" <t_katrina on 02/22/2003 01:24 PM To: cc: Subject: [matlab] autocorrelation problem Hi everyone, I need help in autocorrelation. 
I have generated a particular sequence code, C of period 15 and need to find its autocorrelation. I need to get the values at given lags from i = 0 to i = 15. For example when i = 0, it's a perfect autocorrelation (no lag) so I should get the peak value of the autocorrelation here. Theoretically, this value should be equal to 1. But what I got was more that. Subsequently, I got a constant value (which was lower than the peak) when there is lag. Theoretically, I should be getting a constant 0 value... What I did was multiply C by a cyclic shift of it for each lag. What did I do wrong or forgot? Really need suggestions.... Thanks in advance! Kat~ < http://rd.yahoo.com/M$6920.2960106.4328965.2848452/D=egroupweb/S05083376:HM\ /A64858/R=0/*http://www.gotomypc.com/u/tr/yh/cpm/grp/300_Cquo_1/g22lp?Target=\ mm/g22lp.tmpl > < http://us.adserver.yahoo.com/l?M$6920.2960106.4328965.2848452/D=egroupmail/S=:\ HM/A64858/rand69070115 > _____________________________________ Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group. _____________________________________ About this discussion group: To Join: To Post: To Leave: Archives: http://www.yahoogroups.com/group/matlab < http://www.yahoogroups.com/group/matlab> More DSPRelated Groups: http://www.dsprelated.com/groups.php3 < http://www.dsprelated.com/groups.php3> ">http://docs.yahoo.com/info/terms/>. 
Reply by ●February 25, 200320030225
Hi Katrina! It is all about a LFSR(4) sequence (I suppose) that (theoretically) has properties you are describing: normalized ACF(0) = 1, ACF(<>0) = 0. But it is only for bipolar sequence of infinit length. If the sequence is bipolar, length N then normalized ACF(<>0) = 1/N. If ACF(0) <> 1, then you have to normalize it. Predrag  Original Message  From: <> To: <> Sent: 2003. velja 22 20:24 Subject: [matlab] autocorrelation problem > Hi everyone, > > I need help in autocorrelation. I have generated a particular > sequence code, C of period 15 and need to find its autocorrelation. > I need to get the values at given lags from i = 0 to i = 15. For > example when i = 0, it's a perfect autocorrelation (no lag) so I > should get the peak value of the autocorrelation here. > Theoretically, this value should be equal to 1. But what I got was > more that. Subsequently, I got a constant value (which was lower > than the peak) when there is lag. Theoretically, I should be > getting a constant 0 value... > > What I did was multiply C by a cyclic shift of it for each lag. > What did I do wrong or forgot? Really need suggestions.... > > Thanks in advance! > Kat~ > > _____________________________________ > Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group. > > _____________________________________ > About this discussion group: > > To Join: > > To Post: > > To Leave: > > Archives: http://www.yahoogroups.com/group/matlab > > More DSPRelated Groups: http://www.dsprelated.com/groups.php3 > > ">http://docs.yahoo.com/info/terms/ > > 