dear Sir, Thank you for your replay, you are correct sir, this cross-correlation computation between two digital sequences, 1-The purpose of this calculation to calculate the highest value found form the correlation, but later I want to use the position of the highest value(for example: the highest value comes from first or second or third or???. Calculation). 2-The original input data is digital signal with 1 or 0 , but later I will change this to 1 or -1 only, so all the input values are 1 or -1 only. 3- Yes Sir, as I mentioned before I want to know the position in the M sequence where the highest vale of Ax occurs. 4-the speed is important, but I do not understand the meaning of size of the code. 5- Unfortunately I am not familiar with any programming language except a little knowledge of C, and I am start already to Study a little about C++. I sue personal computer. 6- The sequence M is known already, I mean the element of this sequence is decided already, and should be written in the program body as a known sequence, in other hand, the sequence S is changeable so should be inter by the user as sequence consist of nine element or number. I hope my explanation is enough >From: Bob Ammerman >Reply-To: pic microcontroller discussion list >To: PICLIST@MITVMA.MIT.EDU >Subject: Re: M and S sequences >Date: Mon, 29 Apr 2002 07:15:23 -0400 > >This looks like some sort of correlation computation. > >A couple of questions: > >1. Do you need to keep the individual values of A0, A1, A2, A3, etc. or >only >the highest value found? > >2. Are your inputs all either -1 or 1 (as shown in your example) with no >other values possible? > >3. Do you need to know the position in the sequence M where the highest >value of Ax occurs? > >4. What is most important in this calculation? Speed? Size of RAM? Size of >code? > >5. What language do you want to implement it in? On a PIC? on a PC? > >Bob Ammerman >RAm Systems > > >----- Original Message ----- >From: "ali dawji" >To: >Sent: Monday, April 29, 2002 5:14 AM >Subject: M and S sequences > > > > Dear Sir > > I have a question and I hope if you can answer me. > > > > I have tow one dimensional sequence or matrix., > > The first sequence is called M with length 511 number: > > M[511]={ 1, 1, 1, 1, 1, 1, 1,-1,-1,1,1,-1,-1,1,-1,-1,-1,1?????..}; > > And the second sequence is called with nine numbers > > S [9] = {1, 1, 1,-1,-1, 1,-1,-1,-1}; > > > > I want to > > 1- Multiplied the sequences S with first ninth number of the sequence M, >and > > keep the result as A0, > > 2- The multiplication procedure as flows > > The first number of M * the first number of S, if the answer is negative >=> > > (-1), and if the answer is positive => 1 > > Example: > > M= {1, 1, 1, 1, 1, 1, 1,-1,-1}; /*(first ninth number of M sequence)*/ > > S= {1, 1, 1,-1,-1, 1,-1,-1,-1}; > > 1*1=1 > > 1*1=1 > > 1*1=1 > > 1*-1=-1 > > 1*-1=-1 > > 1*1=1 > > 1*-1=-1 > > -1*-1=1 > > -1*-1=1 > > The answer is 3= (6*1- (3*-1)) > > 3-shift the sequence M on number to the right and multiplied the >sequence >S > > by the M sequence (from number 2-10). And keep the multiplication result >in > > A1, > > repeat the procedure of the multiplication till the end of the sequence >M. > > 4- Calculate the maximum value multiplication result A0, A1, A2? > > > > > > sorry since my language is not so good > > thank you > > > > > > > > _________________________________________________________________ > > Join the world s largest e-mail service with MSN Hotmail. > > http://www.hotmail.com > > > > -- > > http://www.piclist.com hint: The PICList is archived three different > > ways. See http://www.piclist.com/#archives for details. > > > > > >-- >http://www.piclist.com hint: The PICList is archived three different >ways. See http://www.piclist.com/#archives for details. > > _________________________________________________________________ MSN Photos is the easiest way to share and print your photos: http://photos.msn.com/support/worldwide.aspx -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.