 Hello everyone, welcome to lecture on paratic check matrix. At the end of this session, students will be able to describe paratic check matrix and also able to describe and illustrate syndrome testing. Now before starting with the actual session, let's pause the video and think about what is the structure for generator matrix in systematic linear block code. If you remember generator matrix have formed for systematic linear block code as generator matrix equals to combination of parity array and identity matrix. So if you see over here, p is nothing but the parity array and i is nothing but the identity matrix which is having a size of k by k. So total size of generator matrix is having a much k by n by k by n means k rows and n columns. Now let's start with the parity check matrix. It is denoted by h and it is used for the decoding the received vectors which is having a error. So for every k by n generator matrix, there is always exist n minus k by n matrix h which is nothing but the parity check matrix. So basically the rows of generator matrix and rows of h matrix are orthogonal to each other. Orthogonal means what? If you take a dot product of matrix and transpose of that matrix that gives you the identity matrix. So here g is nothing but the generator matrix and h of t is nothing but the transpose of h matrix that is transpose of parity check matrix. So dot product of this is gives you the 0. Now to fulfill the orthogonality requirement of a systematic code, the components of h are written as h equals to i combination of identity matrix and transpose of parity array. If you recall the equation of generator matrix, generator matrix is having combination of parity array and identity matrix. Here it is similar to that one only the difference is what? Identity matrix is combined with the transpose of parity matrix. So this is nothing but the h that is parity check matrix. So now the transpose of this parity check matrix is written as rows becomes columns and columns becomes rows. So this identity matrix comes over here and below that we can write transpose of parity check matrix, right. So if you see this is an identity matrix and this is a parity array. Now it is very easy to verify that the product u h t here u is nothing but the code word set of code word, h t is nothing but the transpose of your parity check matrix of each code word which is generated by g and using of transpose of parity check matrix. So you get the equation is what u h t equals to p 1 plus p 1 p 2 plus p 2 up to how many bits you are having in that array that is up to n minus k plus n minus k which is equals to 0. We already seen the equation that is generated matrix is multiplied with the transpose of parity check matrix that else to 0. So where p 1 p 2 up to p n minus k are nothing but your parity bits. So thus once the parity check matrix h is constructed by using that we can test whether received vector is valid member of the code word set or not. Now it has two conditions for parity check matrix. No columns of h can be all zeros means if the any column is having zero so the same effect on the syndrome testing what is the syndrome testing we going to see in the next slides. So at the receiver side if you going to for the syndrome testing if you having all zeros in the column so that is that does not affect on the syndrome. So you won't be able to detect any error at the receiver side. And second one is all columns of h must be unique it means if it has same column values at the receiver end you will getting the same response or same member at the syndrome testing and you won't be able to distinguish between these two which one is for which received code word. Now what is the syndrome testing? It is denoted by r r is nothing but the received vector at the receiver side which is generated by the transmitting of the u u is nothing but your code word. So you transmitted u code word which is having u1 u2 up to un and you are at the receiver side you are receiving r which is having r1 r2 up to rn. Now therefore at the receiver side you are because of the channel there may be some error gets added in the received code word. So the equation of the r equals to r equals to u plus e where u is your code word and e is nothing but the your error vector which is having components e1 up to en. And that error pattern is generated by the channel as I said when you transmitted the data at the receiver side through the channel some error gets added. So the equation for that receiver code word vector is nothing but the u plus e u is nothing but the transmitted vector and e is nothing but your error vector. So that syndrome is nothing but the s equals to r into h of t where r is your received vector h of t is nothing but the transpose of your parity check matrix. Now it is a result of a parity check matrix performed on the r to determine whether the received r is having a valid member of the code word set or not. So it has two conditions on that if r is a member valid member of the code word set the syndrome s equals to 0 and if it r contains any error which is detected by the syndrome s equals to non-zero value. So from equation 1 and 2 we can say that s equals to u plus e into h of t because we know that r equals to nothing but your transmitted code word plus error vector. So you can multiply this in the bracket you will get u into h of t plus e into h of t. So we already defined for this u into h of t is nothing but equals to 0 in the previous slides. So the equation becomes s equals to e into h of t. So the syndrome test whether perform on the whether either a corrupted code vector which is nothing but the e into h of t or the error pattern that caused it is the same result. So you can say same syndrome. So important property of this of linear block code is what? The mapping between correctable error patterns and syndrome is 1 to 1. How let us we can see using one example. Suppose that you are having a code word u which is nothing but the 1 0 1 1 1 0 is transmitted and the vector r is received which is having 0 0 1 1 0. That means you are having error in the left most list you are transmitted 1 and you are received 0 over here. So means this bit is having error. So we have to find the syndrome vector for this using s equals to r into h of t and also you going we going to verify whether we are getting the same syndrome vector by using the equation s equals to e into h of t. So for that they given the generator matrix g equals to this r a right. So from this we can see that here n is 6 and k is 3. So you are having 2 raise to 3 equals to 8 data word possible. So these are the 8 possible data words from 0 0 0 to 1 1 1. Now we have equation for syndrome that is s equals to r into h of t. So we have r in the example they given. So this r and for h of t we can generate this h of t by using generator matrix we have h of t equals to the nothing but the identity matrix and below that we can add a parity. So if you perform this multiplication so rho into column this rho into this column and this rho into this column we will get 1 for this multiplication 1 plus 1 for this multiplication and again 1 plus 1 for this multiplication. Final result we are getting is what 1 0 0. So this syndrome we are getting from the equation s equals to r into h of t. Now this we going to verify using equation s equals to e into h of t. So e is what received error bit vector that is 0 0 1 1 1 0 and h of t is what this one. So if you multiply this again you are getting the same result. So this is how we going to verify the syndrome at the receiver side. These are the references. Thank you.