ormation is there at the decoder that can be used to create a predicted value x,P?" He answered by realizing that, while Xn-1 , Xn-K isn't available there, X,---, can be found there. "I'll just change my equation slightly to this," he said and wrote (4.56) (4.57) The final question that remains to be answered is this: What do we use for the values A2, up to ak? This takes a bit of math. To start, we'll decide we want to choose a1 A2,..., ak such that is as close as possible to Xn; specifically, we'll decide we want to choose A1,A2,...,AK such that (4.58) where arg min means "the value of an that minimizes." Chapter Four Substituting in the x, equation in (4.56) leads to (4.59) Assuming that xn-k is close to Xn-k, we can rewrite this as (4.60) From now on, we will use mathematics to come up with a more explicit equation for the a to ak values. We'll start by trying to find the best a1 , by taking the deriva- tive of the above term with respect to a and setting it to 0. This leads to (4.61) And now we just do some simple math to come up with (4.62) (4.63) Using correlation notation, we can rewrite this according to (4.64) ,(1-k)=R,(1) (4.65) That's the equation we get when we try to get the best a . When we repeat this to get an equation for the best a up to AK, we get a total of K equations. Putting these equations together by using simple matrix notation, we get the following: R (0) (4.66) R,(K-1) R,(K-2) Rx(K) Source Coding and Decoding: Making it Digital So, if someone gives you the statistics Rx (0), ,R,(1),...,R,(K), , you get the best a, by simply solving the above matrix. Example 4.7 Figure out the best choice for the values of a1 and a2 in a 2-tap predictor when the input has a correlation function given by Rx(0)=1 (E4.25) R,(1)=R,(-1)=0.5 (E4.26) Rx(2)=0.25 (E4.27) Solution: We turn to equation (4.66), and use K=2, which leads us to Rx(0) Rx(1) (E4.28) (E4.29) (E4.30) (E4.31) (E4.32) (E4.33) The Block Diagram This section provides you with a block diagram of the DPCM, and its source decoder, SO that you c