g at Figure 4.16(a), we see that px(x) is 0 outside the interval [a,b], and we use this information in equation (4.19) to get to: mse : (4.22) Now, we use two pieces of information to change the integral. We'll use the definition of a cell, R, to help us out here (so if you don't remember, flip back and have a quick peek). (1) First, the interval [a,b] can be broken up into cells R1, R2, ..., ,RN. . (2) Second, for all values of x that fall into the cell R, the output of the quantizer, x, corresponds to the value y. Applying these two pieces of information to the integral creates mse = (4.23) (4.24) (4.25) Now, consider the ith cell Rj. In this cell, the term x - yi is simply the difference between the input to the quantizer x and the output of the quantizer yi; that is, x - yi is simply the error of the quantizer, which we'll call err; Using the substitution = - in this integral leads to: Chapter Four (4.26) The limits of the integral erri,min and err i,max refer to the smallest value of the quantizer error and the largest value of the quantizer error in a cell R. Let's calculate these values with the help of the graph of Figure 4.16(b). Take a look at the first cell, which maps all the values between a and a+ A into the value y1 a+ A /2. It follows that the largest value of error occurs when the input x = a+ A is mapped to the output = a + A /2; in this case the error is A /2. Similarly, the smallest value of the error occurs when the input X = a is mapped to the output = a + A /2. In this case, the error value is - A /2. It's easily shown that for any cell the largest value of error is A /2 and the smallest value of error is - A /2. Using this maximum value of error and minimum value of error in the integral leads to (4.27) Since all N of the integrals are identical, this equation can be rewritten as mse (err) (4.28) (err) (4.29) (4.30) where the last equation comes about because, looking at Figure 4.16(b), we see N steps A in the range B, which tells us NA = B, or N/B = 1/A Finally, evaluating the integral an