1、 Thus, f (0 = M , f (0 = (1/nb (Yi Y1 1, and M i=2 (Yi Y1 , EY1 ,···,YM f (0 = 0, f (0 = (1/n2 b M i=2 (Yi Y1 1 EY1 ,···,YM f (0 = EY1 ,···,YM 2 nb = M Yi2 2Yi Y1 + Y12 (Yi Y1 i=2 M 1 2 (nb + n2 b 2nb nb + (nb + nb 0 n2 b 2(M 1 nb = Also, g (ns = f (ns /(

2、f (ns ln 2, and g (ns = (f (ns f (ns f (ns 2 /(f (ns 2 ln 2. Finally, we note that h (ns = y1 1 h(ns n s + nb y1 + (ns + nb 2 y1 1 ns + nb 2 h (ns = h(ns Using these denitions, we may write C (ns = log2 M EY1 ,···,YM g (ns = log2 M EY2 ,···,YM EY1 g (ns = log2 M EY2 ,&#

3、183;··,YM y1 =0 g (ns Y1 =y1 h(ns where in the second step we used the fact that Y1 is independent of Y2 , . . . , YM , and in the last step g (ns is evaluated at Y1 = y1 (notation is dropped in the following. It follows that C (ns = EY2 ,···,YM y1 =0 g (ns h(ns + g (ns h (n

4、s = EY2 ,···,YM y1 =0 g (ns + g (ns y1 1 n s + nb h(ns = EY1 ,···,YM g (ns + g (ns Y1 1 ns + nb Y1 1 ns + nb = EY1 ,···,YM f (ns + log2 f (ns f (ns ln 2 26 From the above, it follows that limns 0 C (ns = 0. Proceeding to the second derivative, C (ns = EY2

5、 ,.,YM y1 =0 g (ns h(ns + 2g (ns h (ns + g (ns h (ns = EY1 ,···,YM f (ns f (ns f (ns 2 f (ns +2 2 f (ns ln 2 f (ns ln 2 Y1 + (ns + nb 2 Y1 1 ns + nb 2 Y1 1 ns + nb + log2 f (ns Note that Y1 + (ns + nb 2 Y1 1 ns + nb 2 ns 0 lim E = nb Y2 2Y1 1 nb + n2 b +E 1 +1 = + 2+1=0 2 2 nb nb nb n

6、b n2 b Also, E f (02 = 1 E n2 b M 2 (Yi Y1 i=2 = 1 E (Yi Y1 (Yj Y1 n2 b i=2 j =2 M M M M 1 1 = 2E Yi Yj Yi Y1 Yj Y1 + Y12 = 2 (M 12nb + (M 1(M 2nb nb nb i=2 j =2 M (M 1 nb = where in the second to last step we partitioned the double summation into the part where i = j and where i = j . From the abov

7、e, it follows that 2(M 1 M (M 1 2 nb nb lim EY1 ,···,YM ns 0 M nb ln 2 M 2 ln 2 M ns 0 M lim C (ns = (Yi Y1 i=2 Y1 1 nb = M 1 2 nb + n2 b · (M 1 nb nb + nb M nb ln 2 M nb ln 2 nb (B-1 = M 1 M nb ln 2 =0 27 This establishes the following. Theorem B-2. For M -PPM signaling on a Poisson channel with non-signal slot-count average nb > 0 and signal slo