2 Since S N (Xn - q)2 = S N X n - 2qS N Xn + Nq 2, the pdf could be factorized into n=1 n=1 n=1

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N 1 1 N 2 exp - Nq 2 - 2q X n exp - X n N 2 2 4 =1 3 n =1 (2 444444 2444444 3 14422n444 ) 1p 4 4

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f (X; q ) =

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g (T ( X ) , q )

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h ( X)

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Hence, T(X) = S N Xn is a suf cient statistic for q because g is a function n=1 depending on X only through T. In fact, any one-to-one mapping of T(X) is also a suf cient statistic for q. Example 1.4 (Suf cient Statistics) Let X be Gaussian distributed with zero mean and unknown variance q, with X1, . . . , XN drawn independently according to this distribution. The pdf of the data is given by f (X; q ) = 1 1 1 N 2 1 N 2 exp - X n = exp - X n { 1 N 2 n =1 (2pq ) 2 2 n =1 h ( X) 14444244443

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g (T ( X ) , q )

(2pq )

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N n=1 2 n

Hence, T(X) = S 1.3.3

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X is a suf cient statistic for q.

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A digital transmitter can be modeled as a device with input as information bits and continuous-time analog signals matched to the channel or the medium as output. A string of k information bits is passed to a channel encoder (combined with modulator) where redundancy is added to protect the raw information bits. N encoded symbols {x1, . . . , xN} (where xn X ) are produced at the output of the channel encoder. The N encoded symbols are mapped to the analog signal (modulated) and transmitted out to the channel. The channel input signal x(t) is given by x(t ) = xn g(t - nTs )

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(1.32)

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where Ts is the symbol duration and g(t) is the low pass equivalent transmit pulse with two-sided bandwidth W and pulse energy - |g(t)|2dt = 1. Equation (1.32) is a general model for digitally modulated signals, and the signal set X is called the signal constellation. For example, X = {ej2pm/M : m = {0, 1, . . . , M - 1}} represents MPSK modulation. X = {xR + jxI : xR , xI { 1 , 3 , . . . , M }} 2 2 4 represents MQAM modulation. The signal constellations for MPSK and MQAM are illustrated in Figure 1.8. From Equation (1.28), the low pass equivalent received signal through at fading channel can be expressed as y(t ) = h(t ) x(t ) + z(t ) = xn h(t )g(t - nTs ) + z(t )

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(1.33)

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where z(t) is the low pass equivalent white complex Gaussian noise, h(t) is a zero-mean unit variance complex Gaussian random process, and only x(t) contains information.

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(a) Q 16QAM signal set

(b) Figure 1.8. Illustration of signal constellations: (a) MPSK constellation; (b) MQAM constellation.

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At the receiver side, the decoder produces the string of k decoded information bits based on the observation y(t) from the channel. Hence, the receiver problem can be modeled as a detection problem; that is, given the observation y(t), the receiver has to determine which one of the 2k hypothesis is actually transmitted. It is dif cult to gain very useful design insight if we look at the problem from continuous-time domain. In fact, as we have illustrated, the continuous-time signal y(t) can be represent equivalently as vectors in signal space y. Hence, the receiver detection problem [assuming knowledge of channel fading H(t)] is summarized below. Problem 1.1 (Detection) Let w {1, 2k} be the message index at the input of the transmitter. The decoded message w at the receiver is given by w = argmaxw p(y w , h(t )) Such a receiver is called the maximum-likelihood (ML) receiver, which minimizes the probability of error if all the 2k messages are equally probable. Due to the white channel noise term in (1.33), we need in nite dimension signal space in general to represent the received signal y(t) as vector y. However, since only h(t)x(t) contains information and h(t)x(t) can be represented as vector in a nite-dimensional signal subspace, not all components in y will contain information. Hence, the detection problem in Problem 1.1 could be further simpli ed, and this is expressed mathematically below. Let y = {Y1(t), . . . , Yn(t), . . .} denote the in nite-dimensional signal space that contains y(t) and s = {Y1(t), . . . , YDs(t)} denote a Ds-dimensional subspace of y that contains s(t) = h(t)x(t); that is y(t ) = y j Yj (t )

j =1

(1.34)

and s(t ) = s j Yj (t )