(b)Find the variance of Uand that of V. (e)Use pa (b)todetermine the varianc f(the average energy of thePAMwith uniform distribution.) ■PAM Modulator A PAM modulator converts a discrete-time sequencex kEZ of real symbols into a continuous-time real waveform x0=∑xP1-k灯) where T is the symbol interval,p(r)is the basic modulation pulse,and xA. A PAM system is said to be orthonormal if (p-kT),p-jT)》=d- that is)=ip(t-kT) With the assumption that the underlying channel model is an ideal AWGN channel. the received basebar d waveform is given by )=x0+0 In an orthonormal PAM system,the sequenceof matched-filter outputs is given by y=((I),p1-kT)》=x4+n we obtain without loss of optimality an cquivalent discrete-time GN channel y=x+n 2.4.2 Orthonormal OAM ■QAM Signal Set (MxM)-QAM signal set consists of the Cartesian product of two A=a=(a'+ja")la'EA'.a"EA') whereA'=(-d(M-1)/2.-d/2,d/2.d(M-1)/2).Thus the size of A is Al=M, and the number of bits per symbol is 2log2M=2b.See Fig.2.8 for examples.Assuming that the symbols in A are equiprobable,the average energy per complex symbol is F=2×dM-dM- 2 (2.7) 2.92-9 (b) Find the variance of U and that of V . (c) Use part (b) to determine the variance of S. (i.e., the average energy of the PAM with uniform distribution.) ◼ PAM Modulator A PAM modulator converts a discrete-time sequence {xk, k} of real symbols into a continuous-time real waveform ( ) ( ) k k x t x p t kT  = −  where T is the symbol interval, p(t) is the basic modulation pulse, and xk  . A PAM system is said to be orthonormal if ( ), ( ) k j p t kT p t jT  − − = − that is {k(t)} = {p(t-kT)}. With the assumption that the underlying channel model is an ideal AWGN channel, the received baseband waveform is given by y(t) = x(t) + n(t) In an orthonormal PAM system, the sequence {yk} of matched-filter outputs is given by ( ), ( ) k k k y y t p t kT x n = − = + Thus, we obtain without loss of optimality an equivalent discrete-time AWGN channel model y = x + n 2.4.2 Orthonormal QAM ◼ QAM Signal Set A standard (MM)-QAM signal set consists of the Cartesian product of two standard M-PAM d-spaced signal set; i.e., = = +   a a ja a a ( ' '') | ' , '' ' '  where ' = − − − − { ( 1) / 2,., / 2, / 2,., ( 1) / 2} d M d d d M . Thus the size of  is || = M2 , and the number of bits per symbol is 2log2M = 2b. See Fig. 2.8 for examples. Assuming that the symbols in  are equiprobable, the average energy per complex symbol is 2 2 2 2 2 ( 1) ( 1) 2 12 6 d M d M A − − =  = (2.7)