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# Detection, Estimation modulation theory part 1 - Vantress H.

Vantress H. Detection, Estimation modulation theory part 1 - Wiley & sons , 2001. - 710 p.
ISBN 0-471-09517-6 Previous << 1 .. 179 180 181 182 183 184 < 185 > 186 187 188 189 190 191 .. 254 >> Next Errors in the Presence of White Noise. We assume that n(t) is white with spectral height N0/2. Although the result was first obtained by Yovits and Jackson, appreciably simpler proofs have been given by Viterbi and Cahn, Snyder , and Helstrom. We follow a combination of these proofs. From (128)
SM = S«(«) + (129)
and
G+(»= [s.CA*) + Ů+- (130)
From (78)
ßîÎ) = [5e(w) + No/2]+ {[Se(w) + Ëî/2]“} + (131)t
fTo avoid a double superscript we introduce the notation
G~(joi) = [G4ja>)]\
Recall that conjugation in the frequency domain corresponds to reversal in the time domain. The time function corresponding to G + (JW) is zero for negative time. Therefore the time function corresponding to G~(j<*)) is zero for positive time.
Closed-Form Error Expressions 499

„a,, 1 f s„(q>) + N0I2_____________N0/l ]
°U) [5.Č + N0/2]+ 1(5» + vVo/2]- [SM + N0/2]-J+'
(132)
Now, the first term in the bracket is just [?ŕ(ŕ>) 4- N0/2] +, which is realizable. Because the realizable part operator is linear, the first term comes out of the bracket without modification. Therefore
H°Uw) = 1 “ [ŮŇâä {[Sa(<o)N+N0/2]-}+’ (133)
We take V N0/2 out of the brace and put the remaining V N0/2 inside the [ • ] “. The operation [ • ]" is a factoring operation so we obtain N0/l inside.
Ho(jw) = 1 _ isMrkiw ' (134)
The next step is to prove that the realizable part of the term in the brace equals one.
Proof. Let Sa(oj) be a rational spectrum. Thus
“ ?)(ř2ó Ń35)
where the denominator is a polynomial in a»2 whose order is at least one higher than the numerator polynomial. Then
Sa(w) + No/1 N(a,2) + (No/2) D(co2) No/2 (No/1) D(a>2)
(136)
D(w2) + (2/N0) N(a>2)
~ D{ui2) K >
Observe that there is no additional multiplier because the highest order term in the numerator and denominator are identical.
The a j and ft may always be chosen so that their real parts are positive. If any of the a{ or Pi are complex, the conjugate is also present. Inverting both sides of (138) and factoring the result, we have
500 6.2 Wiener Filters
The transform of all terms in the product except the unity term will be zero for positive time (their poles are in the right-half ë-plane). Multiplying the terms together corresponds to convolving their transforms. Convolving functions which are zero for positive time always gives functions that are zero for positive time. Therefore only the unity term remains when we take the realizable part of (140). This is the desired result. Therefore
VNo/2 [,Sa(j°>) + N0/2Ă'
(141)
The next step is to derive an expression for the error. From Property 4C (27-28) we know that
~ lim h0(t, r) = ^
^ x—*t~ Z e —î4
for the time-invariant case. We also know that
(P = ^-u lim h0{t, r) = ^ lim A0(e) A ^
Ëî(0+)
Ă ho(0+) + ho(0-) h0( 0+)
J_„ Úň =--------------------2-------- = —’
dio
Úň’
(142)
because Ŕ0(ň) is realizable. Combining (142) and (143), we obtain Using (141) in (144), we have
ü-×*.(-{Đ^Ă)?
Using the conjugate of (139) in (145), we obtain
(143)
(144)
(145)
(146)
(147)
Expanding the product, we have
Closed-Form Error Expressions 501
The integral in the first term is just one half the sum of the residues (this result can be verified easily). We now show that the second term is zero. Because the integrand in the second term is analytic in the right half of the s-plane, the integral [ — 00,00] equals the integral around a semicircle with infinite radius. All terms in the brackets, however, are at least of order \s\~2 for large \s\. Therefore the integral on the semicircle is zero, which implies that the second term is zero. Therefore
The last step is to find a closed-form expression for the sum of the residues. This follows by observing that
(To verify this equation integrate the left-hand side by parts with č = In [(oj2 + + Ä2)] and dv = d^llir.)
Comparing (150), (151), and (138), we have
which is the desired result. Both forms of the error expressions (150) and
(152) are useful. The first form is often the most convenient way to actually evaluate the error. The second form is useful when we want to find the Sa(co) that minimizes ?P subject to certain constraints.
It is worthwhile to emphasize the importance of (152). In conventional Wiener theory to investigate the effect of various message spectra we had to actually factor the input spectrum. The result in (152) enables us to explore the error behavior directly. In later chapters we shall find it essential to the solution for the optimum pre-emphasis problem in angle modulation and other similar problems. Previous << 1 .. 179 180 181 182 183 184 < 185 > 186 187 188 189 190 191 .. 254 >> Next 