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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
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488 6.2 Wiener Filters
We can denote the transform of Kdz(r) for r > 0 by the symbol
[Sd2(»]+ Ä Ã êéã{ò)å~ø dr = Ã" h'0(r)e-^ dr.t (74)
Jo Jo
Similarly,
[Sd*0)l- é f° Ê³ã(ò)å-²ò dr. (75)
J — 00
Clearly,
SJJ*) = [Sd2(»]+ + [ÇÄ")]-, (76)
and we may write
H'oijoS) = [SJJ»)U = [W*{jv) Sar(Jw)] + = +- (77)
Then the entire optimum filter is just a cascade of the whitening filter and
"•<» - [cMi#w-L' I (78)
We see that by a series of routine, conceptually simple operations we have derived the desired filter. We summarize the steps briefly.
1. We factor the input spectrum into two parts. One term, C/ + (.s), contains all the poles and zeros in the left half of the .s-plane. The other factor is its mirror image about the yo>-axis.
2. The cross-spectrum between d(t) and z(t) can be expressed in terms of the original cross-spectrum divided by [G + (yw)]*.This corresponds to a function that is nonzero for both positive and negative time. The realizable part of this function (t > 0) is È'0(ò) and its transform is H'0(ju)).
3. The transfer function of the optimum filter is a simple product of these two transfer functions. We shall see that the composite transfer function corresponds to a realizable system. Observe that we actually build the optimum linear filter as single system. The division into two parts is for conceptual purposes only.
Before we discuss the properties and implications of the solution, it will be worthwhile to consider a simple example to guarantee that we all agree on what (78) means.
Example 3. Assume that
r(/) = Vpait) + n(t\ (79)
t In general, the symbol [~]+ denotes the transform of the realizable part of the inverse transform of the expression inside the bracket.
Sn(«) = ~ (81)
Solution of Wiener-Hopf Equation 489 where a(t) and n(t) are uncorrelated zero-mean stationary processes and
s*<"> = (80> [We see that a(t) has unity power so that P is the transmitted power.]
The desired signal is
d(t) = a{t + a), (82)
where a is a constant.
By choosing a to be positive we have the prediction problem, choosing a to be zero gives the conventional filtering problem, and choosing a to be negative gives the filtering-with-delay problem. ‘
The solution is a simple application of the procedure outlined in the preceding section:
0 , 4 2kP , N0 No <*>* + k2( 1 + 4P/kNo) /0„
«“) = + T = T------------------------- (83)
It is convenient to define
ë = Ø (84>
(This quantity has a physical significance we shall discuss later. For the moment, it can be regarded as a useful parameter.) First we factor the spectrum
ñ, ÷ No at2 + k\ 1 + A) ^ + w^ + /. 41* ,OC4
SM = ~2-----------à>¿~+ "ê2---- = G W [O' 0")J • (85)
so
= <“>
Now
Êë,(T) = E[d(t)r(t - T)] = E{a(t + î)WPa(t - t) + «(/ - T)]}
= Vp E[a(t + a) a(t - t)] = VpKa(ò + a). (87)
Transforming,
/— 2k VP e + j(OCC
S«U") = VP Sa(o>)e * (88)
and
t StrQw) _ 2kVPe+i™ {-²Ø + k)
W) - [G+0<u)]* «“ + Ë» V]Vo/2(-y« +/fcVTTA)'
To find the realizable part, we take the inverse transform:
4. 7+ tvi-ãë)]-<M)
The inverse transform can be evaluated easily (either by residues or a partial fraction expansion and the shifting theorem). The result is
2 Vp l
+ ò + a > 0,
b,M.(V'"2, + '/l+A (9.)
7VP 1 _
J-fcVi + Ad + a) ã -f a < 0.
LViVo/2 1 + Vl + Ë The function is shown in Fig. 6.10.
490 6.2 Wiener Filters
Kdz(r)
0 ò = -cr
Fig. 6.10 Cross-covariance function.
r
Now h'0(r) depends on the value a. In other words, the amount of Kdz(r) in the range r > 0 is a function of a. We consider three types of operations:
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