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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 .. 112 113 114 115 116 117 < 118 > 119 120 121 122 123 124 .. 254 >> Next Fredholm Equations of the First Kind 317
consider a simple example. The first step is to show how goo(t) enters the picture. Assume that
s-w - Ř) (253>
and recall that
B(t - u) = J°° eia,t-u) ~ (254)
Differentiation with respect to t gives
p 8(t - u) = J (255)
where p é. djdt. More generally,
N(-p2) 8(t — u) = Ă (256)
J — oo 2tt
In an analogous fashion
D(-P2) Kn(t - u) = J*e D(a>2) Sn(co) (257)
From (253) we see that the right sides of (256) and (257) are identical. Therefore the kernel satisfies the differential equation obtained by equating the left sides of (256) and (257) :
N(-p2) S(t - u) = D(-p2) Kn(t - u). (258)
Now the integral equation of interest is
VI s(t) = f' Kn(t - č) g(u) du, T(<t< Tf. (259)
JTi
Operating on both sides of this equation with D(—p2), we obtain D(-p2)VE s(t) = f' D(-p2) Kn(t - u) g(u) du, T{<t< Tf. (260)
JTi
Using (258) on the right-hand side, we have
D(—P2)Ves(t) = N(-P2) g(t), Tt < t < Tf, (261)
but from our previous results (234) we know that if the observation interval were infinite,
D(oj2)Ve S(jcj) = N(cj2) G^(jco), (262)
or
D(—p2)VEs(t) = N(~p2)gm(t), -oo <<<oo. (263)
318 4.3 Detection and Estimation in Nonwhite Gaussian Noise
Thus goo(0 corresponds to the particular solution of (261). There are also homogeneous solutions to (261):
0 = N(-p2) ghi(t\ ł = 1, 2,..., 2q. (264)
We now add the particular solution gao(t) to a weighted sum of the 2q homogeneous solutions gh{(t% substitute the result back into the integral equation, and adjust the weightings to satisfy the equation. At this point the discussion will be clearer if we consider a specific example.
Example. We consider (246) and use limits [0, T] for algebraic simplicity.
Kn(t - u) = Kn(r) = an2e-*'*1, - oo < r < oo (265)
or
0 / 4 2kan2 (266)
5"(“) =«»+ ę»•
Thus
N(w2) = 2k a „2 (267)
and
Z>(<o2) = Ř2 + k2. (268)
The differential equation (261) is
Ve(-s'V) + k2s(l)) = 2kan2g{t). (269)
The particular solution is
gut) = 2^ [-S'V) + k*s(0] (270)
and there is no homogeneous solution as
q = 0. (271)
Substituting back into the integral equation, we obtain
VEs(t) = <r„2 jT exp (~k\t - u\)g(u) du, 0 < t < T, (272)
For #(0 to be a solution, we require,
+ e + kt JJ g ~ ~ S "(U2k" duy 0 < t <T. (273)
Because there are no homogeneous solutions, there are no weightings to adjust. Integrating by parts we obtain the equivalent requirement,
0 = e'kt{jic [/(0) ~ Ü(0)1}
_ Ş + Ř-Ň,^±_ + ks(J)]j> 0 < I < T. (274)
Clearly, the two terms in brackets must vanish independently in order for g^(t) to satisfy the integral equation. If they do, then our solution is complete. Unfortunately, the signal behavior at the end points often will cause the terms in the brackets to be
Fredholm Equations of the First Kind 319
nonzero. We must add something to goo(t) to cancel the e~kt and ekit~T) terms. We denote this additional term by ga(t) and choose it so that
<Tn2 Jo exp ( k\t - u\)g6(u) du
= -j-k [-s'(O) - fe(0)]e-'“ + ^ [*'(Ď + ks(T)]e*m~T\ 0 < / < Ă. (275)
To generate an e~kt term g6{u) must contain an impulse Ńł 8(w). To generate an ĺ + ř- t) term must contain an impulse c2 8(w — T). Thus
gi(u) = Cl 8(tt) + c2 8(č - T), (276)
where
ę 5(0) - s'(0)
Cl bn2 ’
(277)
. _ ks(T) + s'(T)
2 " ęŕă ’ to satisfy (274).t Thus the complete solution to the integral equation is git) = g»it) + g6{t\ 0 <t<T.
From (153) and (154) we see that the output of the processor is
/ = Jo I'd) g(t) dt
=+cir(T)+
Thus the optimum processor consists of a filter and a sampler.
Observe that g(t) will be square-integrable only when Ci and c2 are zero the significance of this point in Section 4.3.7. Previous << 1 .. 112 113 114 115 116 117 < 118 > 119 120 121 122 123 124 .. 254 >> Next 