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

ISBN 0-471-09517-6

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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 - Ø) (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) = Ã (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 - è) 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\ ³ = 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"(“) =«»+ ê»•

Thus

N(w2) = 2k a „2 (267)

and

Z>(<o2) = Ø2 + 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}

_ ª + Ø-Ò,^±_ + 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-'“ + ^ [*'(Ï + ks(T)]e*m~T\ 0 < / < Ã. (275)

To generate an e~kt term g6{u) must contain an impulse Ñ³ 8(w). To generate an å + ø- t) term must contain an impulse c2 8(w — T). Thus

gi(u) = Cl 8(tt) + c2 8(è - T), (276)

where

ê 5(0) - s'(0)

Cl bn2 ’

(277)

. _ ks(T) + s'(T)

2 " êàã ’ 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.

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