mohatta2015.pdf

THE MATH 39
correlation function f(t) and a decision variable given by
Zr = Reirr{t)r(t)dt\. (2.34)
We want to find f(t) that maximizes the SNR of z r .
We start by substituting the model for r(t) from (2.22) into (2.34, giving
z T \= Reif f*(t)[y/F s h(t)s]dt+ f f*(t)n(t)dt\
\= As + e. (2.35)
Let E denote the variance of e. Since n(t) is zero-mean, complex Gaussian and the
filtering is linear, e is a real Gaussian random variable with zero mean. The output
SNR is given by
SNR 0 = A 2 /E. (2.36)
Next, we determine E as a function of f(t). Let's take a closer look at e. We will
use the facts that for complex numbers x — a+jb and y = c+jd, Re{x*y} = ac+bd
and \x\ 2 = a 2 + b 2 . First, from (2.35),
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/
fr(t)n r {t) dt+ j fi(t)ni(t) dt = ei+ e 2 , (2.37)
where subscripts r and i denote real and imaginary parts. As a result, E becomes
|/(t)| 2 dt. (2.38)
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Observe that E depends on the energy in f(t), not its shape in time.
Now let's look at the signal power, A 2 . From (2.35),
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A 2 = E b ( Γ Re{f*{t)h(t)} dt) . (2.39)
where we've replaced E s with Eb because we are considering BPSK. Substituting
(2.38) and (2.39) into (2.36) gives
SNR 0 = pft/ΛΓοΛ
r j f { t W d t
J - ( 2 - 4 °)
It is convenient at this point to introduce a form of the Schwartz inequality, which
states
ÍRe{a*{t)b(t)} dt< J f \a{t)\ 2 dt J Í \b(t)\ 2 dt, (2.41)
where equality is achieved when a(t) = b(t). Applying the inequality to (2.40) gives
f°° }f(t)\ 2 dtr \h(t)\ 2 dt
SNR 0 < (2£ b /Aro) J -° ol - / yj ) {f{^dt (2-42)