mohatta2015.pdf
signal processing from power amplifier operation control point of view
signal processing from power amplifier operation control point of view
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THE MATH 39<br />
correlation function f(t) and a decision variable given by<br />
Zr = Reirr{t)r(t)dt\. (2.34)<br />
We want to find f(t) that maximizes the SNR of z r .<br />
We start by substituting the model for r(t) from (2.22) into (2.34, giving<br />
z T \= Reif f*(t)[y/F s h(t)s]dt+ f f*(t)n(t)dt\<br />
\= As + e. (2.35)<br />
Let E denote the variance of e. Since n(t) is zero-mean, complex Gaussian and the<br />
filtering is linear, e is a real Gaussian random variable with zero mean. The output<br />
SNR is given by<br />
SNR 0 = A 2 /E. (2.36)<br />
Next, we determine E as a function of f(t). Let's take a closer look at e. We will<br />
use the facts that for complex numbers x — a+jb and y = c+jd, Re{x*y} = ac+bd<br />
and \x\ 2 = a 2 + b 2 . First, from (2.35),<br />
oo<br />
/>oo<br />
/<br />
fr(t)n r {t) dt+ j fi(t)ni(t) dt = ei+ e 2 , (2.37)<br />
where subscripts r and i denote real and imaginary parts. As a result, E becomes<br />
|/(t)| 2 dt. (2.38)<br />
/<br />
-oo<br />
Observe that E depends on the energy in f(t), not its shape in time.<br />
Now let's look at the signal power, A 2 . From (2.35),<br />
OO<br />
A 2 = E b ( Γ Re{f*{t)h(t)} dt) . (2.39)<br />
where we've replaced E s with Eb because we are considering BPSK. Substituting<br />
(2.38) and (2.39) into (2.36) gives<br />
SNR 0 = pft/ΛΓοΛ<br />
r j f { t W d t<br />
J - ( 2 - 4 °)<br />
It is convenient at this point to introduce a form of the Schwartz inequality, which<br />
states<br />
ÍRe{a*{t)b(t)} dt< J f \a{t)\ 2 dt J Í \b(t)\ 2 dt, (2.41)<br />
where equality is achieved when a(t) = b(t). Applying the inequality to (2.40) gives<br />
f°° }f(t)\ 2 dtr \h(t)\ 2 dt<br />
SNR 0 < (2£ b /Aro) J -° ol - / yj ) {f{^dt (2-42)