A.3 SNR in second-order pole regime

Substituting Eqs. (528View Equation) and (527View Equation) into Eq. (526View Equation), we obtain that
2 Ω ∫ ∞ dy Ω 1 σ2 = ---0- ---2----2-2--------2-= √--0--∘----------------√-------------------, (556 ) π Ωq −∞ (4y − λ ) + 1 + 2s 2Ωq (1 + 2s2 + λ4)( 1 + 2s2 + λ4 − λ2)
where
ν Λ Ω0 y = ---, λ = --, s = ---ξtech. (557 ) Ωq Ωq Ωq
If λ = 0 (the sub-optimal case), then
2 --Ω0-- 2 −3∕4 σ = √ -- (1 + 2s ) . (558 ) 2Ωq
The maximum of (556View Equation) in λ corresponds to
∘ -------- 2 1 +-2s2 λ = 3 , (559 )
and is equal to
( 2)− 3∕4 σ2 = -√Ω0--- 1 +-2s- . (560 ) 2 2 Ωq 3
In both cases (558View Equation) and (560View Equation), we should minimize the expression
2 −3∕4 [ ( 2 )3∕4]− 1 (1-+-2s-)---- = Ωq 1 + 2Ω-0ξ2 (561 ) Ωq Ω2q tech
in Ωq. It is easy to see that the optimal Ωq is equal to
Ωq = Ω0 ξtech, (562 )
which gives the SNR (530View Equation) for the optimal case (529View Equation) and (531View Equation) – for the sub-optimal one Λ = 0.


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