sM−ASK(t)=Ac∑k=−∞∞aIkp(t−kTs)cos(ωct)aIk={0,A,+2A,...}
yR(t)=AR∑k=−∞∞aIkp(t−kTs)+nI(t)aIk={0,A,2A,...M−1}mIk=0,ATs,2ATs,...Es=Esc2→1M⋅Q(|VT−m|σ2)M−2→1M⋅2Q(|VT−m|σ2)
|VT−m|=ATs2σ2=η2TsPs≃2×1M⋅Q(|VT−m|σ2)+∑i=1M−21M⋅2Q(|VT−m|σ2)=2×1M⋅Q(|VT−m|σ2)+(M−2)1M⋅2Q(|VT−m|σ2)=2⋅M−1MQ(|VT−m|σ2)=2⋅M−1MQ(ATs2η2Ts)=2⋅M−1MQ(A2Ts24η2Ts)=2⋅M−1MQ(A2Ts2η)
Es=∑i=0M−1pr(M)⋅Ei=∑i=0M−11M⋅Ei=1M∑i=0M−1EiEi={0,A2Ts,4A2Ts,9A2Ts,..}Es=1M∑i=0M−1Ei=1M∑i=0M−1i2A2Ts=1MA2Ts∑i=0M−1i2∑i=1ni2=n(n+1)(2n+1)6→∑i=0M−1i2=(M−1)M(2(M−1)+1)6=(M−1)M(2M−1)6
∑i=0M−1i2=(M−1)M(2M−1)6Es=1MA2Ts∑i=0M−1i2=1MA2Ts⋅(M−1)M(2M−1)6=A2Ts6(M−1)(2M−1)Es=A2Ts2⋅3(M−1)(2M−1)↔A2Ts2=3Es(M−1)(2M−1)Ps=2⋅M−1MQ(A2Ts2η)⇒Ps=2⋅M−1MQ(3(M−1)(2M−1)Esη)
Comprobación:
4−ASK→Ps≃32Q(Es7η)Ps=2⋅M−1M|M=4Q(3(M−1)(2M−1)Esη)|M=4=32Q(33⋅7Esη)=32Q(17Esη)→c.q.d.
