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2PSK- 同 步 检 波012 2( 0) exp( ) / 2 )pe1= p v< = ∫ −(v−a σ dv−∞ 2πσ1= erfc( r )2∞12 2( 0) exp( ) /2 )∫pe2= pv> = −(v+a σ dv0 2πσ1= erfc( r )21pe = p(1) pe1 + p(0) pe2= erfc( r)21 − r在 大 信 噪 比 下 , pe≈ e2 π r2011-9-11 76
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