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cad作业题 随机信号分析基础作业题作业(大部分)--桂电

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D(X )E( =X2 )− E2 X( )=σx2 π6、知随机已量变与 XY 有=,E (X) ,1= E() Y3, = ( XD )4,= DY( )16, =ρ XY 0 .5, 令 U= +3X YV= X,− 2Y , 试求 E( U)、(EV、 D(U ))、 D( V)和 oC(vU V,) 。解题路:思察考机随变量函的数数字特征协方:差 o (CvX , Y )=E X(Y) −EX ()⋅E (Y) 相系数关ρ:X= Y Cv(oX , Y) (DX ) D(Y )E aX (+ bY = aE )(X )+b EY() D a(X + Yb= ) a2D (X +)b 2D Y() + 2ab oCv(X , Y) :E (U解)= (3XE +Y )= 3E (X )+EY(= )3×1 +3 =E 6( )V= E(X−Y 2)= E( )X− 2E (Y )=1−× 32= −5 10 =ρY CXo(vX ,Y )= .05 ,D (X)=,4 D(Y) =1 6,D (X) D(Y )∴ ovC X(, Y)= 4又Co (Xv, Y )=E XY ()−E (X)⋅E Y( )∴E(Y X)= D7 =(U) (3 XD + =Y ) 32 D(X +) 21D( )+Y2××3 1× oC(vX Y ),7=6 (= DV) ( DX −2= Y)D X )((− 2+) 2D ( Y)2+×1 ×−()2C× vXo( , Y= ) 52 E( X2 )=D X() E+ (2X )=+ 412 =5, E(Y2 )= D( )Y+ E2 Y(= )61 +23 =2 5∴ E(V U) =E[(3 X +Y) (X −2 Y)]= E 3(2X− 5YX 2− Y2 ) =3E (2)X−E5 X( Y)−2 (EY2 =)− 7

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