A random process X(t) is applied to a network with impulse response h(t)=te−α(t), while α is a positive constant. The cross correlation of X(t) with the output Y(t) is RXY(τ)=te−ατu(τ). For τ≤0;
A
|RXY(τ)|≤√RX(0).RY(0)
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B
RY(α)=e−α|τ|(1+ατ)4α3
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C
Output power is 4α2
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D
If mean of X(t)=α then output mean 1/α.
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Solution
The correct option is BRY(α)=e−α|τ|(1+ατ)4α3 RY(τ)=RXY(τ).h(−τ) =∫∞−∞u(τ+r)(τ+x)e−α(+x).xe−αxdx =e−ατ∫∞−∞x(τ+x)e−2xαdx=e−ατ∫∞−∞(τx+x2)e−2xαdx =eαx(1−ατ)4α3 RY(τ)=e−α|τ|(1+∞)4α3 h(t)↔H(s) te−αtu(t)↔1(s+α)2⇒H(0)=1α2 My=H(0).Mx =1α2.α=1α ⇒{E[X(t).Y(t+τ)]}2≤E[X2(t)]×E[Y(t+τ)]←Sctwartz unequality theorem ∴|RXY(τ)|≤√RX(0).RY(0)