推 aj0938:看懂了，感謝 12/16 08:35
※ 引述《aj0938 ()》之銘言：
: Let X(t) be a zero-mean, stationary, Gaussian process with autocorrection
: function Rx(τ). The process is applied to a square-law device,which is
: 2
: defined by the input-output relation Y(t) = X(t)
: Find the autocovariance function of Y(t) (i.e.,in terms of Rx(τ))?
^^^^^^^^^^^^^^
: 2 2
: 這題算一算需要算到E[X(t1)X(t2)]就卡住不會算了
: 請高手指點一下，感謝
Cov[Y(t1),Y(t2)]
= E[Y(t1)Y(t2)] - E[Y(t1)]E[Y(t2)]
2 2 2 2
= E[X(t1)X(t2)] - E[X(t1)]E[X(t2)]
2 2 2
= E[X(t1)X(t2)] - [R(0)]
2 2
X(t1) ~ Gauss(0,a ) where a = R(o)
2 2
X(t1),X(t2) ~ jointGauss(0,0,a, a ,ρ)
Cov(X(t1),X(t2)) E[X(t1)X(t2)] R(τ)
and ρ = -------------------- = --------------- = -------
R(0) R(0) R(0)
the MGF of X(t1),X(t2) is
1 2 2 2 2 2
M (t1,t2) = exp---[a t1 + a t2 + 2ρa t1t2]
X1,X2 2
(by Tayler Expansion)
2 2 2 2
a t1 a t2 2
= [1+ ----- + ... ][1 + ------ + ...][1 + ρa t1t2 + ...]
2 2
2 2
we want to determine the value E[X(t1)X(t2)]
相當於算
2
δ M(t1,t2)
-------------- |
δt1δt2 | t1=t2=0
2 2 1 2 2 2 1 2 2
比較係數 E[X(t1)X(t2)] = ---ρa a (2)(2) + ----(2)(2)a a
2 2!2!
2 2 2 2 2
= 2ρa a + a a
2
R(τ) 2 2
= 2(--------)R(0) + R(0)
2
R(0)
2 2
= 2R(τ) + R(0)
2 2 2
∴ Cov(Y(t1),Y(t2))= E[X(t1)X(t2)] - R(0)
2 2 2
= [2R(τ) + R(0)] - R(0)
2
= 2R(τ)
--
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※ 編輯: ting301 來自: 118.160.68.40 (12/15 23:47)