written 8.4 years ago by | • modified 8.4 years ago |
Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis
Marks: 10M
Year: Dec 2014
written 8.4 years ago by | • modified 8.4 years ago |
Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis
Marks: 10M
Year: Dec 2014
written 8.4 years ago by | • modified 8.4 years ago |
To prove X(t) is WSS random process
i.e. E(X(t))=μ
& $E(X(t)×X(t-τ))=R(τ$)
Given: X(t) = 10 cos (100t+6)
$$∴ E(X(t))=10E(cos(100t+θ)) $$
$$=10E(cos100t.cosθ-sin100t.sinθ)$$
$$E(X(t)) =10[cos100t×E(cosθ)-sin100t×E(sinθ)]--- (1)$$
Given θ is uniformly distributed $∴ f_θ (θ)=\frac{1}{2π}$
$$∴ E(cosθ)=∫_0^{2π}cosθ.f_θ (θ)dθ$$
$$=∫_0^2πcosθ.\frac1{2π} dθ$$
$$E(cosθ)=0 -------(2)$$
Similarly
$$E(sinθ)=∫_{0}^{2π}sinθ.f_θ (θ)dθ$$
$$=∫_0^{2π}sinθ.\frac1{2π} dθ$$
$$E(sinθ)=0---(3)$$
From eqn(1), eqn(2) and eqn(3) we get
$$∴ E(X(t))=0$$
R(τ)=$E(X(t)×X(t+τ)) \\ =E(10 cos(100t+θ).10cos(100t+100τ+θ))\\ =\frac{100}2 E[2 cos(100t+θ).cos(100t+100τ+θ)]\\ =50E[cos(200t+100τ+θ)+cos(100τ))] =50[∫_0^{2π}cos(200t+100τ+θ).\frac1{2π} dθ+ ∫_0^{2π}cos(100τ).\frac1{2π} dθ] $
$R(τ)=50cos(100τ)$
Correlation Ergodic
A stationary random process is said to be correlation ergodic if the Time average tends to the ensemble average .i.e.,
$$(\bar{R_T }) =\frac1{2T} ∫_{-T}^T X(t) X (t+τ) dt$$
$$(\bar{R_T} ) =\frac1{2T} ∫_{-T}^T100 cos(100t+θ)cos(100t+100τ+θ) dt$$
$$(\bar{R_T} ) =\frac{25}T ∫_{-T}^Tcos(200t+100τ+θ)+cos(100τ) dt$$
$$(\bar{R_T}) =\frac{25}T ∫_{-T}^Tcos(200t+100τ+θ)dt+\frac{25}T ∫_{-T}^Tcos(100τ) dt$$
$$(\bar{R_T} ) =\frac{25}T ∫_{-T}^Tcos(200t+100τ+θ)dt+50cos(100τ) dt$$
$$Now \lim_{T→∞}R_T=50cos{100τ}=R(τ)$$
∴ {X(t)} is correlation ergodic.