Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis

Marks: 10M

Year: Dec 2014

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.