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

**Marks:** 10M

**Year:** May 2015

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Define an ergodic process. Determine whether the stochastic process X(t) = Asin(t) + Bcos(t) is ergodic. Here A& B are normally distributed independent r.vs with zero mean and equal standard deviation

written 8.2 years ago by | • modified 8.2 years ago |

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

**Marks:** 10M

**Year:** May 2015

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written 8.2 years ago by | • modified 8.2 years ago |

**Ergodic Processes**

A random process is said to be ergodic if the time averages of the process tend to the appropriate ensemble averages.

This definition implies that with probability 1, any ensemble average of {X(t)} can be determined from a single sample function of {X(t)}.

Clearly, for a process to be ergodic, it has to necessarily be stationary. But not all stationary processes are ergodic.

**Mean Ergodic**

A stationary random process is said to be ergodic in mean if the Time average tends to the ensemble average .i.e.,

$$\bar{X_T}=\frac{1}{2T}\int_{-T}^TX(t)dt→μ=E(X(t)) \ \ as \ \ T→\infty$$

Given A, b are normally distributed with zero mean and equal standard deviation.

∴ E(A)=0, E(B)=0 $σ_A=σ_B=1$

**Given: X (t) = Asin (t) + Bcos (t)**

Ensemble average is given by:

E[X(t)] = E (Asint+Bcost)= sintE(A)+ costE(B) = 0 -------- (1)

Time- average of {X(t)}is given by:

$$(\bar{X_T} ) =\frac{1}{2T} ∫_{-T}^TX(t) dt$$

$$(\bar{X_T} ) =\frac{1}{2T} ∫_{-T}^TAsint+Bcost dt$$

$$=\frac{1}{2T} (-Acost+Bsint)_{-T}^T$$

$$ =\frac{1}{2T}.2BsinT=\frac{BsinT}T$$

Since the function sinT is bounded (|sint| ≤1) as T→∞, the limit of $(\bar{X_T} )$ will be zero.

$$∴lim_{T→∞}\frac{1}{2T} ∫_{-T}^TX(t) dt=0$$

**Since the time average $\bar{X_T}$ is equal to the ensemble average E[X(t)], {X(t)} is mean ergodic.**

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