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Prove that if input LTI system is WSS the output is also WSS. What is Ergodic process?

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

Marks: 10M

Year: Dec 2014

1 Answer
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Linear Time Invariant System:

Linear systems: Let T be a continuous – time system which is at rest – i.e. all of its energy storage elements are devoid of any stored energy. Let an input $x_1(t)$ to T give rise to an output and an input to T give rise to an output . Then the system is said to be linear, if for any pair of arbitrary constants $a_1$ and $a_2$ an input $[a_1x_1(t)+a_2x_2(t)]$ to T gives rise an output $[a_1y_1(t)+a_2y_2(t)]$.

Time Invariant Systems: Let be the response of a continuous –time system T to an arbitrary input signal x(t). Then the system is said to be time invariant, if for any value of the real constant $\tau$ it gives a response of $y(t-\tau)$ for an input of $x(t-\tau)$.

We have the output

$$y(t)=h(t)*x(t)=\int_{\infty}^{\infty}h(\tau) x (t-\tau)d\tau$$

$$\therefore E(y(t))=\int_{\infty}^{\infty}h\tau E(X(t-\tau))d\tau$$

$$=\mu_x\int_\infty^{\infty}h(\tau)d\tau....[since \ \ \ \{x(t)\}] \ \ is \ \ WSS$$

⇒ E(y(t)) is constant. ........(i)

Now, the autocorrelation of the output process is given by

[Since {x(t)} is WSS]

Now the RHS is a function of $(t_1-t_2).$ Therefore the LHS is also a function only of $(t_1-t_2).$ Hence $R_{yy}(t_1,t_2)$ is a function of only Hence the output {Y(t)} is also WSS.

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.

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