Re: stationary process pass through an LTI system still stationary?

From: Tim Wescott (tim_at_wescottnospamdesign.com)
Date: 12/18/04 

Date: Sat, 18 Dec 2004 09:44:25 -0800

Randy Yates wrote:

> Andor,
>
> What about startup transients? I agree in the steady-state.
>
> --RY
>
> "Andor" <an2or@mailcircuit.com> writes:
>
>
>>Tim wrote:
>>
>>"
>>Well, think about it. A stationary process is one who's parameters are
>>independent of time, although the value of the signal resulting from
>>the
>>process can be correlated to past or future values of itself.
>>
>>A time-invariant system is one that modifies a signal in a
>>time-independent way, although the output signal at any given time may
>>depend on past (or future, if you don't care about causality) values of
>>
>>the input signal.
>>
>>So you tell me -- is it still stationary, or has it acquired a time
>>dependence?
>>"
>>
>>Your explanation seems to make sense.
>>
>>However, if you have an LTI system with poles on the unit circle, one
>>can still say that it modifies the signal in a time-independent way,
>>but the output is not stationary. Can you refine your argument?
>>Regards,
>>Andor
>>
>
>
I missed Andor's post, so I'm answering it here.

You're probably thinking of the Wiener process, where you integrate
white noise starting at time t = zero. You are correct that this is not
stationary, but if you have a system that holds its output at zero until
time t = zero then the system is not time invariant. If your system
_is_ time invariant then you must have a signal that is zero for t < 0
and has energy for t > 0 -- and that's not a stationary signal.

Actually any "LTI" that is held at zero and only released at time t=0
will have an output that's non-stationary because you must be violating
one of the conditions above. To keep the overall system time invariant
you have to start things up at -infinity.

So I think I'm still right.

Now, granted, if you have an unstable or metastable system your output
signal in such a case will have infinite variance -- this is probably
why you want to hold things to zero until t = 0.

I suspect that trying to do the analysis with any sort of rigor while
allowing unstable or metastable systems would require many extra reams
of paper, but I think my assertions would hold true. 

-- 
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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