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Causality with time invariance

by tutumar
Tags: causality, invariance, time
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tutumar
#1
Jun30-12, 01:29 PM
P: 2
Assume u:R[itex]\rightarrow[/itex] C^n and define shift operator S([itex]\tau[/itex]) with

S([itex]\tau[/itex])u(t)=u(t-[itex]\tau[/itex])

and truncation operator P([itex]\tau[/itex]) with

P([itex]\tau[/itex])u(t)=u(t) for t[itex]\leq[/itex][itex]\tau[/itex] and 0 for t>[itex]\tau[/itex]

Then P([itex]\tau[/itex])S([itex]\tau[/itex])=S([itex]\tau[/itex])P(0) for every [itex]\tau[/itex]>=0.

Can someone please prove last statement..
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HallsofIvy
#2
Jun30-12, 02:39 PM
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Looks like pretty direct computation. If u(t) is any such function, then what is[itex]SD(\tau)u[/itex]? What is [itex]P(\tau)S(\tau)u[/itex]? Then turn around and find [itex]S(\tau)P(0)u[/itex].
tutumar
#3
Jun30-12, 02:54 PM
P: 2
Yes, I tried that, and it just doesn't fit..

P([itex]\tau[/itex])S([itex]\tau[/itex])u(t)=P([itex]\tau[/itex])u(t-[itex]\tau[/itex])=u(t-[itex]\tau[/itex]) if t-[itex]\tau[/itex]<=[itex]\tau[/itex] and 0 for t-[itex]\tau[/itex]>[itex]\tau[/itex]

S([itex]\tau[/itex])P(0)u(t)=S([itex]\tau[/itex])u(t) for t<=0 and 0 otherwise=u(t-[itex]\tau[/itex]) if t<=0 and 0 otherwise..

Well, something's got to be wrong here, but I can't see what..

Einj
#4
Jul1-12, 02:58 AM
P: 305
Causality with time invariance

I think your last equation is wrong. As, if we have:

$$P(0)u(t)=u(t) \mbox{ if } t\leq 0 \mbox{ and } 0 \mbox{ otherwise }$$

than:

$$S(\tau)P(0)u(t)=u(t-\tau) \mbox{ if } t-\tau\leq 0 \mbox{ and } 0 \mbox{ if } t-\tau>0$$

Still, I'm not able to prove the statement as in the first case you have $$t-\tau\leq\tau$$ and in this case there is $$t-\tau\leq 0$$. I'm sorry...


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