
#1
Jun3012, 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.. 



#2
Jun3012, 02:39 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,902

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].




#3
Jun3012, 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.. 



#4
Jul112, 02:58 AM

P: 235

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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