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..
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].
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..
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 $$t-\tau\leq 0$$. :tongue: I'm sorry...