## Causality with time invariance

Assume u:R$\rightarrow$ C^n and define shift operator S($\tau$) with

S($\tau$)u(t)=u(t-$\tau$)

and truncation operator P($\tau$) with

P($\tau$)u(t)=u(t) for t$\leq$$\tau$ and 0 for t>$\tau$

Then P($\tau$)S($\tau$)=S($\tau$)P(0) for every $\tau$>=0.

Can someone please prove last statement..
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 Recognitions: Gold Member Science Advisor Staff Emeritus Looks like pretty direct computation. If u(t) is any such function, then what is$SD(\tau)u$? What is $P(\tau)S(\tau)u$? Then turn around and find $S(\tau)P(0)u$.
 Yes, I tried that, and it just doesn't fit.. P($\tau$)S($\tau$)u(t)=P($\tau$)u(t-$\tau$)=u(t-$\tau$) if t-$\tau$<=$\tau$ and 0 for t-$\tau$>$\tau$ S($\tau$)P(0)u(t)=S($\tau$)u(t) for t<=0 and 0 otherwise=u(t-$\tau$) if t<=0 and 0 otherwise.. Well, something's got to be wrong here, but I can't see what..

## 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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