Signals and Systems, system properties

[freydawg56]

I know that y(t) = t x(t) is unstable, for bounded inputs yielding unbounded outputs, but would

y(t) = x(t)/ t also be unstable? when t is going in the negative direction? please help. test Monday.

 

[kovachattack]

As t approaches zero, you get unbounded outputs from bounded inputs. So, it is unstable.

Also, please somebody correct me if I am wrong, but I believe y(t) = t x(t) does in fact have a bounded output for a bounded input. If you don't let x go to infinity, then y will also not go to infinity, and is therefore bounded and stable.

 

[CEL]

As t approaches zero, you get unbounded outputs from bounded inputs. So, it is unstable.

Also, please somebody correct me if I am wrong, but I believe y(t) = t x(t) does in fact have a bounded output for a bounded input. If you don't let x go to infinity, then y will also not go to infinity, and is therefore bounded and stable.

No, even with a bounded x, y will grow with time, so it is unbounded.
As, for the original question, x(t)/t is unstable for t equal zero, so no real system can have such characteristics.