Signals and Systems, system properties

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SUMMARY

The discussion centers on the stability of the systems defined by the equations y(t) = t x(t) and y(t) = x(t)/t. It is established that y(t) = t x(t) can be considered stable under certain conditions where x(t) does not approach infinity. However, as t approaches zero, both systems yield unbounded outputs from bounded inputs, confirming their instability. The consensus is that y(t) = x(t)/t is inherently unstable due to its behavior at t equal to zero.

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freydawg56
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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.
 
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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.
 
kovachattack said:
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.
 

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