SInce not many people did this one i thought i should give it a try.

Stable System

I remember her saying that a system is stable if and only if bounded inputs yield bounded outputs, or which inputs lead into outputs that do not diverge.

$ \,\ y(t) = e^{x(t)} $

To prove that this system is stable we need to prove that for all bounded inputs there will be a bounded outputs. So what we do is to let B be a random positive number, and let x(t) be bounded by B because x(t) is the input. That is to say:

$ \,\ |x(t)| < B $ or $ \,\ -B < x(t) < B $

which means that

$ \,\ e^{-B} < |y(t)| < e^B $

and now you see that y(t) is also bounded which means that the system is stable.

Unstable System

If a system is not stable then it is unstable, which means that bounded inputs lead to unbounded outputs.

$ \,\ y(t) = te^{x(t)} $

It is a little easier to prove that a system is not stable because you need to find just one bounded input that proves that the output that is unbounded. To prove that this system is unstable i will make

$ \,\ x(t) = 0 $

which means

$ \,\ y(t) = te^0 = t $

now it proves that it is unbounded because no matter what finite constant we pick, $ \,\ |y(t)| $ will exceed that constant by t.

Hope that helps

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang