# Exponential signal Stable?

1. Mar 7, 2014

### shawrix

Is exponential function e to the power x[t] stable? My book uses BIBO and says its stable but for -x[t] it says its not stable. Is my book wrong?

2. Mar 7, 2014

### MathematicalPhysicist

Yes, it does seem that exp(+-x(t)) is bounded whenever x(t) is bounded.

3. Mar 7, 2014

4. Mar 7, 2014

### MathematicalPhysicist

There's no difference between the two.

If $$|x[t]| \leq M$$ then $$0<e^{-x[t]}< e^M$$ thus $$|e^{-x[t]}|<e^M< \infty$$

So you still have BIBO criterion satisified.

5. Mar 7, 2014

### shawrix

I have two books as reference and they both say e to the power -x[n] is unstable. They take x[n] and y[n] in terms of impulse and impulse response and prove that for n=0 we have output e^-1 and for n/=0 it is 1. Then it uses the bibo stability condition for causal and stable system and prove that the system never converges and hence it is unstable.

Last edited: Mar 7, 2014
6. Mar 7, 2014

### shawrix

Yes :tongue: i finally proved it myself, as t-> infinite the output will also become bounded ie 1. Both books are wrong... :yuck:

Last edited: Mar 7, 2014
7. Mar 7, 2014

### MathematicalPhysicist

I am not sure I understand. BIBO criterion says that is you have bounded input then your output will also be bounded, as you've seen what I wrote, from bounded input also the output of e^x(t) is bounded.

BTW in this case it doesn't matter if your system is discrete or in the continuum, either way the same BIBO condition is satisified. The difference is that the domain of the input in one is the natural numbers and on the other is real numbers.

8. Mar 7, 2014

### shawrix

Wait, Can we apply BIBO for non-linear functions like this one?

Ps i have edited my prev post, i mentioned incorrectly what i had found.

Last edited: Mar 7, 2014