Digital Signal Processing - Unstable Systems


Unstable systems do not satisfy the BIBO conditions. Therefore, for a bounded input, we cannot expect a bounded output in case of unstable systems.


a) $y(t) = tx(t)$

Here, for a finite input, we cannot expect a finite output. For example, if we will put $x(t) = 2 \Rightarrow y(t) = 2t$. This is not a finite value because we do not know the value of t. So, it can be ranged from anywhere. Therefore, this system is not stable. It is an unstable system.

b) $y(t) = \frac{x(t)}{\sin t}$

We have discussed earlier, that the sine function has a definite range from -1 to +1; but here, it is present in the denominator. So, in worst case scenario, if we put t = 0 and sine function becomes zero, then the whole system will tend to infinity. Therefore, this type of system is not at all stable. Obviously, this is an unstable system.