Correct Answer - Option 3 : Unstable
Stable and Unstable system:
The system is sold to be stable only when the bounded output for the input.
For a bounded input if output is unbounded then system is said to be UNSTABLE.
Condition:
\(\mathop \smallint \nolimits_{ - \infty }^\infty \left| {h\left( t \right)} \right|dt < \infty \) or finite.
Where h(t) is impulse response of system.
Suppose input x(t) = cos 4t
\(y\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^t {\cos ^2}4\tau dt = \mathop \smallint \nolimits_{ - \infty }^t \left( {\frac{{1 + \cos 8\tau }}{2}} \right)d\tau \;\)
\(y\left( t \right) = [\left. {\frac{1}{2}t\tau } \right|_{ - \infty }^t + \frac{1}{2}\sin \left. {\frac{{8\tau }}{8}} \right|_{ - \infty }^t\)
Since y(t) → ∞, it is unbounded of P so this is an unstable system.
TRICK:
→ check for stability: try to find at least one input value for which output will be unbounded.
e.g. \(y\left( t \right) = \frac{{{e^x}\left( {t - 5} \right)}}{{t - 5}}\) at t = 5 y(t) → ∞ - unbounded.
