In this topic, you study the Stable and Unstable Systems theory, definition & solved examples.
Let $x(t)$ and $y(t)$ be the input and output signals, respectively, of a system shown in Figure 1. Then the transformation of $x(t)$ into $y(t)$ is represented by the mathematical notation
$y(t) = {\mathbf{T}}x(t)$
where $\mathbf{T}$ is the operator which defined rule by which $x(t)$ is transformed into $y(t)$.
Figure 1: System with a single input and output signal.
A system is bounded-input/bounded-output (BIBO) stable if for any bounded input $x(t)$ results in the bounded output $y(t)$.
Mathematically,
if
$|x(t)| \leqslant {m_x} < \infty $
then
$|y(t)| \leqslant {m_y} < \infty $
So system is said to be stable. And where $m_x$ , and $m_y$, are finite real constants. Bounded means amplitude is finite and some examples of bounded inputs as sine function, cosine function, dc signal, etc.
Example : Determine whether or not each of the following systems are stable with input $x(t)$ and output $y(t)$.
(i) \[y(t) = tx(t)\]
(ii) \[y(t) = \frac{{dx(t)}}{{dt}}\]
(iii) \[y(t) = u[x(t)]\]
(iv) \[y(t) = \cos [x(t)]\]
Solution : (i) \[y(t) = tx(t)\]
Let
\[x(t) = 2\]
Here 2 is the dc signal which is bounded input so,
\[y(t) = 2.t\]
The output $y(t)$ is unbounded and the bounded input produces unbounded output hence system is unstable.
(ii) \[y(t) = \frac{{dx(t)}}{{dt}}\]
Let
\[x(t) = 2\]
Here 2 is the dc signal which is bounded input so,
\[y(t) = \frac{{d2}}{{dt}} = 0\]
The output $y(t)$ is bounded. Let
\[x(t) = u(t)\]
Here unit step function is bounded input so,
\[y(t) = \frac{{du(t)}}{{dt}} = \delta (t)\]
The unit impulse function $\delta (t)$ is unbounded output and the bounded input produces unbounded output hence the system is unstable.
(iii) \[y(t) = u[x(t)]\]
Let
\[x(t) = 2\]
Here 2 is the dc signal which is bounded input so,
\[y(t) = u[2] = 1\]
The output $y(t)$ is bounded and the bounded input produces bounded output hence system is stable.
(iv)\[y(t) = \cos [x(t)]\]
Let
\[x(t) = 2\]
Here 2 is the dc signal which is bounded input so,
\[y(t) = \cos (2)\]
The output $y(t)$ is bounded and the bounded input produces bounded output hence system is stable.
