Signals and Systems: Stable and Unstable System

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Stable System or BIBO Stable System

A system is called a BIBO (bounded input bounded output) stable system or simply stable system, if and only if every bounded input produces a bounded output. The output of a stable system does not change unreasonably.

The stability of a system indicates the usefulness of the system. The stability of a system can be determined from the impulse response of the system. The impulse response of the system is nothing but the output of the system for a unit impulse input.

If the impulse response of the system is absolutely integrable for a continuoustime system or absolutely summable for a discrete time system, then the system is a stable system.

Let an input signal x(t) is bounded signal, i.e.,

$$\mathrm{|x(t)| \: \lt \: k_x \:\lt\: \infty \:\: for \: - \infty \: \lt \: t \: \lt \: \infty}$$

Where, k_x is a positive real number. Then, if

$$\mathrm{|y(t)| \: \lt \: k_y \:\lt \: \infty}$$

That is, the output of the system y(t) is also bounded, then the system is called BIBO stable system.

Unstable System

If a system does not satisfy the BIBO stability condition, the system is called the unstable system. Therefore, for a bounded input, it is not necessary that the unstable system produces a bounded output. Thus, we can say that a system is unstable even if one bounded input generates an unbounded output.

Solved Example

Find whether the given systems are stable or unstable −

$$\mathrm{y(t) \:=\: e^{x(t)}\: \:for\: |x(t)|\: \leq \: 6}$$

$\mathrm{h(t) \:=\: \frac{1}{RC}e^{\frac{-t}{RC}}u(t)}$

$$\mathrm{y(t) \:=\: (t \:+\: 7)u(t)}$$

$$\mathrm{}$$h(t) \:=\: e^{3t}u(t)

Solution

The output of the given system is,

$$\mathrm{y(t) \:=\: e^{x(t)} \:\:for\: |x(t)|\: \leq \: 6}$$

The input of the given system is bounded, i.e.,

$$\mathrm{|x(t)| \: \leq \: 6}$$

Therefore, to the system be stable, the output must be bounded.

For the given system, the output y(t) becomes,

$$\mathrm{e^{-6} \: \leq\: y(t)\: \leq\: e^6}$$

Thus, the output y(t) is also bounded. Hence the system is stable.

The given system is

$$\mathrm{h(t) \:=\: \frac{1}{RC}e^{\frac{-t}{RC}}u(t)}$$

For stability of the system,

$$\mathrm{\int_{-\infty }^{\infty}\: \left | h(t) \right |dt \:\lt \: \infty }$$

For the given system,

$$\mathrm{\int_{-\infty }^{\infty}\: \left | h(t) \right |dt \:=\: \int_{-\infty }^{\infty}\:\left | \frac{1}{RC} e^{\frac{-t}{RC}}u(t) \right |dt \:=\: \int_{0}^{\infty}\:\left | \frac{1}{RC}e^{\frac{-t}{RC}} \right |dt } \:=\: 1 \: \lt\: \infty $$

Therefore, the given system is a stable system.

The given system is

$$\mathrm{y(t) \:=\: (t \:+\: 7)u(t)}$$

$$\mathrm{\Rightarrow\: y(t) \:=\: (t \:+\: 7);\: t \:\geq\: 0}$$

Hence,

for t → ∞; y(t) → ∞

Thus, the output of the system increases without any bound. Therefore, the given system is an unstable system.

The output of the given system is

$$\mathrm{h(t) \:=\: e^{3t}u(t)}$$

For stability of the system,

$$\mathrm{\int_{-\infty }^{\infty}\: \left | h(t) \right |dt \:=\:\int_{-\infty }^{\infty}\left | e^{3t}u(t) \right |dt \:=\: \int_{0 }^{\infty}\left | e^{3t} \right |dt}$$ $$\mathrm{\Rightarrow \: \int_{-\infty }^{\infty}\: \left | h(t) \right |dt \:=\: \left [ \frac{e^{3t}}{3} \right ]_{0}^{\infty } \:=\: \left [ \frac{e^{\infty }}{3} \:-\: \frac{e^{0}}{3}\right ]=\infty }$$

The impulse response of the given system is not absolutely integrable. Therefore, the given system is an unstable system.