To solve the given problem in a clear, step-by-step classical form, we will prove that a Linear Time-Invariant (LTI) system in state-space form, if asymptotically stable, is also BIBO (Bounded Input, Bounded Output) stable.

Step 1: Understanding Asymptotic Stability

An LTI system in state-space form is given by: $\dot{x}(t) = Ax(t) + Bu(t), \quad y(t) = Cx(t) + Du(t)$ Here:

• $x(t)$ is the state vector,
• $u(t)$ is the input,
• $y(t)$ is the output,
• $A$, $B$, $C$, and $D$ are constant matrices.

Asymptotic Stability means that all eigenvalues of the matrix $A$ have negative real parts. This ensures that the homogeneous solution (response to initial conditions) of the system decays to zero as $t \to \infty$.

Step 2: BIBO Stability Definition

An LTI system is BIBO stable if every bounded input $u(t)$ results in a bounded output $y(t)$. This implies: $\| u(t) \| \leq a \implies \| y(t) \| \leq b,$ where $a$ and $b$ are finite bounds.

We need to show that asymptotic stability guarantees BIBO stability.

Step 3: Bounded Output for Bounded Input

To analyze the system, we examine the state $x(t)$ and the output $y(t)$: $x(t) = e^{At}x(0) + \int_{0}^{t} e^{A(t-\tau)}Bu(\tau) \, d\tau$ $y(t) = Cx(t) + Du(t)$

1. Contribution of Initial State

The term $e^{At}x(0)$ in $x(t)$ decays to zero as $t \to \infty$ due to the asymptotic stability of $A$. Therefore, the effect of the initial state is bounded for all $t$.

2. Contribution of Input $u(t)$

For bounded input $u(t)$, we need to show that $\int_{0}^{t} e^{A(t-\tau)}Bu(\tau) \, d\tau$ remains bounded.

• Since $A$ is asymptotically stable, $e^{At}$ is exponentially decaying, meaning there exists a constant $M > 0$ such that: $\|e^{At}\| \leq M e^{-\alpha t}, \quad \text{for some } \alpha > 0.$
• The integral term becomes: $\int_{0}^{t} e^{A(t-\tau)}Bu(\tau) \, d\tau.$ Since $u(t)$ is bounded ($\|u(t)\| \leq a$), and $e^{A(t-\tau)}$ is bounded, the overall contribution of this term is also bounded.

3. Boundedness of Output $y(t)$

Finally, substituting into $y(t)$: $y(t) = C\left(e^{At}x(0) + \int_{0}^{t} e^{A(t-\tau)}Bu(\tau) \, d\tau\right) + Du(t).$ Each term on the right-hand side is bounded:

• $Ce^{At}x(0)$ is bounded since $e^{At}x(0)$ decays exponentially.
• $C \int_{0}^{t} e^{A(t-\tau)}Bu(\tau) \, d\tau$ is bounded because $u(t)$ is bounded and $e^{A(t-\tau)}$ is exponentially decaying.
• $Du(t)$ is bounded because $u(t)$ is bounded.

Therefore, $y(t)$ is bounded for all bounded $u(t)$.

Step 4: Conclusion

We have shown that if an LTI system is asymptotically stable, then its output $y(t)$ remains bounded for all bounded inputs $u(t)$. Hence, the system is BIBO stable.

Summary

• Asymptotic stability ensures the exponential decay of $e^{At}$, which bounds the state response $x(t)$.
• For bounded input $u(t)$, the boundedness of $x(t)$ leads to bounded output $y(t)$.

Do you have any questions, or would you like me to elaborate on any specific step?

5 Relative Questions

1. What is the difference between BIBO stability and asymptotic stability?
2. How does the eigenvalue condition for $A$ relate to stability in practical systems?
3. Can a system be BIBO stable but not asymptotically stable? Why or why not?
4. How does the boundedness of $u(t)$ influence the stability of $y(t)$?
5. What role does the matrix $C$ play in determining the output $y(t)$?

Tip

When proving stability, always start with the system definitions and use mathematical bounds systematically to ensure clarity and rigor.