Proof of Impulse response of a BIBO stable system

Question

I was wondering if anyone here could expand upon or provide the proof of the boxed theorem which I have shown in the image below?

Any help would be greatly appreciated.

Answer 1

To be concrete (and to side step many technicalities), I will assume the system is a real, finite dimensional, LTI system described by the usual $(A,B,C,D)$ matrices, that is, $\dot{x} = Ax+Bu, y = Cx+Du$.

If we let $G(t) = Ce^{At}B+D$ (impulse response), and $\hat{G} = {\cal L}G$ (frequency domain), then we have that the system is BIBO stable iff $G$ is integrable iff $\hat{G}$ is analytic on $\bar{\mathbb{C}}_+ = \{z \in \mathbb{C}| \operatorname{re} z \ge 0 \}$.

Suppose the system is BIBO stable and the input is some continuous $u$, then the zero initial state response is given by $y_u(t) = \int_0^t G(t-\tau) u(\tau) d \tau = \int_0^t G(\tau) u(t-\tau) d \tau$.

If $u(t) = \alpha$ for all $t\ge 0$, then $y_u(t) = (\int_0^t G(\tau) d \tau)\alpha$. Since the system is BIBO stable, $G$ is integrable, and we have $\lim_{t \to \infty} y_u(t) = (\int_0^\infty G(\tau) d \tau)\alpha$. Since the system is BIBO stable, $\hat{G}$ is analytic on $\bar{\mathbb{C}}_+$ and so $\int_0^\infty G(\tau) d \tau = ({\cal L}G)(0) = \hat{G}(0)$, and so $\lim_{t \to \infty} y_u(t) = \hat{G}(0) \alpha$.

If $u(t) = e^{i \omega_0 t}$ for all $t\ge 0$, then $y_u(t) = e^{i\omega_0 t} \int_0^t G(\tau) e^{-i\omega_0 \tau} d \tau$. Let $z(t) = e^{i\omega_0 t} \int_0^\infty G(\tau) e^{-i\omega_0 \tau} d \tau = \hat{G}(i \omega_0)e^{i\omega_0 t}$ (using the fact that $\hat{G}$ is analytic on $\bar{\mathbb{C}}_+$), then we have $\|y_u(t)-z(t)\| = \| e^{i\omega_0 t} \int_t^\infty G(\tau) e^{-i\omega_0 \tau} d \tau \| \le \int_t^\infty \|G(\tau) \| d \tau$, and since $G$ is integrable, we have $\lim_{t \to \infty}\int_t^\infty \|G(\tau) \| d \tau = 0$. Hence it follows that $\lim_{T \to \infty} \sup_{t \ge T} \|y_u(t)-z(t)\| = 0$.

If $u(t) = \sin \omega_0 t = {1 \over 2 i} (e^{i\omega_0 t} -e^{- i\omega_0 t} )$, we can use linearity to see that $y_u$ approaches (in the above sense) the function $z(t) = {1 \over 2 j} (\hat{G}(i \omega_0)e^{i\omega_0 t} - \hat{G}(-i \omega_0)e^{-i\omega_0 t})$. Since the system is real, we have $\hat{G}(-i \omega_0) = \overline{\hat{G}(i \omega_0)}$, and so $z(t) = \operatorname{im} (\hat{G}(i \omega_0)e^{i\omega_0 t}) = \operatorname{im} (|\hat{G}(i \omega_0)|e^{i(\omega_0 t + \operatorname{arg} \hat{G}(i \omega_0)})= |\hat{G}(i \omega_0)| \sin (\omega_0 t + \operatorname{arg} \hat{G}(i \omega_0))$.

Note that I need to assume that we are dealing with the zero initial state response, it doesn't follow from the assumptions that the system is internally stable (just add some unstable, unobservable dynamic to the system to see why).