How to prove for a system with rational stable transfer function, the output is square integrable?

Question

I want to know for a system with rational stable transfer function, i.e. H(jw)=1/[(a1+jw)(a2+jw)...(an+jw)] (a1,a2,..,an>0), why a square integrable (L2 integrable) input must generate a square integrable output? Please help me. Thanks very much.

Answer 1

If you consider a minimal state space representation of $G$ $$\dot{x}=Ax+Bu\\y=Cx+Du$$ then $A$ must be a stable (Hurwitz) matrix for stable $G$.

If $u\in\mathcal{L}_2$ and $A$ is stable then $x\in\mathcal{L}_2$

Proof: If $A$ is stable there exists $P=P^T>0$ such that $$PA+A^TP=-I$$ Define now the nonnegative function $V=x^TPx$. Then $$\dot{V}=-\|x\|^2+2x^TPBu$$ Using the inequality $$2x^TPBu\leq \frac{1}{2}\|x\|^2+2\|PB\|^2\|u\|^2$$ we obtain $$\dot{V}\leq -\frac{1}{2}\|x\|^2+2\|PB\|^2\|u\|^2$$ Integrating over $[0,t]$ we have $$-V(0)\leq V(t)-V(0)\leq-\frac{1}{2}\int_0^{t}{\|x(s)\|^2ds}+2\|PB\|^2\int_0^t{\|u(s)\|^2ds}$$ that yields in the limit $t\rightarrow\infty$ $$\int_0^{\infty}{\|x(s)\|^2ds}\leq 4\|PB\|^2\int_0^{\infty}{\|u(s)\|^2ds}+2V(0).$$ The square integrability of $y$ follows then directly from $y=Cx+Du$.

Edit: An alternative path would be a frequency response approach. The $L_2[0,\infty)$ space is linearly isomorphic to $\mathcal{H}_2$ the space of analytic in $Re(s)>0$ functions. As $u(t)\in L_2[0,\infty)$ then $U(s)\in\mathcal{H}_2$. Combining this with the fact that $G(s)\in\mathcal{H}_2$ we have $Y(s)=G(s)U(s)\in\mathcal{H}_2$ and therefore $y\in L_2[0,\infty)$.