Is it true that a rational transfer function without RHP pole must be square integratable? I am stuck with a problem:
For a rational transfer function without RHP pole, for example $$H(s)=\frac{(-z_0 + s)(-z_1+s)\cdots(-z_n+s)}{(-p_0+s)(-p_1+s)\cdots(-p_m+s)}$$ where $Re\{p_i\}<0$ and $n<m$, could we prove that it is square integratable in frequency domain, i.e. $$\int_0^\infty H^*(j\omega)H(j\omega) \,d\omega<\infty \text{ ?}$$ 
Please help me, thx!
 A: I may be thinking about a setting that is not yours, but what I understand is this: You are starting with $u : [0,\infty)\rightarrow\mathbb{C}$ with $\int_{0}^{\infty}|u(t)|^2dt < \infty$, processing with a system through a transfer function $H(s)$, meaning that the output satisfies
$$
              v(t) = \mathcal{L}^{-1}\left( H(s)\mathcal{L}\{u\}\right).
$$
That is, you Laplace transform the signal, multiply by $H(s)$ and inverse transform to obtain the output. And you want the output to be square integrable as well.
A function $u(t)$ is square integrable for $t \ge 0$ iff the Laplace transform $\mathcal{L}\{u\}$ of $u$ is analytic (holomorphic) in the right half plane $\Re s > 0$, and satisfies the following for some constant $M$
$$
            \int_{-\infty}^{\infty}|\mathcal{L}\{u\}(u+iv)|^2dv \le M,\;\;\;\; 0 < u < \infty.
$$
In order for the output to also be square integrable in time, you must also have the following for some constant $M'$:
$$
        \int_{-\infty}^{\infty}|H(u+iv)\mathcal{L}\{u\}(u+iv)|^2dv \le M',\;\;\; 0 < u < \infty. \tag{*}
$$
This would definitely hold if $H(s)$ if there is a constant $C$ such that
$$
                     |H(s)| \le C,\;\;\; \Re s > 0.
$$
I'm not quite sure this boundedness condition is necessary, but it is definitely sufficient, and probably close to necessary when you require $(*)$ to hold for all square integrable signals $u$.
So I don't see any reason to believe that condition $(*)$ will imply that $H$ satisfies an $L^2$ condition on the real axis of the form you suggest:
$$
               \int_{0}^{\infty}|H(u+i0)|^2du < \infty.  \tag{**}
$$
For example, $H(s)=\frac{s-1}{s+1}$ will transform square integrable signals to other square integrable signals because $H$ is uniformly bounded in the right half plane, but this $H$ does not satisfy your desired condition $(**)$.
If I'm thinking about a setting that is not at all what you had in mind, then just ignore me. :)
