Doubt about $H_{\infty } $ norm of a transfer function

Question

I have a doubt about the $H_{\infty } $ norm of a transfer function. Suppose I have a transfer function:

$T(s)=C(sI-A)^{-1}B+D$

I have seen that

$\lim_{t->\infty } T(j\omega ) = D$

so, it means that the $||T||_{\infty } = D$. But I don't understand how this result comes out.

Infact, if I do the limit of the transfer function $T(s)=C(sI-A)^{-1}B+D$ I have :

$\frac{CB}{A}+D$

can somebody please help me?

Answer 1

I assume that you meant to define the infinity norm as follows

$$ \|T\|_\infty = \lim_{\omega\to\infty} T(j\,\omega), $$

so taking the limit in $\omega$ instead of $t$. In this case it can be noted that when the absolute value of $s$ is really large it holds that that $s\,I - A \approx s\,I$, so

\begin{align} \|T\|_\infty &= \lim_{\omega\to\infty} C\,(j\,\omega\,I - A)^{-1} B + D, \\ &= \lim_{\omega\to\infty} C\,(j\,\omega\,I)^{-1} B + D, \\ &= \lim_{\omega\to\infty} \frac{C\,B}{j\,\omega} + D, \\ &= D. \end{align}

However, it can be noted that the infinity norm can also refer to

$$ \|T\|_\infty = \sup_\omega \bar{\sigma}(T(j\,\omega)), $$

with $\bar{\sigma}(X)$ the largest singular value of $X$. This definition in general does not give the same value as the one defined in my first equation.

The equation that you obtained is the limit of $s$/$\omega$ to zero

$$ \lim_{\omega\to0} T(j\,\omega) = C\,A^{-1} B + D. $$