]> 7.2.3 The step response of an abstract system

#### 7.2.3 The step response of an abstract system

In this section we deﬁne the step response of the system

 $\left\{\begin{array}{ccc}\hfill ẋ\left(t\right)& \hfill =\hfill & \mathcal{𝒜}\left[x\left(t\right)+du\left(t\right)\right]\hfill \\ \hfill x\left(0\right)& \hfill =\hfill & 0\hfill \\ \hfill y\left(t\right)& \hfill =\hfill & {c}^{#}x\left(t\right)\hfill \end{array}\right\},\phantom{\rule{2em}{0ex}}{c}^{#}x={h}^{\ast }\mathcal{𝒜}x={〈\mathcal{𝒜}x,h〉}_{\text{H}}$ (7.4)

i.e., the response of the system for the Heaviside function $u\left(t\right)=1l\left(t\right)$ under null initial conditions.

Theorem 7.2.3. Suppose that the semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ generated by $\mathcal{𝒜}$ is EXS. Assume also that either ${c}^{#}$ is an admissible observation functional or $d$ is an admissible factor control vector, and the compatibility assumption

 $d\in D\left({c}^{#}\right)$ (7.5)

holds.

If ${c}^{#}$ is an admissible observation functional then for almost all $t\ge 0$ the step response has the form

 $y\left(t\right)=\left(\overline{P}d\right)\left(t\right)-{c}^{#}d$ (7.6)

where $\overline{P}\in L\left(\text{H},{\text{L}}^{2}\left(0,\infty \right)\right)$ denotes the closure of the observability map $P$,

$P:\text{H}\supset D\left(\mathcal{𝒜}\right)\ni x↦{h}^{\ast }\mathcal{𝒜}S\left(\cdot \right)x\in {\text{L}}^{2}\left(0,\infty \right).$

If $d$ is an admissible factor control vector then for almost all $t\ge 0$ the step response can be represented as

 $y\left(t\right)=\left(\overline{Q}h\right)\left(t\right)-{c}^{#}d$ (7.7)

where $\overline{Q}\in L\left(\text{H},{\text{L}}^{2}\left(0,\infty \right)\right)$ denotes the closure of the observability map $Q$,

$Q:\text{H}\supset D\left({\mathcal{𝒜}}^{\ast }\right)\ni x↦{d}^{\ast }{\mathcal{𝒜}}^{\ast }{S}^{\ast }\left(\cdot \right)x\in {\text{L}}^{2}\left(0,\infty \right).$

Proof. Since [69, p. 5], [18, Theorem 2.1.10, p. 21]

${\int }_{0}^{t}S\left(\tau \right){x}_{0}d\tau \in D\left(\mathcal{𝒜}\right),\phantom{\rule{2em}{0ex}}\mathcal{𝒜}{\int }_{0}^{t}S\left(\tau \right){x}_{0}d\tau =S\left(t\right){x}_{0}-{x}_{0}\phantom{\rule{2em}{0ex}}\forall t\ge 0\phantom{\rule{1em}{0ex}}\forall {x}_{0}\in \text{H}$

then by (7.3)

$x\left(t\right)=\mathcal{𝒜}{\int }_{0}^{t}S\left(\tau \right)dd\tau =S\left(t\right)d-d$

satisﬁes (7.2) for almost all $t\ge 0$ and all $w\in D\left({\mathcal{𝒜}}^{\ast }\right)$, i.e., in the weak sense,

$\frac{d}{dt}{〈x\left(t\right),w〉}_{\text{H}}={〈x\left(t\right)+d,{\mathcal{𝒜}}^{\ast }w〉}_{\text{H}}⇔{\mathcal{𝒜}}^{-1}ẋ\left(t\right)=\frac{d}{dt}\left[{\mathcal{𝒜}}^{-1}x\left(t\right)\right]=x\left(t\right)+d.$

Thus the Laplace transform $\stackrel{̂}{x}$ of $x$ is

$\stackrel{̂}{x}\left(s\right)={\left(sI-\mathcal{𝒜}\right)}^{-1}d-\frac{1}{s}d,\phantom{\rule{2em}{0ex}}s\in \overline{{\Pi }^{+}}\setminus 0.$

The ﬁrst term belongs to $D\left(\mathcal{𝒜}\right)$ while the second one is in $D\left({c}^{#}\right)$ due to (7.5). Now

$ŷ\left(s\right)={h}^{\ast }\mathcal{𝒜}{\left(sI-\mathcal{𝒜}\right)}^{-1}d-\frac{1}{s}{c}^{#}d={d}^{\ast }{\mathcal{𝒜}}^{\ast }{\left(sI-{\mathcal{𝒜}}^{\ast }\right)}^{-1}h-\frac{1}{s}{c}^{#}d=\overline{ŷ\left(\overline{s}\right)}\phantom{\rule{2em}{0ex}}s\in {\Pi }^{+}$

If ${h}^{\ast }\mathcal{𝒜}$ is an admissible observation functional with respect to ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ then (7.6) holds, while if $d$ is an admissible factor control vector then, due to duality Theorem 7.2.2, ${d}^{\ast }{\mathcal{𝒜}}^{\ast }$ is an admissible observation functional with respect to ${\left\{{S}^{\ast }\left(t\right)\right\}}_{t\ge 0}$ and thus the representation (7.7) holds. □

Lemma 7.2.1. Let all assumptions of Theorem 7.2.3 hold and additionally the operator $\mathcal{𝒜}$ has a Riesz basis of eigenvectors ${\left\{{x}_{n}\right\}}_{n\in \mathbb{ℕ}}$, corresponding to the set of its eigenvalues ${\left\{{\lambda }_{n}\right\}}_{n\in \mathbb{ℕ}}$, with the biorthogonal system ${\left\{{v}_{n}\right\}}_{n\in \mathbb{ℕ}}$ being the system of eigenvectors of ${\mathcal{𝒜}}^{\ast }$ corresponding to its eigenvalues ${\left\{\overline{{\lambda }_{n}}\right\}}_{n\in \mathbb{ℕ}}$. Then the step response has the spectral representation

 $y\left(t\right)=\sum _{n=1}^{\infty }{\lambda }_{n}\left({e}^{{\lambda }_{n}t}-1\right){〈d,{v}_{n}〉}_{\text{H}}{〈{x}_{n},h〉}_{\text{H}}$ (7.8)

Proof. The result follows from Theorem 7.2.3 and the proof of Lemma 2.6.1. □