Dynamical observations and finite system realization — Examples and Counterexamples

Since $h>0$, ${\omega }_1=2\pi \mathrm{/}h>0={\omega }_0$, so the frequencies are distinct. By [F1], the $k$th samples occur at $t_k=kh$. The two continuous outputs are $\cos ({\omega }_0{t}_k)=\cos 0=1$ and $\cos ({\omega }_1{t}_k)=\cos (2\pi k)=1$ by [F2]. Thus their entire noiseless sample sequences agree.

The distinct admissible frequencies in step 1.1 have identical complete noiseless data. Therefore the data do not uniquely determine the frequency, which refutes the stated claim.

Direct multiplication gives $T{T}^{-1}=T^{-1}T=I_2$, so $T$ is invertible. Multiplying the displayed matrices gives $T^{-1}AT=\widehat{A}$, $T^{-1}B=\widehat{B}$, $CT=\widehat{C}$, and $T^{-1}{x}_0={\widehat{x}}_0$. Also $\widehat{A}\mathrm{\ne }A$ and $\widehat{C}\mathrm{\ne }C$, so these are distinct coordinate descriptions.

The hypotheses of [F2] hold by step 1.1, so every finite common-input experiment has the same output sequence in both descriptions.

Steps 1.1 and 2.1 exhibit distinct realization matrices and initial coordinates with identical data, as claimed.

The first initial state remains zero because $B=0$, so its outputs are all zero. For the difference $e_2=(0,1{)}^{\mathsf{T}}$, $A{e}_2=2e_2$; induction gives $A^k{e}_2=2^k{e}_2$ for each $k\ge 0$. Since $C{e}_2=0$, every output difference is $C{A}^k{e}_2=0$. Thus the two distinct initial states produce identical outputs at all samples.

For each $j\ge 0$, $C{A}^j=[1,0]$, so $O_q$ consists of $q$ copies of $[1,0]$ and has rank $1$. Since $1<d=2$, [F2] implies that the initial-state output map is noninjective for every $q\ge 1$.

The initial states in step 1.1 are different but have the same complete observed sequence, so no data-only rule can recover both states exactly in this experiment.

The function $z_{\theta }(t)=\theta t^2\mathrm{/}2$ has $z_{\theta }(0)=0$ and, by [F1], derivative ${\dot{z}}_{\theta }(t)=\theta t$. Any other solution is continuous on $[0,1]$ and has the same derivative on $(0,1)$, so [F8] makes its difference from $z_{\theta }$ constant; the common initial value makes this constant zero. Thus the solution is unique and gives the exact endpoint map $A(\theta )=z_{\theta }(1)=\theta \mathrm{/}2$. At $\theta =1$ the exact observation is $1\mathrm{/}2$.

The stated datum $b=0$ belongs to $Y=\mathbb{R}$. By [F2]–[F4], $X=Y=\mathbb{R}$ with the standard inner product is a real Hilbert space. The maps $A(\theta )=\theta \mathrm{/}2$ and $A_h(\theta )=0$ are linear and bounded by [F5], with bounds $1\mathrm{/}2$ and $0$; their ranges have bases $(1)$ and the empty basis, respectively. Hence $A$ is compact by [F6], and $A_h$ has finite-dimensional range. The adjoints are multiplication by $1\mathrm{/}2$ and $0$, since $⟨A\theta ,y⟩=\theta y\mathrm{/}2=⟨\theta ,y\mathrm{/}2⟩$ and $⟨A_h\theta ,y⟩=0$. For $\lambda =1$, the shifted normal operators are multiplication by $5\mathrm{/}4$ and $1$, so both are bijective. All hypotheses of [F7] are satisfied.

The stated forward-Euler formula at $(t_0,z_0)=(0,0)$ gives $z_1=0+1\cdot \theta \cdot 0=0$, so $A_h(\theta )=0$ and $A_h(1)=0$. Thus the two endpoint maps differ even though the observation perturbation is zero.

Set $x^{†}=1$, $e=0$, $\delta =0$, and $\lambda =1$. Then [F7] gives $d_h=(A-A_h)x^{†}=1\mathrm{/}2-0=1\mathrm{/}2$. This is a nonzero discretization contribution with no observation noise.

For the first system, $M_0=CB=\epsilon$ and $M_1=CAB=0$ by [F1]. Thus the $q=1$ blocks are $H_0=[\epsilon ]$ and $H_1=[0]$ by [F3], and [F2] gives the singular value of $H_0$ as $\sqrt{{\epsilon }^2}=\epsilon$. Here $O_1=[1]$ and $R_1=[\epsilon ]$ both have rank $1$. By [F4] and the given successful-call hypothesis, the exact algorithm succeeds and its scalar transition is similar to $A=0$; scalar similarity leaves that transition equal to $0$.

For the perturbed data, the scalar system with $\tilde{A}=1\mathrm{/}\sqrt{\epsilon }$, $\tilde{B}=\epsilon$, and $\tilde{C}=1$ has ${\tilde{M}}_0=\epsilon$ and ${\tilde{M}}_1=\sqrt{\epsilon }$ by [F1]; [F3] identifies these as its $q=1$ Hankel blocks. Its $O_1=[1]$ and $R_1=[\epsilon ]$ again have rank $1$, so [F4] and the given successful-call hypothesis imply that the exact algorithm returns a scalar transition similar to $\tilde{A}$. A one-dimensional similarity does not change a scalar, hence $\tilde{A}$ is returned.

Take ${\epsilon }_n=1\mathrm{/}{n}^2$ for positive integers $n$. The perturbation size is $\sqrt{{\epsilon }_n}=1\mathrm{/}n\to 0$, while the change in transition value is $1\mathrm{/}\sqrt{{\epsilon }_n}=n\to \mathrm{\infty }$ and the singular value of $H_0$ is $1\mathrm{/}{n}^2\to 0$. This is an exact finite example of unbounded sensitivity as the Hankel singular value approaches zero; it makes no claim of noisy rank-selection correctness.