CAD.5:5.4 - What an internal-path ablation does and does not establish

For the scalar noiseless plant (x_{k+1}=a x_k+u_k), with a one-tick delayed observation (y_k=x_{k-1}), the predictor

\hat x_k=a y_k+u_{k-1},\qquad u_k=-a\hat x_k

uses the executed-input history. Removing that input term while retaining the remaining gains gives (u_k=-a^2x_{k-1}). At (a=1.2), the resulting characteristic polynomial is (\lambda^2-1.2\lambda+1.44); its roots have modulus (1.2), so this ablation is unstable.

It does not prove that every controller lacking that input-history path is unstable. Retuning the memoryless delayed-output law to (u_k=-0.6x_{k-1}) gives roots of (\lambda^2-1.2\lambda+0.6), both with modulus (\sqrt{0.6}<1). The latter establishes stabilization of this noiseless model, not the original predictor’s performance or a bounded-disturbance constraint guarantee.

Use the first comparison to explain why a particular realization depends on its internal path. Use an impossibility argument such as :5.2–:5.3 when claiming that the available information cannot meet a requirement under any controller in the stated class.