Large Signal ModelReading edition

13. A complete small signal example

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13. A complete small signal example

Consider the rational linear signal dynamics

\[ x_{t+1}=Lx_t,\qquad L=\begin{pmatrix}\tfrac12&1\\[2pt]\tfrac14&\tfrac13\end{pmatrix}, \qquad x_0=\begin{pmatrix}1\\0\end{pmatrix}. \]

The spatial archive stores \(x_t\). A second atlas uses the orthogonal Haar matrix

\[ U=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}. \]

Its coefficients \(c_t=Ux_t\) reconstruct as \(x_t=U^*c_t\). The same dynamics in that atlas are \(ULU^*\), not a separate physical law. Thus coordinate transport preserves the full dynamics.

Now retain only the first spatial coordinate. With \(P=\operatorname{diag}(1,0)\), the retained scalar \(z_t\) and discarded scalar \(q_t\) obey

\[ z_{t+1}=\tfrac12z_t+q_t,\qquad q_{t+1}=\tfrac14z_t+\tfrac13q_t. \]

The exact first-coordinate sequence begins

\[ z_0=1,\qquad z_1=\tfrac12,\qquad z_2=\tfrac12, \qquad z_3=\tfrac{11}{24}. \]

The compressed one-step matrix predicts \(z_2=1/4\) instead of \(1/2\). The lost return path contributes exactly

\[ BC=1\cdot\tfrac14=\tfrac14. \]

There are two exact repairs. One is state refinement: start with \(C_0=(1,0)\) and add \(C_0L=(1/2,1)\). The resulting row space is the full two-dimensional space, so it closes. The other is memory:

\[ z_{t+1}=\tfrac12z_t+ \tfrac14\sum_{k=0}^{t-1}\left(\tfrac13\right)^kz_{t-1-k}, \]

since \(q_0=0\). This is the exact same system expressed without retaining \(q_t\) as a separate current-state variable. At \(t=2\), it yields

\[ z_3=\tfrac12\cdot\tfrac12 +\tfrac14\cdot\tfrac12 +\tfrac1{12}\cdot1 =\tfrac{11}{24}. \]

To see a nontrivial reduced closure, adjoin a third independent coordinate with multiplier \(5/6\). The smallest closed refinement of the first coordinate then contains the first two coordinates, not the third. Exact prediction of the chosen summary does not require reconstructing every component of the enlarged signal.

Finally, low energy is not the same as low predictive importance. For

\[ L'=\begin{pmatrix}0&1000\\0&0\end{pmatrix}, \qquad x=(0,1/1000)^\top, \]

omitting the second coordinate loses norm \(1/1000\), but changes the next first coordinate by exactly one. The correct certificate is the product of omitted magnitude and transfer amplification.

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