Laplace Transforms — Question 5

Question 5 – Solution

Strategy. Separate the input contribution from the initial-state contribution before cancelling rational factors.

Step 1: Calculate the zero-state response. The determinant is $s^2-s-2=(s-2)(s+1)$, and $$(sI-A)^{-1}B= \frac 1{(s-2)(s+1)} \begin{pmatrix}s-1&1\\2&s\end{pmatrix}\binom 1{-1} =\frac 1{s+1}\binom 1{-1}.$$ The constant input contributes an additional $1/s$. Inverting yields $\boxed{Z_0(t)=(1-e^{-t})(1,-1)^T}$, which is bounded. Its transform converges for $\operatorname{Re}s>0$; the cancelled expression extends through $s=2$, where the full matrix resolvent itself is undefined.

Step 2: Identify why the pole disappears. We have $AB=-B$, so the input lies entirely in the stable eigendirection. The other eigenpair is $2,(1,2)^T$. From zero, this input never excites that direction. The pole cancellation describes the chosen input and data; it does not change either eigenvalue of $A$.

Step 3: Include all initial conditions. Write $(p,q)^T=c_+(1,2)^T+c_-(1,-1)^T$, where $c_+=(p+q)/3$ and $c_-=(2p-q)/3$. The transform contains $$\frac{c_+}{s-2}\binom 12+ \frac{c_-}{s+1}\binom 1{-1}+\frac 1{s(s+1)}\binom 1{-1}.$$ Therefore $$Z=c_+e^{2t}\binom 12+ \bigl[1+(c_--1)e^{-t}\bigr]\binom 1{-1}.$$ The growing vector cannot be cancelled by a bounded term, so boundedness holds exactly when $\boxed{p+q=0}$. Then the limit is $(1,-1)^T$. A common convergence half-plane for the general family is $\operatorname{Re}s>2$.

Step 4: Test sensitivity to the omitted term. For the specified perturbation, $c_+=\varepsilon$ and $c_-=0$. The exact difference is $\boxed{Z-Z_0=\varepsilon e^{2t}(1,2)^T}$, whose Euclidean norm is $\sqrt 5|\varepsilon|e^{2t}\to\infty$. Thus arbitrarily small initial perturbations in that direction destroy boundedness. The student’s inference confuses a bounded zero-state response with boundedness for all initial states.