Laplace Transforms — Question 6

Question 6 – Solution

Strategy. Solve the coupled algebra first; vector numerators can either retain or cancel the repeated oscillator poles.

Step 1: Derive the rational responses. The transformed equations are $sF+G=s/(s^2+1)$ and $-F+sG=\sigma/(s^2+1)$. Solving gives, on the sufficient half-plane $\operatorname{Re}s>0$, $$\boxed{F=\frac{s^2-\sigma}{(s^2+1)^2},\qquad G=\frac{(1+\sigma)s}{(s^2+1)^2}.}$$ The numerator, not the unsimplified denominator alone, decides actual pole order.

Step 2: Invert the co-rotating input. For $\sigma=1$, differentiating the standard transforms with respect to $s$ gives $\mathcal L\{t\cos t\}=(s^2-1)/(s^2+1)^2$ and $\mathcal L\{t\sin t\}=2s/(s^2+1)^2$. Hence $$\boxed{x=t\cos t,\qquad y=t\sin t,\qquad \|Z(t)\|=t.}$$ Indeed $x'=\cos t-t\sin t=-y+\cos t$ and $y'=\sin t+t\cos t=x+\sin t$, with zero initial state. The response is unbounded even though each forcing component is bounded.

Step 3: Invert the opposite rotation. For $\sigma=-1$, cancellation gives $F=1/(s^2+1)$ and $G=0$. Consequently $\boxed{x=\sin t,\ y=0}$, with norm $|\sin t|\le 1$. The first equation reads $\cos t=0+\cos t$; the second reads $0=\sin t-\sin t$. Both initial conditions hold. No polynomial factor survives in this response.

Step 4: Interpret the distinction. For $\sigma=1$, both transforms have double poles at $s=\pm i$; for $\sigma=-1$, the first has simple poles and the second is zero. The free system rotates counterclockwise. The input $(\cos t,\sin t)$ rotates with it, whereas $(\cos t,-\sin t)$ rotates oppositely. More explicitly, write $Z=R(t)V$, where $R$ is the counterclockwise rotation matrix. Then $V'=(1,0)^T$ in the first case and $V'=(\cos 2t,-\sin 2t)^T$ in the second. Their integrals respectively grow linearly and remain bounded. A frequency match alone therefore does not force an unbounded response; the vector direction and resulting pole cancellations matter.