Laplace Transforms — Question 9

Question 9 – Solution

Strategy. Shift a single step response, then check both state components at switch-off; displacement alone is not a stopping condition.

Step 1: Transform and invert by shifting. Since $sF=G$ and $sG=-4F+(1-e^{-Ts})/s$, $$F=\frac{1-e^{-Ts}}{s(s^2+4)},\qquad G=\frac{1-e^{-Ts}}{s^2+4},\qquad \operatorname{Re}s>0.$$ Let $h(t)=(1-\cos 2t)/4$ and $j(t)=\sin 2t/2$. Then $$\boxed{x=h(t)-H(t-T)h(t-T),\quad y=j(t)-H(t-T)j(t-T)}.$$ Both delayed terms begin at zero, proving state continuity.

Step 2: Enforce complete rest at the endpoint. At $T$, the state is $((1-\cos 2T)/4,\sin 2T/2)^T$. Both components vanish exactly when $\cos 2T=1$, giving $\boxed{T=k\pi,\ k=1,2,\ldots}$; the earliest is $\pi$. Zero endpoint data then yield the identically zero unforced continuation. Conversely, rest for all later times requires these endpoint values by continuity.

Step 3: Compute the residual and its spectral cancellation. After the pulse, differentiation gives $(y^2+4x^2)'=2y(-4x)+8xy=0$. Evaluation at $T$ yields $$\boxed{y^2+4x^2=\tfrac 14\sin^22T+\tfrac 14(1-\cos 2T)^2 =\sin^2T.}$$ The finite-pulse transform $U_T=(1-e^{-Ts})/s$ has zeros at $s=\pm 2i$ exactly when $e^{2iT}=1$. These are precisely the stopping durations; they cancel the oscillator poles. At $s=0$ its singularity is removable, with value $T$, so a nonzero total input area does not prevent this cancellation.

Step 4: Compare two durations. For $T=\pi$, $x=y=0$ after $\pi$. For $T=\pi/2$, writing $\tau=t-\pi/2\ge 0$, the endpoint state is $(1/2,0)$ and $\boxed{x=\tfrac 12\cos 2\tau,\ y=-\sin 2\tau}$ afterward. During either pulse, $x=h(t)$ and $y=j(t)$; the plotted curves coincide up to $\pi/2$. A full-period pulse lets the forced trajectory return to rest before switch-off; accumulated state contributions have different oscillatory phases and can cancel even when the input never becomes negative. Equal input sign is not equal state phase.