Laplace Transforms — Question 3

Question 3 – Solution

Strategy. First invert the undelayed rational factors, then translate the entire response using the second shifting theorem.

Step 1: Transform both equations. With $F=\mathcal L\{x\}$ and $G=\mathcal L\{y\}$, $(s+1)F=e^{-2s}/s$ and $(s+2)G=F$. Hence $$F=\frac{e^{-2s}}{s(s+1)},\qquad G=\frac{e^{-2s}}{s(s+1)(s+2)},\qquad \operatorname{Re}s>0.$$ It is $e^{-2s}$, rather than an exponential in time, that encodes the delay.

Step 2: Shift the inverse transforms. The undelayed inverses are $1-e^{-t}$ and $\tfrac 12-e^{-t}+\tfrac 12e^{-2t}=\tfrac 12(1-e^{-t})^2$. Writing $\tau=t-2$, the solution is zero for $t<2$, and for $t\ge 2$, $$\boxed{x=1-e^{-\tau},\qquad y=\tfrac 12(1-e^{-\tau})^2.}$$ These formulas give zero at the joining point. Multiplication by $e^{-2t}$ would change decay rates while generally leaving a nonzero response before $t=2$; it is a different transform operation.

Step 3: Check the switch regularity. Both states are continuous. Before the switch their derivatives vanish; after it $x'=e^{-\tau}$ and $y'=e^{-\tau}-e^{-2\tau}$. Thus $x'(2^-)=0$, $x'(2^+)=1$, whereas $y'(2^-)=y'(2^+)=0$. Afterward $y''=-e^{-\tau}+2e^{-2\tau}$, so $\boxed{y''(2^-)=0,\ y''(2^+)=1}$. In particular $y$ is continuously differentiable at the switch but is not twice differentiable there.

Step 4: Establish shape and limits. For $\tau>0$, $x'>0$ and $y'=e^{-\tau}(1-e^{-\tau})>0$. Also $0<x<1$ and $0<y<1/2$, with limits $1$ and $1/2$. These inequalities prove the absence of overshoot for all later times. The dotted horizontal guides in the sketch are limits; the solid and dashed curves are the components. Both remain zero throughout the waiting interval.