Laplace Transforms — Question 3

Question 3 – Solution

Strategy. Solve a zero-state ramp response, then translate time and check the switching jet.

Step 1: Transform the shifted input. All initial terms vanish. The second shifting theorem gives $$Y(s)=\frac{e^{-s}}{s^2(s+2)^3},\qquad \operatorname{Re}s>0.$$

Step 2: Invert before translating. The unshifted fraction decomposes as $$\frac 1{s^2(s+2)^3}=-\frac 3{16s}+\frac 1{8s^2} +\frac 3{16(s+2)}+\frac 1{4(s+2)^2}+\frac 1{4(s+2)^3}.$$ Thus, with $r=t-1$, $$g(r)=\frac r8-\frac 3{16}+e^{-2r}\left(\frac 3{16}+\frac r4+\frac{r^2}{8}\right), \qquad \boxed{y(t)=\begin{cases}0,&0\leq t\leq 1,\\g(t-1),&t>1.\end{cases}}$$

Step 3: Verify forcing and regularity. The exponential term is annihilated by $(D+2)^3$; applying it to $r/8-3/16$ gives $r$. Also $g(r)=r^4/24+O(r^5)$ as $r\downarrow 0$. Consequently $y,y',y'',y'''$ match zero at switching, while $y^{(4)}(1+)-y^{(4)}(1-)=1$. The solution is $C^3$ but not $C^4$; the original equation also holds at $t=1$.

Step 4: Establish shape and asymptote. Convolution gives $g(r)=\int_0^r (r-v)v^2e^{-2v}/2\,dv>0$ for $r>0$. The exponential correction tends to zero, so the asymptote is $y=(t-1)/8-3/16$, approached from above. It is drawn only for $t\geq 1$.