IVP’s With Step Functions — Question 2

Question 2 – Solution

Strategy. A window does not reset the ramp’s clock. At each endpoint rewrite $t$ as delayed time plus the endpoint.

Step 1: Shift the whole input. The input is $H_1(t)[(t-1)+1]-H_3(t)[(t-3)+3]$, so $$\boxed{F=e^{-s}\left(\frac 1{s^2}+\frac 1s\right) -e^{-3s}\left(\frac 1{s^2}+\frac 3s\right).}$$ The proposed expression instead transforms $H_1(t)(t-1)-H_3(t)(t-3)$, which has a constant tail of $2$ after $3$. It describes neither the given ramp window nor its shutoff.

Step 2: Invert the response. The zero datum gives $Y=F/(s+1)$. For a delayed input $\tau+c$, the zero-start response solves $r_c'+r_c=\tau+c$, and is $$r_c(\tau)=\tau+c-1+(1-c)e^{-\tau}.$$ Its transform is $(1/s^2+c/s)/(s+1)$, as direct transformation confirms. In particular $r_1=\tau$ and $r_3=\tau+2-2e^{-\tau}$. Thus $$\boxed{y=H_1(t)(t-1)-H_3(t)[(t-3)+2-2e^{-(t-3)}].}$$

Step 3: Read the intervals and switches. Simplifying after the second switch gives $$y(t)=\begin{cases}0,&0\le t<1,\\t-1,&1\le t<3,\\ 2e^{-(t-3)},&t\ge 3.\end{cases}$$ The pieces meet at $0$ and $2$, and satisfy $y'+y=0,t,0$. Their slopes are $0,1,-2e^{-(t-3)}$, so $[y']_1=1$ and $[y']_3=-3$. The unique global maximum is $\boxed{y(3)=2}$, by increase during the window and strict decay afterward.

Step 4: Check the total response. Directly, $$\int_0^\infty y\,dt=\int_1^3(t-1)\,dt+ \int_3^\infty 2e^{-(t-3)}\,dt=2+2=\boxed 4.$$ Integrating the equation gives the same answer because $y(\infty)-y(0)=0$ and $\int_1^3t\,dt=4$. The compact input has a transform for every real $s$, with its apparent $s=0$ poles canceling. The output’s nonzero exponential tail gives exact domain $\boxed{s>-1}$, and the removable value $Y(0)=4$ agrees with the area.