Dirac Delta Function — Question 9

Question 9 – Solution

Strategy. Infer the required impulses from velocity jumps after enforcing displacement continuity. Then independently check the recovered forcing by inversion.

Step 1: Enforce continuity at the second impulse. The target displacement just before $3$ is $2e^{-2}$, while the right-hand value is $C$. Hence the permitted actuator requires $$\boxed{C=2e^{-2}.}$$ The first junction is already continuous because $(t-1)e^{-(t-1)}$ vanishes at $1$.

Step 2: Recover the impulse strengths. For $h(\tau)=\tau e^{-\tau}$, $h'(0)=1$ and $h'(2)=-e^{-2}$. Thus the first velocity jump is $1$. At the second switch the new tail has velocity $-C=-2e^{-2}$, so $$\boxed{J_1=1,\qquad J_3=-2e^{-2}-(-e^{-2})=-e^{-2}.}$$ Because $y$ is continuous, integrating the equation across a junction gives $J_j=[y']_j$. Direct differentiation of each piece gives $y''+2y'+y=0$ on all three open intervals.

Step 3: Verify by transformation and inversion. The recovered input gives $$Y=\frac{e^{-s}-e^{-2}e^{-3s}}{(s+1)^2},\qquad y=H_1(t)(t-1)e^{-(t-1)} -e^{-2}H_3(t)(t-3)e^{-(t-3)}.$$ For $t=3+\tau$, the last expression reduces to $$e^{-2}[(\tau+2)-\tau]e^{-\tau} =2e^{-2}e^{-\tau},$$ which is the prescribed tail. Setting $C=0$ instead would require a jump of $-2e^{-2}$ in displacement at $3$, forbidden for the stated impulse-only actuator. Canceling velocity cannot instantaneously erase nonzero displacement.

Step 4: Check the area balance. Direct integration gives $$\int_0^\infty y\,dt =\int_0^2\tau e^{-\tau}\,d\tau+2e^{-2}\int_0^\infty e^{-\tau}\,d\tau =1-3e^{-2}+2e^{-2}=\boxed{1-e^{-2}}.$$ The integrated derivative terms are zero when their impulse jumps are included: initial and limiting displacement and velocity are zero. Thus $\int y=J_1+J_3=1-e^{-2}$ independently confirms the result. The nonzero $e^{-t}$ tail gives exact real domain $\boxed{s>-1}$.