Dirac Delta Function — Question 1

Question 1 – Solution

Strategy. Preserve pulse area rather than height. Compare the two exact responses before making any claim about convergence.

Step 1: Test the pulse and transform. Its action on a continuous $\phi$ is $$\frac J\varepsilon\int_a^{a+\varepsilon}\phi(t)\,dt \longrightarrow J\phi(a).$$ The error is at most $J\sup_{[a,a+\varepsilon]}|\phi(t)-\phi(a)|$, which tends to zero. Taking $\phi=e^{-st}$ gives $F_\varepsilon=Je^{-as}(1-e^{-\varepsilon s})/(\varepsilon s)\to Je^{-as}$, with the removable value $F_\varepsilon(0)=J$.

Step 2: Solve both equations. The impulse gives $Y=Je^{-as}/(s+2)$, so $\boxed{y=JH_a(t)e^{-2(t-a)}}$ and $[y]_a=J$. The finite-pulse solution is continuous and equals $$y_\varepsilon(t)=\begin{cases} 0,&t<a,\\ \dfrac{J}{2\varepsilon}(1-e^{-2(t-a)}),&a\le t<a+\varepsilon,\\ \dfrac{J}{2\varepsilon}(1-e^{-2\varepsilon}) e^{-2(t-a-\varepsilon)},&t\ge a+\varepsilon. \end{cases}$$ Each piece satisfies its ordinary equation, and the endpoints match.

Step 3: Bound the tail error. Put $\tau=t-a\ge\varepsilon$. The explicit tail can be rewritten as $y_\varepsilon=(J/\varepsilon)\int_0^\varepsilon e^{-2(\tau-u)}\,du$. Since $x\mapsto e^{-2x}$ is decreasing and has derivative magnitude at most $2$ for $x\ge 0$, $$0\le y_\varepsilon-y =\frac J\varepsilon\int_0^\varepsilon [e^{-2(\tau-u)}-e^{-2\tau}]\,du \le\frac J\varepsilon\int_0^\varepsilon 2u\,du =\boxed{J\varepsilon}.$$ This representation follows directly from the finite-pulse formula.

Step 4: Distinguish the kinds of convergence. For each fixed $t\ne a$, $y_\varepsilon(t)\to y(t)$. At $a$, $y_\varepsilon(a)=0$ but the right-hand representative has $y(a)=J$. Uniform convergence on an interval containing $a$ is impossible: the approximants are continuous and the limit has a jump. Changing only $y(a)$ does not remove that jump or change its transform. The graph uses $a=J=1$; open and filled dots mark the impulsive response.