Convolution Integrals — Question 2

Question 2 – Solution

Strategy. For unit-height rectangles, the integral measures overlap length.

Step 1: Track the overlap. Using commutativity, integrate $f(u)g(t-u)$. Both factors are $1$ on $[1,3]\cap[t-5,t-2]$, so $$w=\max\{0,\min(3,t-2)-\max(1,t-5)\} =\begin{cases}0,&t<3,\\t-3,&3\le t<5,\\ 2,&5\le t<6,\\8-t,&6\le t<8,\\0,&t\ge 8.\end{cases}$$ Endpoints of the intervals do not change their length.

Step 2: Transform and recover the ramps. For $s\ne 0$, $$W=\frac{e^{-3s}(1-e^{-2s})(1-e^{-3s})}{s^2}.$$ The inverse is $w=(t-3)H_3-(t-5)H_5-(t-6)H_6+(t-8)H_8$. Activating these four ramps successively gives slopes $1,0,-1,0$, exactly as in the piecewise formula.

Step 3: Describe the geometry and regularity. The support (closure of the nonzero set) is $[3,8]$; $w>0$ precisely on $(3,8)$. Its maximum is $2$, attained at every $t\in[5,6]$. It is continuous everywhere. The first derivative has jumps $+1,-1,-1,+1$ at $3,5,6,8$, respectively, and is undefined there in the ordinary two-sided sense.

Step 4: Check area and convergence. Two triangles and the plateau have total area $2+2+2=6$. Independently, the convolution area is $(\int f)(\int g)=2\cdot 3=6$, by integrating over the rectangle in the two integration variables. Compact support makes the transform finite for every real $s$. The displayed expression has removable value $W(0)=6$.