The Definition — Question 9

Question 9 – Solution

Strategy. A time translation by one period multiplies the weighted area by a fixed factor, producing a geometric series.

Step 1: Build the series and its domain. For $s>0$, the $n$th nonzero interval contributes $$\int_{2n}^{2n+1}e^{-st}\,dt=e^{-2ns}\frac{1-e^{-s}}s.$$ The ratio $e^{-2s}$ lies in $(0,1)$, so the sum converges. At $s=0$, each interval contributes one and the sum diverges. For $s<0$, $e^{-st}\ge 1$ on each such interval, so it also diverges. The real convergence set is exactly $(0,\infty)$, with absolute convergence since $f\ge 0$.

Step 2: Simplify the geometric sum. Summing gives $$F(s)=\frac{1-e^{-s}}{s(1-e^{-2s})} =\boxed{\frac 1{s(1+e^{-s})},\qquad s>0.}$$ The integral over only one period is the first term of this sum, not the transform of the endlessly repeated signal.

Step 3: Interpret the two normalized limits. The formula gives $$\boxed{\lim_{s\to 0^+}sF(s)=\tfrac 12,\qquad \lim_{s\to\infty}sF(s)=1.}$$ The first value is the period average $(1/2)\int_0^2f(t)\,dt=1/2$. The second agrees with $f(0^+)=1$: large $s$ weights short times most strongly. These normalized limits do not assert that $F(0)$ exists; it does not.

Step 4: Quantify finite-window error. Writing $F_N=\int_0^{2N}e^{-st}f(t)\,dt$, the finite geometric sum gives $$F_N=F(s)(1-e^{-2Ns}),\qquad \boxed{\frac{F(s)-F_N}{F(s)}=e^{-2Ns}.}$$ At $s=0.2$, the requirement is $e^{-0.4N}\le 10^{-3}$, so $N\ge\ln(1000)/0.4\approx 17.27$. The least integer is $\boxed{N=18}$, corresponding to a cutoff time of $36$. Endpoint assignments do not affect the integrals; the graph uses the values specified in the question.