Laplace Transforms — Question 7

Question 7

For real $s$, write $\mathcal L\{f\}(s)=F(s)=\int_0^\infty e^{-st}f(t)\,dt$ where the integral converges.

An unknown signal belongs to the family $f(t)=e^{-at}\cos bt$, with $a\ge 0$ and $b>0$. Its transform satisfies the exact observations $$\lim_{s\to\infty}s[sF(s)-1]=-2,\qquad \lim_{s\to\infty}s^3[F(s)-1/s+2/s^2]=-5.$$

Tasks

1. Derive the transform in terms of $a,b$ and expand it through the term in $s^{-3}$.

2. Recover $a,b$ from the observations and state the resulting transform with its convergence interval.

3. Compute $f(0)$, $f'(0)$ and $f''(0)$ directly. Explain how the expansion coefficients encode these initial quantities.

4. For $q(t)=t^3e^{-t}$, compute its transform and show that $f+\lambda q$ has the same two observed limits for every real $\lambda$. Explain the limit of the parameter recovery claim.

Question 7 – Solution

Strategy. Large transform parameters emphasize early time. Within a specified family, the first few coefficients can determine parameters, but they do not determine an arbitrary signal.

Step 1: Transform and expand. Exponential shifting in the cosine pair gives $$F(s)=\frac{s+a}{(s+a)^2+b^2},\qquad s>-a.$$ Expanding this rational expression for large positive $s$ yields $$\boxed{F(s)=\frac 1s-\frac a{s^2}+\frac{a^2-b^2}{s^3}+O(s^{-4}).}$$ For example, factor $s^2$ from the denominator and expand its reciprocal through order $s^{-2}$ before multiplying by $1/s+a/s^2$.

Step 2: Recover the parameters. The first limit equals $-a$, so $a=2$. With this value, the second equals $a^2-b^2=-5$, giving $b^2=9$. Since $b>0$, $$\boxed{a=2,\quad b=3,\qquad F(s)=\frac{s+2}{(s+2)^2+9},\quad s>-2.}$$ The boundary has an undamped cosine integral that does not converge; smaller parameters produce growing oscillations. Thus the interval is exact.

Step 3: Read the initial information. Direct differentiation gives $$\boxed{f(0)=1,\qquad f'(0)=-a=-2,\qquad f''(0)=a^2-b^2=-5.}$$ These are precisely the coefficients of $1/s$, $1/s^2$ and $1/s^3$ in the expansion. Repeated integration by parts explains the same pattern when the needed derivative bounds hold.

Step 4: Show what the measurements cannot identify. Three integrations by parts give $\mathcal L\{q\}(s)=6/(s+1)^4$ for $s>-1$. Thus $f+\lambda q$ has transform $F(s)+6\lambda/(s+1)^4$ on their common domain. The added term is $O(s^{-4})$, so it changes neither observed limit. Also $q(0)=q'(0)=q''(0)=0$. Distinct $\lambda$ therefore give different signals with the same recorded information. The parameters are uniquely recovered only within the stated exponential-cosine family.