Laplace Transforms: Question 4-2.7 & Solution

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

Derive the transform in terms of $a,b$ and expand it through the term in $s^{-3}$ .
Recover $a,b$ from the observations and state the resulting transform with its convergence interval.
Compute $f(0)$ , $f'(0)$ and $f''(0)$ directly. Explain how the expansion coefficients encode these initial quantities.
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.

Original worksheet page 1: question and worked solution for 4-2-007

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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.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 4-2-007

Laplace Transforms — Question 7

Question 7

Question 7 – Solution