Table Of Laplace Transforms: Question 4-10.2 & Solution

Question 2

Use causal one-sided Laplace transforms. Write $H_a(t)=0$ for $t<a$ and $1$ for $t\ge a$ . Ordinary functions are zero for $t<0$ . Justify the table entries and operational rules you use; give exact expressions.

For $a>0$ , define the delay and time-multiplication operations $(D_af)(t)=H_a(t)f(t-a),\qquad (Mf)(t)=t f(t).$ Take $f(t)=e^{-t}$ , and compare $p=D_a(Mf)$ with $q=M(D_af)$ .

Tasks

Derive the transform rules for $D_a$ and $M$ from their defining integrals. State a domain where the derivations are justified for this $f$ .
Find $p,q$ and their transforms. Explain exactly why applying the two operations in opposite orders changes the answer.
Prove the general identity $MD_af-D_aMf=aD_af$ whenever the expressions are defined, and verify its transform.
For $a=2$ , invert $e^{-2s}/(s+1)^2$ and $-\frac{d}{ds}[e^{-2s}/(s+1)]$ . Compute the area of each inverse and use it to distinguish them.

Original worksheet page 1: question and worked solution for 4-10-002

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Question 2 – Solution

Strategy. A delayed clock measures $t-a$ ; multiplication after the delay still uses the original clock $t$ .

Step 1: Derive the rules. The substitution $u=t-a$ gives $\mathcal L\{D_af\}=e^{-as}F(s)$ . Differentiating the transform integral with respect to $s$ gives $\mathcal L\{Mf\}=-F'(s)$ . For $f=e^{-t}$ and its delayed polynomial multiples, $s>-1$ makes these integrals absolutely convergent and permits this differentiation, locally dominated by an integrable exponential times a polynomial.

Step 2: Keep the clocks explicit. Here $p=H_a(t)(t-a)e^{-(t-a)},\qquad q=H_a(t)t e^{-(t-a)},$ so $P=\frac{e^{-as}}{(s+1)^2},\qquad Q=-\frac{d}{ds}\frac{e^{-as}}{s+1} =e^{-as}\left[\frac a{s+1}+\frac 1{(s+1)^2}\right].$ The extra term comes from differentiating the delay factor itself.

Step 3: Prove the operation identity. For $t<a$ both sides vanish. For $t\ge a$ , subtraction gives $t f(t-a)-(t-a)f(t-a)=a f(t-a)$ . Therefore $\boxed{MD_af-D_aMf=aD_af.}$ In transform language, $-(e^{-as}F)'-e^{-as}(-F')=ae^{-as}F$ , the same identity by the product rule.

Step 4: Invert and distinguish the two expressions. For $a=2$ the first inverse is $H_2(t)(t-2)e^{-(t-2)}$ , while the second is $H_2(t)t e^{-(t-2)}$ . Their areas, after putting $u=t-2$ , are $\int_0^\infty u e^{-u}\,du=\boxed{1},\qquad \int_0^\infty(u+2)e^{-u}\,du=\boxed{3}.$ Their transforms at zero have these same values. The first starts continuously from zero at the delay; the second has right-hand value $2$ there. Isolated endpoint conventions do not change either area.

Original worksheet page 2: question and worked solution for 4-10-002

Table Of Laplace Transforms — Question 2

Question 2

Question 2 – Solution