Question

Question 1a:

Compute the Laplace Transforms of

f(x) = { x² +2x +1 , x < 3
{ e⁻⁵ˣ , x ≥ 3


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Answer 1

This is made easy if you know the Laplace transform pair

$u(t - c) y(t - c) \mapsto e^{-cs} Y(s)$

where u(t) is the Heaviside step function and Y(s) is the Laplace transform of y(t).

We rewrite f, g, and h in terms of u :

$f(x) = (x^2+2x+1) (u(x) - u(x-3)) + e^{-5x} u(x-3)$

$g(x) = \sin(3x) (u(x-5) - u(x-10))$

$h(x) = (u(x) - u(x-4)) + 2(u(x-4)-u(x-8)) + 3 (u(x-8)-u(x-16))$

and we also rearrange terms to get the right shift in argument:

$f(x) = (x^2+2x+1) u(x) - ((x-3)^2+8(x-3)+16) u(x-3) + e^{-15} e^{-5(x-3)} u(x-3)$

$g(x) = [\cos(15) \sin(3(x-5)) + \sin(15) \cos(3(x-5))] u(x-5) \\\\ ~~~~~~~~~~- [\cos(30) \sin(3(x-10)) + \sin(30) \cos(3(x-10))] u(x - 10)$

$h(x) = u(x) + u(x-4) + u(x-8) - 3 u(x - 16)$

Now we can apply the known transform pair, along with

$x^n \mapsto \dfrac{n!}{s^{n+1}}$

$e^{ax} \mapsto \dfrac1{s-a}$

$\sin(ax) \mapsto \dfrac{a}{s^2+a^2}$

$\cos(ax) \mapsto \dfrac{s}{s^2+a^2}$

The transforms of f, g, and h are then

$F(s) = e^{-s} \left(\dfrac2{s^3} + \dfrac2{s^2} + \dfrac1s\right) - e^{-3s} \left(\dfrac2{s^3} + \dfrac8{s^2} + \dfrac{16}s\right) + \dfrac{e^{-15} e^{-3s}}{s + 5} \\\\ = \boxed{e^{-s} \left(\dfrac2{s^3} + \dfrac2{s^2} + \dfrac1s\right) - e^{-3s} \left(\dfrac2{s^3} + \dfrac8{s^2} + \dfrac{16}s\right) + \dfrac{e^{-3(s+5)}}{s + 5}}$

$G(s) = \dfrac{3\cos(15) e^{-5s}}{s^2+9} + \dfrac{s \sin(15) e^{-5s}}{s^2+9} - \dfrac{3\cos(30)e^{-10s}}{s^2+9} - \dfrac{s \sin(30) e^{-10s}}{s^2+9} \\\\ ~~~~~~~ = \dfrac{3(\cos(15)-\cos(30)) + s(\sin(15) - \sin(30))}{s^2+9} \\\\ ~~~~~~~ = \boxed{\dfrac{6\sin\left(\frac{45}2\right)\sin\left(\frac{15}2\right) - 2s \cos\left(\frac{45}2\right)\sin\left(\frac{15}2\right)}{s^2+9}}$

$H(s) = \boxed{\dfrac{e^{-s} + e^{-4s} + e^{-8s} - 3 e^{-16s}}s}$