Respuesta :
Answer
(a) [tex]F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}[/tex]
(b) [tex]F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}[/tex]
Step-by-step explanation:
(a) [tex]f(t) = 20.5 + 10t + t^2 +[/tex] δ(t)
where δ(t) = unit impulse function
The Laplace transform of function f(t) is given as:
[tex]F(s) = \int\limits^a_0 f(s)e^{-st} \, dt[/tex]
where a = ∞
=> [tex]F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt[/tex]
where d(t) = δ(t)
=> [tex]F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt[/tex]
Integrating, we have:
=> [tex]F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.[/tex]
Inputting the boundary conditions t = a = ∞, t = 0:
[tex]F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}[/tex]
(b) [tex]f(t) = e^{-t} + 4e^{-4t} + te^{-3t}[/tex]
The Laplace transform of function f(t) is given as:
[tex]F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt[/tex]
[tex]F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt[/tex]
[tex]F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt[/tex]
Integrating, we have:
[tex]F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.[/tex]
Inputting the boundary condition, t = a = ∞, t = 0:
[tex]F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}[/tex]
