Answer:
hello attached is the detailed solution
Step-by-step explanation:
Laplace transform of the functions and the sketches are attached below
A) f(t) = u(t) + 3u(t-1) - 2u(t-2) - 2u(t-3)
F(s) = [tex]\frac{1 + 3e^{-s} -2e^{-2s}-2e^{-3s} }{s}[/tex]
B) f(t) = 2u(t+2) - 3u(t+1) +4u(t-1) - 3u(t-2)
F(s) = [tex]\frac{2e^{2s} - 3e^s + 4e^{-s} -3e^{2s} }{s}[/tex]
attached below is the remaining part of the answer
