a) Derive an analytical expression for the τs component of the solution (equivalently, execution or
trajectory) starting from τs(0, 0) = T ∗ s 2 and ms(0, 0) = 0. Write it in terms of hybrid time (t, j).
• The solution initially flows until τs = T ∗ s . When τs reaches T ∗ s , the solution jumps to zero.
• Since τs(0, 0) = T ∗ s 2 , the first jump occurs at t = T ∗ s 2 .
• m˙ s = 0 and ms(0, 0) = 0 =⇒ ms stays at zero during the time [0, T ∗ s 2 ].
• At j = 1, the timer τs is reset to zero and vs is used to update ms (i.e., τ + s = 0, m+ s = vs).
b1)
What are the values of the ms components of the solution from τs(0, 0) = 0 and ms(0, 0) = 0 at
the following specific hybrid times:
(t, j) = (0.5, 0),(1, 0),(1, 1)
• when t = 0.5, j = 0, ms(0.5, 0) = 0
• when t = 1, j = 0, ms(1, 0) = 0
• when t = 1, j = 1, ms(1, 1) = vs(1, 1) = sin(2πt) = 0.
b2):
Plot as a function of (t, j) the component τs of the
execution from τs(0, 0) = 0 and ms(0, 0) = 0
