Consider the function f t t 0 lessthanorequalto t SolutionN

Consider the function f (t) = {t 0 lessthanorequalto t

Solution

Now f(0) = t = 0

f(1) = 4-2t = 4-2 = 2

f(2) = 1

Plotting the given point on the graph we will get our figure using MATLAB which will be somewhat overlapping figure.

(b). f(t) = t[u(t-0) -u(t-1)] + (4-2t)[u(t-1) - u(t-2)] +1[u(t-2) - u(t-)]

which becomes

f(t) = t[u(t) - u(t-1)] + (4-2t)[u(t-1) - u(t-2)] + [u(t-2)]

(c) We use {u(ta)}=e^-as/s

We will not use linear property since there is only a single function.

(d) Similarly we can find e^tf

Using  {u(ta)}=e^-as/s

We will use the Time Displacement Theorem:

{u(ta)g(ta)}=e^-asG(s)