Consider the function f(t). Suppose that f\"\'(t) is continuous on [0, infinity)
ID: 3123544 • Letter: C
Question
Consider the function f(t). Suppose that f"'(t) is continuous on [0, infinity) and is of exponential order. Also suppose that lim t rightarrow infinity f"(t) e^-st = 0 for s > 0. Then show that L{f"'(t)} = s^3 F(s) - s^2 f(0) - s f'(0) - f"(0) by following these steps: (i) Start to evaluate the integral L {f"'(t)} = integral^infinity _0 f"'(t) e^-st dt using integration by parts ONLY ONE TIME and by setting u = e^-st, dv = f"'(t) dt. (ii) Then use the already computed formula L{f"(t)} = s^2 F(s) - s f(0) - f'(0). (iii) Also use the assumption mentioned above that lim t rightarrow infinity f"(t) e^-st = 0 for s > 0. Prove that F(s) = s^2/3s^2 + s + 1 is NOT the Laplace transform of any piecewise continuous function f(t) of exponential order.Explanation / Answer
A function f is said od exponential order if there exists a constant a and positive constants t0 and M such that.
Let f(t) begiven for t>=0 and assume the function satisfy certain conditions to be stated later on. The laplace transform of f(t), that it is denoted by L{f(t)} or F(s), it is defined by the equation
L{f(t)}= F(s) = e !st f(t)dt= o
wheneve rthe improper integral converges.