Laplace transforms?

Question

If f(t) is continuous for t >= 0, the Laplace transform of f is the function F defined by

F(s) = int[0,infinity] f(t) e^(-st) dt

and the domain of F is the set consisting of all numbers s for which the integral converges. Find the Laplace transforms of the following functions:

1. f(t) = 1
2. f(t) = e^t
3. f(t) = t

(I know the answers, but I need to know HOW to get the answers.)

Answer 1

1)1. f(t) = 1

∫1.e^(-st) dt = [(-1/s)*e^(-st)] |0,∞

=(-1/s)*0 - (-1/s)*1

=1/s


2)f(t) = e^t

∫e^t.e^(-st) dt |0,∞ = ∫e^[-t(s-1)] *dt

=[ {-1/(s-1)} * {e^-(s-1)t} ] |0,∞

= 1/(s-1)


3) f(t) = t

∫t.e^(-st) dt

put st = x

L(t) = ∫(x/s)*e^(-x) dt |0,∞

= 1/s²

Answer 2

The Laplace Transform is a mathematical technique used to convert a real-valued function into a complex valued one. It has applications in many, mnay areas of physics and engineering such as optics, control theory, signal processing and electrical engineering. For a function f(t) the Laplace transform is F(s) = integral from 0* to infinity of exp(-s t) f(t) dt For example, the Laplace Transform of sin(t) is 1/(1+s²) s is usually a complex number. The reason this is useful is because integration and differentiation become multiplication and division in the Laplace domain. This very often simplifies the analysis of physical or mathematical systems. * (the lower limit is not strictly 0 but rather the limit as an aribitrarily samll quantity tends to zero).

Answer 3

Allrighty then, the main problem is that trying to analytically integrate a function with limits 0 to infinity is complicated. To get around this change infinity for any literal, say b, integrate and then calculate the limit of the resulting function as b tends to infinity. Simply put, instead of doing this:

F(s) = int[0,infinity] f(t) e^(-st) dt, which is the definition of the Laplace transform,

do this:

F(s) = lim(int[0,b] f(t) e^(-st) dt), where b tends to infinity.

During this process you'll realize that many things are easily simplified, just remember that e^0=1 and that e^-inf=0 and you'll be fine.

Answer 4

For the first, F(s) = ∫[0 - ∞] 1 * e^(-st)dt = -(1/s)*e^-(st)| 0,∞ = -(1/s)*[0 - 1] = 1/s

For the second F(s) = ∫e^t * e^-(st)dt[0 - ∞] = ∫e^(-st+t)dt = ∫e^(1-s)t = 1/(1-s)*e^(1-s)t [∞ - 0] = 1/(s-1)

For the third F(s) = ∫te^-(st)dt = (t+1/s)(-1/s)*e^-(st)[∞ - 0] = 1/s^2

Note: lim[t->∞] t*e^-(st) = 0

Answer 5

1.
int[0,infinity] f(t)e^(-st) dt
= (-1/s) int[0,infinity] 1* e^(-st) d(-st)
= 1/s

2.
int[0,infinity] e^t e^(-st) dt
= int[0,infinity] e^(1-s)t dt
= 1/(s-1)

3.
int[0,infinity] t e^(-st) dt
= int[0,infinity]- t/s de^(-st)
= -(t/s)e^(-st)[0,infinity] + int[0,infinity] (1/s) e^(-st) dt, integration by parts
= 1/s^2

Answer 6

just substitute in the defining formula!

Then, carry out the integration normally.