Question

find the inverse LaPlace transform of the following function:

F(s) = (e^(-s) - e^(-4s))/ s^2

Answer 1

\[F(s) = (e ^{-s} - e ^{-4s})/ s ^{2}\]

Answer 2

F(s) = (e^(-s) - e^(-4s))/ s^2\[F(s)=e^{-s}/s^2-e^{-4s}/s^2\]
we're going to do this as a heaviside function right? just thinking how

Answer 3

The Heaviside function/unit step function u(x) has Laplace transform 1/s.

The function u(x-a) has Laplace transform e^(-as)/s

So now you should have it.

Answer 4

Just be sure to simplify your answer. I.e., you can simplify and get rid of all the u(.) in the final answer; it is instructive if you do.

Answer 5

How do you get rid of the u in the final answer?

Answer 6

Oh sorry. You have 1/s^2. The Laplace transform of t is 1/s^2

Answer 7

Now the e^(-as) are shift functions

Answer 8

\[F(s)=e^{-s}(1/s^2)-e^{-4s}(1/s^2)\]yeah so we have do do it like e^(-s)G(s)

Answer 9

That is, if a function f(t) has Laplace transform F(s), then the function f(t-a) has Laplace transform e^(-as)F(s)

Answer 10

Hence, for example, the inverse transform of e^(-1s)/s^2 is (t-1)

Answer 11

so I would re write my equation as a step function and then take the LaPlace transform?

Answer 12

No -- now that realize we have 1/s^2 terms, we have no step functions involved.

Answer 13

Look at what I just did with e^(-1s)/s^2

Answer 14

sorry I meant to say shift function

Answer 15

the answer is just (t-1) - (t-4)

Answer 16

Now be careful about domains.

Answer 17

The inverse Laplace transform of e^(-s)/s^2 is t - 1, for t >= 1 and 0 elsewhere

Answer 18

therefore I need to include the Uc(t) function

Answer 19

Likewise the inverse Laplace transform of e^(-4s)/s^2 is
t - 4, t >=4
0, t < 4

So when you add these two functions you get what?

Answer 20

or subtract them actually.

Answer 21

Ok thank you, I do want to use the Uc(t) expression then.

Answer 22

The final answer is
f(t) = 0 , t < 1
= t - 1 , t in [1,4]
= (t-1) - (t-4), t > 4

Now simplify this.

Answer 23

I understand my instructor wants us to use Uc(t) expression he taught us, which gives the domain. Thank you so much

Answer 24

ok

Answer 25

I have another question.
Find the LaPlace transform of \[f(t) = 1/\sqrt{t}\]

Answer 26

I can't do that one, but it's number 4 on this chart:
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
I'm not sure how to use that capital Gamma thing though. I'm sure JamesJ does. You may need to repost the question.

Answer 27

thank you

Answer 28

Just saw this. Look up the gamma function; you'll find that the Laplace (not LaPlace ;-) ) transform is just applying the definition of that function to 1/sqrt(t) and the fact of the matter is except for integral value of gamma function, it's not meaningfully reducible to elementary functions.

Answer 29

blech, sounds ugly. I've heard of the Gamma function. Better study up on it.

Answer 30

I'm sure we all thought sin and cos sounded a bit complicated at first, but we later figured out they're quite elegant. Something is true with the Gamme Function too. It's quite nice.