Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Ok. So you split up your integrand into two parts, right?
I got 2+(2/(1+s)) and it was wrong
@Herp_Derp yes I did

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[L\{f(t)\}=\int_{0}^{inf}e^{-st}f(t)~dt\]
in this case, you split the integral up into 2 sections
I did that and I was wrong :(
can you type up your work? then we can see where the mistake might have happened
theres a shifting method that i can never remember, so i tend to do it the splitted way
yea 2\[2\int\limits_{0}^{2}e^(-st)+\int\limits_{0}^{infinity}4e^(-st)dt\]
it's supposed to be e^(-st)
0,2; then 2,inf for starters; and dont forget your "t" on the left side
f(t)=-2e^(-2s)/s + 2/s
oh sorry i actually have that down on my paper but I just typed it wrong
\[2\int_{0}^{2}e^{-st}t+4\int_{2}^{inf}e^{-st}dt\]
\[\large \int_0^2 2te^{-st}\,dt+\int_0^{\infty}4e^{-st}\,dt=2\int_0^2 te^{-st}\,dt+4\int_0^{\infty}e^{-st}\,dt\]Just making it look better :)
@amistre64 thats what I have. What do I do afterwards but thats where I went wrong
Thanks @Herp_Derp but I have that but afterwards is where I'm confused
2 to inf on the last one, the long right term is simple enough; the left side is by parts
I know I got that but afterwards is where I'm confused
for the firsr part the integral ends up equaling (3(1-e^(-2s)(2s+1))/(s^2)
and the second integral is (e^(-2s))/s
so (3(1-e^(-2s)(2s+1))/(s^2) +(e^(-2s))/s
\[2\int_0^2 te^{-st}\,dt+4\int_2^{\infty}e^{-st}\,dt\] \[2(~\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\frac{1}{s}\left(\int_0^2 e^{-st}\,dt\right)+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\]
this thing aint latex friendly :) \[2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\frac{2}{s}\left(\int_0^2 e^{-st}\,dt\right)+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\] \[2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\left.\frac{2}{s}\frac{e^{-st}}{-s}\right|_{t=0}^{t=2}+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\] \[2\frac{te^{-2s}}{-s}-2\frac{te^{0}}{-s}+\frac{2}{s}\frac{e^{-2s}}{-s}-\frac{2}{s}\frac{e^{0}}{-s}+4\frac{e^{-s(inf)}}{-s}-4\frac{e^{-2s}}{-s}\]
do we stop there?
id clean it up a little if i was you
ok thanks!
\[-\frac{2te^{-2s}}{s}+\frac{2t}{s}-\frac{2e^{-2s}}{s^2}+\frac{2}{s^2}+\cancel{4\frac{e^{-s(inf)}}{-s}}+\frac{4e^{-2s}}{s}\] stuff like that
it's wrong :(
great time for loading issues eh
\[2\int_{0}^{2}e^{-st}t+4\int_{2}^{inf}e^{-st}dt\] \[2(\frac{e^{-2s}-1}{s^2})+4\frac{e^{-2s}}{s}\] or some variation of that should work
i never did get into the heaviside stuff to be competent enough in that method
sorry openstuy froze up! an it's still wrong

Not the answer you are looking for?

Search for more explanations.

Ask your own question