anonymous
  • anonymous
Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Ok. So you split up your integrand into two parts, right?
anonymous
  • anonymous
I got 2+(2/(1+s)) and it was wrong
anonymous
  • anonymous
@Herp_Derp yes I did

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More answers

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

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