anonymous
  • anonymous
laplace transform of \[f(t)=u_1(t)+2u_3(t)-6u_4(t)\]and the answer is \[F(s)=s^{-2}[(1-s)e^{-2s}-(1+s)e^{-3s}]\]but I dont know how to get this
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
All I know is that the laplace of \(u_c(t)\)=\[\frac{e^{-cs}}{s}\]
Mertsj
  • Mertsj
Have to get the smart guys to help you with this.
anonymous
  • anonymous
well, you got the table right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so with the identity I have \[F(s)=\frac{e^{-s}}{s}+2\frac{e^{-3s}}{s}-6\frac{e^{-4s}}{s}\]and how do I make it look like the answer I have above?
anonymous
  • anonymous
and yes i do have the table
anonymous
  • anonymous
alright, let's try something different
anonymous
  • anonymous
i have another identity regarding step functions, it's this \[u_c(t)f(t-c)=>e^{-cs}F(s)\]
anonymous
  • anonymous
do you know how they look like though?
anonymous
  • anonymous
yeah i think
anonymous
  • anonymous
|dw:1364173680082:dw|
anonymous
  • anonymous
guys i apologize sincerely....... it looks like I have read the wrong answer to the problem......
anonymous
  • anonymous
ok, that happens
anonymous
  • anonymous
sorry and thanks for your help!!
UnkleRhaukus
  • UnkleRhaukus
i was gonna say...
UnkleRhaukus
  • UnkleRhaukus
\[f(t)=u_1(t)+2u_3(t)-6u_4(t)\] \[\mathcal L\big\{f(t)\big\}=\mathcal L\big\{u_1(t)\big\}+2\mathcal L\big\{u_3(t)\big\}-6\mathcal L\big\{u_4(t)\big\}\\\qquad\qquad=\frac {e^{-s}}s+2\frac {e^{-3s}}s-6\frac {e^{-4s}}s\\\qquad F(s)=\frac1s\big(e^{-s}+2e^{-3s}-6e^{-4s}\big)\]
nincompoop
  • nincompoop
table looks like this |dw:1364174488647:dw|
nincompoop
  • nincompoop
jk
UnkleRhaukus
  • UnkleRhaukus
|dw:1364174630421:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.