Question

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

Answer 1

All I know is that the laplace of \(u_c(t)\)=\[\frac{e^{-cs}}{s}\]

Answer 2

Have to get the smart guys to help you with this.

Answer 3

well, you got the table right?

Answer 4

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?

Answer 5

and yes i do have the table

Answer 6

alright, let's try something different

Answer 7

i have another identity regarding step functions, it's this \[u_c(t)f(t-c)=>e^{-cs}F(s)\]

Answer 8

do you know how they look like though?

Answer 9

yeah i think

Answer 10

guys i apologize sincerely....... it looks like I have read the wrong answer to the problem......

Answer 11

ok, that happens

Answer 12

sorry and thanks for your help!!

Answer 13

i was gonna say...

Answer 14

\[f(t)=u_1(t)+2u_3(t)-6u_4(t)\]
\[\mathcal L\big\{f(t)\big\}=\mathcal L\big\{u_1(t)\big\}+2\mathcal L\big\{u_3(t)\big\}-6\mathcal L\big\{u_4(t)\big\}\\\qquad\qquad=\frac {e^{-s}}s+2\frac {e^{-3s}}s-6\frac {e^{-4s}}s\\\qquad F(s)=\frac1s\big(e^{-s}+2e^{-3s}-6e^{-4s}\big)\]

Answer 15

table looks like this

Answer 16

jk