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anonymous
 3 years ago
laplace transform of \[f(t)=u_1(t)+2u_3(t)6u_4(t)\]and the answer is \[F(s)=s^{2}[(1s)e^{2s}(1+s)e^{3s}]\]but I dont know how to get this
anonymous
 3 years ago
laplace transform of \[f(t)=u_1(t)+2u_3(t)6u_4(t)\]and the answer is \[F(s)=s^{2}[(1s)e^{2s}(1+s)e^{3s}]\]but I dont know how to get this

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0All I know is that the laplace of \(u_c(t)\)=\[\frac{e^{cs}}{s}\]

Mertsj
 3 years ago
Best ResponseYou've already chosen the best response.0Have to get the smart guys to help you with this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, you got the table right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0and yes i do have the table

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright, let's try something different

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i have another identity regarding step functions, it's this \[u_c(t)f(tc)=>e^{cs}F(s)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do you know how they look like though?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364173680082:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0guys i apologize sincerely....... it looks like I have read the wrong answer to the problem......

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry and thanks for your help!!

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1i was gonna say...

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[f(t)=u_1(t)+2u_3(t)6u_4(t)\] \[\mathcal L\big\{f(t)\big\}=\mathcal L\big\{u_1(t)\big\}+2\mathcal L\big\{u_3(t)\big\}6\mathcal L\big\{u_4(t)\big\}\\\qquad\qquad=\frac {e^{s}}s+2\frac {e^{3s}}s6\frac {e^{4s}}s\\\qquad F(s)=\frac1s\big(e^{s}+2e^{3s}6e^{4s}\big)\]

nincompoop
 3 years ago
Best ResponseYou've already chosen the best response.0table looks like this dw:1364174488647:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1364174630421:dw
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