johnny0929
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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johnny0929
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All I know is that the laplace of \(u_c(t)\)=\[\frac{e^{cs}}{s}\]

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

modphysnoob
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well, you got the table right?

johnny0929
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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?

johnny0929
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and yes i do have the table

modphysnoob
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alright, let's try something different

johnny0929
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i have another identity regarding step functions, it's this \[u_c(t)f(tc)=>e^{cs}F(s)\]

modphysnoob
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do you know how they look like though?

johnny0929
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yeah i think

modphysnoob
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dw:1364173680082:dw

johnny0929
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guys i apologize sincerely....... it looks like I have read the wrong answer to the problem......

modphysnoob
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ok, that happens

johnny0929
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sorry and thanks for your help!!

UnkleRhaukus
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i was gonna say...

UnkleRhaukus
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\[f(t)=u_1(t)+2u_3(t)6u_4(t)\]
\[\mathcal L\big\{f(t)\big\}=\mathcal L\big\{u_1(t)\big\}+2\mathcal L\big\{u_3(t)\big\}6\mathcal L\big\{u_4(t)\big\}\\\qquad\qquad=\frac {e^{s}}s+2\frac {e^{3s}}s6\frac {e^{4s}}s\\\qquad F(s)=\frac1s\big(e^{s}+2e^{3s}6e^{4s}\big)\]

nincompoop
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table looks like this
dw:1364174488647:dw

nincompoop
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jk

UnkleRhaukus
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dw:1364174630421:dw