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## johnny0929 Group Title 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 one year ago one year ago

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1. johnny0929 Group Title

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

2. Mertsj Group Title

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

3. modphysnoob Group Title

well, you got the table right?

4. johnny0929 Group Title

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?

5. johnny0929 Group Title

and yes i do have the table

6. modphysnoob Group Title

alright, let's try something different

7. johnny0929 Group Title

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

8. modphysnoob Group Title

do you know how they look like though?

9. johnny0929 Group Title

yeah i think

10. modphysnoob Group Title

|dw:1364173680082:dw|

11. johnny0929 Group Title

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

12. modphysnoob Group Title

ok, that happens

13. johnny0929 Group Title

sorry and thanks for your help!!

14. UnkleRhaukus Group Title

i was gonna say...

15. UnkleRhaukus Group Title

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

16. nincompoop Group Title

table looks like this |dw:1364174488647:dw|

17. nincompoop Group Title

jk

18. UnkleRhaukus Group Title

|dw:1364174630421:dw|