## johnny0929 2 years ago 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

1. johnny0929

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

2. Mertsj

3. modphysnoob

well, you got the table right?

4. johnny0929

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

and yes i do have the table

6. modphysnoob

alright, let's try something different

7. johnny0929

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

8. modphysnoob

do you know how they look like though?

9. johnny0929

yeah i think

10. modphysnoob

|dw:1364173680082:dw|

11. johnny0929

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

12. modphysnoob

ok, that happens

13. johnny0929

sorry and thanks for your help!!

14. UnkleRhaukus

i was gonna say...

15. UnkleRhaukus

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

16. nincompoop

table looks like this |dw:1364174488647:dw|

17. nincompoop

jk

18. UnkleRhaukus

|dw:1364174630421:dw|