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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

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  1. johnny0929
    • 2 years ago
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    All I know is that the laplace of \(u_c(t)\)=\[\frac{e^{-cs}}{s}\]

  2. Mertsj
    • 2 years ago
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    Have to get the smart guys to help you with this.

  3. modphysnoob
    • 2 years ago
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    well, you got the table right?

  4. johnny0929
    • 2 years ago
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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?

  5. johnny0929
    • 2 years ago
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    and yes i do have the table

  6. modphysnoob
    • 2 years ago
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    alright, let's try something different

  7. johnny0929
    • 2 years ago
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    i have another identity regarding step functions, it's this \[u_c(t)f(t-c)=>e^{-cs}F(s)\]

  8. modphysnoob
    • 2 years ago
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    do you know how they look like though?

  9. johnny0929
    • 2 years ago
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    yeah i think

  10. modphysnoob
    • 2 years ago
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    |dw:1364173680082:dw|

  11. johnny0929
    • 2 years ago
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    guys i apologize sincerely....... it looks like I have read the wrong answer to the problem......

  12. modphysnoob
    • 2 years ago
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    ok, that happens

  13. johnny0929
    • 2 years ago
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    sorry and thanks for your help!!

  14. UnkleRhaukus
    • 2 years ago
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    i was gonna say...

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

  16. nincompoop
    • 2 years ago
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    table looks like this |dw:1364174488647:dw|

  17. nincompoop
    • 2 years ago
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    jk

  18. UnkleRhaukus
    • 2 years ago
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    |dw:1364174630421:dw|

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