Find y(t), having u(t) and transfer function

Discussion in 'Homework Help' started by ellosma, Jan 9, 2014.

  1. ellosma

    Thread Starter New Member

    Jan 9, 2014
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    I have a exercise and i can't do a part of it. I've a transfert function G(s) = 1/(s+1) and u(t), the entering signal:

    u(t) is
    1+2t for 0<=t<1
    t-1 for t>=1

    I have to find y(t), the outcoming signal of the system.

    I find, correctly, the signal y(t) about the first part of u(t), 1+2t:
    y1(t)=2t + e^{-t} -1

    First i use laplace transform on u(t), for have U(s) , only the first part
    Second i know that Y(s)= G(s)U(s) so, i calculate Y(s) because i've G(s) and U(s)
    Antitransform Y(s) for have y(t)

    But now i don't know how to find y(t) for the second part, t-1. Could someone tell me how to find it, i'd like to comprend because it's not an homework but an exercise that i do for preparing me to an exam. Thanks!
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Do the Laplace transform of t-1, the interval of integration is from 1 to infinity.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    The "best" way to do it, unless it introduces too much complexity which it may or may not, is to take the Laplace transform of your u(t). Not the first part of it, but the whole thing.

    So use unit step functions to write a single equation for u(t) and then take the Laplace transform of that one equation and proceed from there.

    The alternative is to treat your signal as two different signals, one from 0 < t < 1 and a second from 1 < t <∞. Keep in mind that to do this, you will need to determine two sets of initial conditions -- one at t=0 and one at t=1.

    A third way it to take the inverse Laplace transform of H(s) to get h(t) and then use convolution.
     
  4. ellosma

    Thread Starter New Member

    Jan 9, 2014
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    For solve the problem in the first way, that you say it's the best, how i could write the intere laplace transformation of the first and second part ?

    I write u(t) as : 1+2t - (1+2t)*1(t-1) + (t-1)*1(t-1) and now i should apply the laplace transform for have U(s). But i don't know if it's correct :/ really Thanks for the help, i'm going crazy with this ex!
     
  5. rsashwinkumar

    New Member

    Feb 15, 2011
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    First i suggest you to plot your u(t) in time domain; Differentiate it w.r.t. time, twice and you will finally end up with some impulses. Then apply differentiation property and find Laplace transform of u(t);

    Now you can easily determine the inverse laplace of U(s)*1/(s+1);
     
    Last edited: Jan 10, 2014
  6. WBahn

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    Mar 31, 2012
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    Please. The first part of your post is just fine -- great, in fact. But then, after giving a good suggestion on a possible path to take, you then proceed to just give out the answer to that part instead of letting the OP struggle through it on their own for awhile first. That is frowned upon.
     
    Last edited: Jan 10, 2014
  7. rsashwinkumar

    New Member

    Feb 15, 2011
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    Point noted.! Edited the post..
     
  8. WBahn

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    Mar 31, 2012
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    Thanks! I've edited my post to also remove the answer from it.
     
  9. ellosma

    Thread Starter New Member

    Jan 9, 2014
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    Really Thanks, now i'm oggi to translate some of your advice because i'm italian and not so good in english. I've the complet solution of this exercise, but the first part i solved ti so i don't look the solution. But for the second part, in that i've problems , my book find the second part of y(t) writing
    y(t) = 2t - 1 + e^{-t} - [ t - 1 + 2 -2e^{-(t-1)}]

    I don't know how is the espression between [....] ?! Sorry for the question and really Thanks for the help , i'd like to comprend and after i'll do some ochre ex like this !!
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Please show your work as far as you were able to get. Then we can see exactly where and what concept is giving you grief and can help you understand that concept better. That will also help with the language issue because it will better focus on onto the same point of the process.
     
  11. ellosma

    Thread Starter New Member

    Jan 9, 2014
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    [​IMG]
    In the post i put the photo of that i wrote. I don't put the photo of the Two inverse laplace transform that i do but if you want i can send also them. The result of the first part is correct but not the second.. Thanks and sorry for my bad english!
     
  12. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Write out how you did second part on paper and post the picture of the paper.
     
  13. ellosma

    Thread Starter New Member

    Jan 9, 2014
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    [​IMG]

    I know that the error is that i study the first part, and the second as Two indipendent part but i should consider the second as the joined with the y(t) of the first,but i don't know how
     
  14. ellosma

    Thread Starter New Member

    Jan 9, 2014
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  15. bertus

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  16. ellosma

    Thread Starter New Member

    Jan 9, 2014
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    Ok ,Sorry i posted the image on dropbox but now it should be ok
     
  17. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    t->1/s^2 when the integral is from 0 to infinity. You don't have integral from 0 to infinity. You have integral from 1 to infinity. That is why this is wrong.

    1->1/s when the integral is from 0 to infinity. You don't have integal from 0 to infinity. You have integral from 1 to infinity. That is why this is also wrong.
     
  18. ellosma

    Thread Starter New Member

    Jan 9, 2014
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    I know i'm a stalker, i'm really Sorry, but i don't know how to solve the problem. May e i had to use the convolution? :( how to join the y(t) of the first part with the u(t) of the second?
     
  19. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Do the integral.
    u(t)=t-1, for t\geq1
    U(s)=\int ^\infty_1 (t-1)e^{-st}dt

    U(s)=\int ^\infty_1 te^{-st}dt-\int ^\infty_1 1e^{-st}dt

    U1(s)=\int ^\infty_1 te^{-st}dt

    U2(s)=\int ^\infty_1 1e^{-st}dt

    Consult Table of Integrals.
    Plug in infinity and 1 to get an actual answer for U1(s) and U2(s).
    U(s)=U1(s)-U2(s)
     
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