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anonymous

  • one year ago

laplace transforms-process control

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  1. anonymous
    • one year ago
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  2. dan815
    • one year ago
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    if you're ever in doubt just write out the laplace transform integral

  3. dan815
    • one year ago
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    |dw:1443058593494:dw|

  4. anonymous
    • one year ago
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    yep but f(t) is like two types of funtions... a ramp function which results in a new steady state value..

  5. dan815
    • one year ago
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    thats okay laplace transform is a linear operator so we can find laplace of each part and add it up

  6. anonymous
    • one year ago
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    so you propose finding the first function i.e the ramp function?

  7. dan815
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    well a ramp function usually has a laplace of 1/s^2

  9. dan815
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    |dw:1443058802969:dw|

  10. anonymous
    • one year ago
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    yep. correct

  11. anonymous
    • one year ago
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    i think we usually define it as A for the step change etc but yeah k is fine

  12. dan815
    • one year ago
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    okay wait we gotta do something else here, to make sure stuff after 30 mins is 0

  13. anonymous
    • one year ago
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    want to find the slope or leave it in variable notation?

  14. anonymous
    • one year ago
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    yep

  15. anonymous
    • one year ago
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    thats what i'm confused about

  16. dan815
    • one year ago
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    |dw:1443059031652:dw|

  17. dan815
    • one year ago
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    okay you will find lapalce of these 3 functions

  18. dan815
    • one year ago
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    one is that first slope

  19. anonymous
    • one year ago
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    i don't get what you did on the bottom of the axis

  20. dan815
    • one year ago
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    the 2nd is a slope line starting from down there with that U_c(t) shift

  21. dan815
    • one year ago
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    the one on the bottom is

  22. dan815
    • one year ago
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    f(t)=kt-b lets say and to that we applied a unit step function L{U_c(t)*f(t-c)}=e^cs*L{f(t)}

  23. dan815
    • one year ago
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    you will add this so that the part after 30 mins gets negated

  24. dan815
    • one year ago
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    then you will simple add the constant function A

  25. dan815
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    |dw:1443059246798:dw|

  26. dan815
    • one year ago
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    basically what i did was rewrite this as 3 other functions which sum to what we want

  27. anonymous
    • one year ago
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    that extended dotted line is your step function right?

  28. dan815
    • one year ago
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    but now all 3 functions have a way to be reprsented from t=0 time

  29. anonymous
    • one year ago
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    let me grasp this on a piece of paper

  30. dan815
    • one year ago
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    |dw:1443059322182:dw|

  31. dan815
    • one year ago
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    okay wait maybe this is just silly lol

  32. dan815
    • one year ago
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    i wonder if we can just simply break out integral like this

  33. anonymous
    • one year ago
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    nah i think you're on the right track

  34. dan815
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    |dw:1443059492791:dw|

  35. dan815
    • one year ago
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    like can we just break the integrall ike this for a piecewise function

  36. anonymous
    • one year ago
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    what i don't get is why you sorta did the reciprocial on the bottom of the axis

  37. dan815
    • one year ago
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    yep u can do this just break it up and solve

  38. anonymous
    • one year ago
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    but why did you draw the function with other parts, thats what i don't get...

  39. dan815
    • one year ago
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    |dw:1443059801163:dw|

  40. dan815
    • one year ago
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    forget that that is a long annoying way this is better

  41. anonymous
    • one year ago
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    lel haha

  42. anonymous
    • one year ago
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    so just worry about the orginal function?

  43. dan815
    • one year ago
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    yes just break it up piecewise like that

  44. dan815
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    |dw:1443059991702:dw|

  45. anonymous
    • one year ago
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    yep

  46. anonymous
    • one year ago
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    so we know the laplace of the first integral is 1/s^2 and the laplace of the second is 1/s

  47. anonymous
    • one year ago
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    A/s

  48. dan815
    • one year ago
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    yep just do these integrals like like normal style

  49. anonymous
    • one year ago
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    we would have to use by parts method on the first one tho?

  50. anonymous
    • one year ago
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    \[\int\limits_{0}^{30}te ^{-st}dt=\ \frac{ -1 }{ s }(30e ^{-30s})-\frac{ 1 }{ s^2 }(e ^{-30s}-1)\]

  51. dan815
    • one year ago
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    yep go on

  52. anonymous
    • one year ago
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    \[\int\limits_{30}^{\infty}e ^{-st}=\frac{ 1 }{ s }e ^{-30s}\]

  53. anonymous
    • one year ago
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    as lim t>infty of e^(-st)=0

  54. anonymous
    • one year ago
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    so grouping...

  55. anonymous
    • one year ago
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    so would that be it?

  56. anonymous
    • one year ago
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    where A=55 and k=(75-20)/(30-0)

  57. dan815
    • one year ago
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    lol maybe the other way was faster since we know the laplace transforms of the 3 functiosn right away hehe

  58. dan815
    • one year ago
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    but whatever good to know both ways

  59. anonymous
    • one year ago
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    whats the other way?

  60. dan815
    • one year ago
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    you can do it the other way and compare the answer u are getting with this integration it should be equal

  61. anonymous
    • one year ago
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    i was confused what you did at first..how would u set the equations up in the first way

  62. anonymous
    • one year ago
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    so how would you go about doing the other way?

  63. dan815
    • one year ago
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    okay so first identify these 3 graphs

  64. dan815
    • one year ago
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    |dw:1443062354574:dw|

  65. dan815
    • one year ago
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    the first 2 are pretty straight forward can you figure out the 3rd one

  66. dan815
    • one year ago
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    laplace of the 3rd graph

  67. anonymous
    • one year ago
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    sorry i had lunch

  68. anonymous
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    isn't the third graph simply the first one just in diff order...

  69. anonymous
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    the first one has laplace k/s^2 the second identity has laplace A/s third.. well thats just a linear line then a slope like the question

  70. anonymous
    • one year ago
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    anyone want to enlighten me?

  71. anonymous
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    sorry, the net just cut out

  72. dan815
    • one year ago
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    yep but now for the 3rd one we can use that step function trick

  73. dan815
    • one year ago
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    the 3rd graph is very similar to some step function line graph that is shifted down on the y axis now

  74. dan815
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    |dw:1443065288309:dw|

  75. dan815
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    |dw:1443065315083:dw|

  76. dan815
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    so its just this one constant we tacked onto the function

  77. anonymous
    • one year ago
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    thats the heavside function right?

  78. dan815
    • one year ago
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    yes

  79. dan815
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    |dw:1443065438565:dw|

  80. anonymous
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    so b would be the shift of 55 downards?

  81. dan815
    • one year ago
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    75 i think because u gotta get rid of the stuff after that http://prntscr.com/8jtl4f

  82. anonymous
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    we aren't starting from the origin tho?

  83. dan815
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    yeah its the heaveside aplied to a line shifted right and down

  84. dan815
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    |dw:1443065665540:dw|

  85. anonymous
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    right

  86. dan815
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    right so its heave side to this function

  87. dan815
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    |dw:1443065735982:dw|

  88. dan815
    • one year ago
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    just math it lol u get the theory now

  89. anonymous
    • one year ago
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    yeah now i see, so T(s) is just the sum of those 3 identities?

  90. dan815
    • one year ago
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    all u gotta think is what kind of functions can i add that i can take the laplace of easily

  91. dan815
    • one year ago
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    yeah

  92. dan815
    • one year ago
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    think of other ways to get starting functions

  93. dan815
    • one year ago
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    do this one

  94. dan815
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    |dw:1443065897265:dw|

  95. dan815
    • one year ago
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    these 3

  96. dan815
    • one year ago
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    the sum of (1) + (2) + (3) = is the function you want

  97. dan815
    • one year ago
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    you can see that right

  98. anonymous
    • one year ago
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    :/ not really,

  99. dan815
    • one year ago
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    see what happens when you add graph 1 and 2

  100. anonymous
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    |dw:1443066102562:dw|

  101. dan815
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    |dw:1443066092625:dw|

  102. dan815
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    |dw:1443066187929:dw|

  103. dan815
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    all the values after 75 keep cancelling out

  104. dan815
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    at every coordinate on x

  105. dan815
    • one year ago
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    or t

  106. dan815
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    thats what it means to add functions right

  107. dan815
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    at every t value we add both of their y values

  108. dan815
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    if they have the same slope and one is starting exactly negative of it, then u have to keep cancelling

  109. anonymous
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    makes much more sense!!!

  110. anonymous
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    ok so those two end up like that ok so we have

  111. dan815
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    okay yes thats what ive been doing

  112. dan815
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    now we add that constant function heavsided to get that constant line

  113. dan815
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    |dw:1443066363555:dw|

  114. anonymous
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    yeeeeeeeeeee

  115. anonymous
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    i see it now

  116. anonymous
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    so graph 2 is just a ramp function shifted 30units of t to the right and shifted down 75 units |dw:1443066491364:dw|

  117. dan815
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    yeah

  118. dan815
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    finding the laplace transform of these functions is much simpler, now u can add them to get the desired effect

  119. anonymous
    • one year ago
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    so T(t) = t + u(t-30) -75 + 55?

  120. anonymous
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    cause the first has slope t, second has that shift of heavside function and the thirs is just a step function?

  121. dan815
    • one year ago
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    http://www.efunda.com/math/laplace_transform/rules.cfm list of some basic laplace properties

  122. dan815
    • one year ago
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    i gotta watch a lecture for my class right now, ill be back, work thru it

  123. anonymous
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    np, i try get my head around it

  124. anonymous
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    @dan815

  125. anonymous
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    you finished?

  126. anonymous
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    @ganeshie8 could you help me with this?

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