## anonymous one year ago laplace transforms-process control

1. anonymous

2. dan815

if you're ever in doubt just write out the laplace transform integral

3. dan815

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4. anonymous

yep but f(t) is like two types of funtions... a ramp function which results in a new steady state value..

5. dan815

thats okay laplace transform is a linear operator so we can find laplace of each part and add it up

6. anonymous

so you propose finding the first function i.e the ramp function?

7. dan815

yes

8. anonymous

well a ramp function usually has a laplace of 1/s^2

9. dan815

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10. anonymous

yep. correct

11. anonymous

i think we usually define it as A for the step change etc but yeah k is fine

12. dan815

okay wait we gotta do something else here, to make sure stuff after 30 mins is 0

13. anonymous

want to find the slope or leave it in variable notation?

14. anonymous

yep

15. anonymous

16. dan815

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17. dan815

okay you will find lapalce of these 3 functions

18. dan815

one is that first slope

19. anonymous

i don't get what you did on the bottom of the axis

20. dan815

the 2nd is a slope line starting from down there with that U_c(t) shift

21. dan815

the one on the bottom is

22. dan815

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

you will add this so that the part after 30 mins gets negated

24. dan815

then you will simple add the constant function A

25. dan815

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26. dan815

basically what i did was rewrite this as 3 other functions which sum to what we want

27. anonymous

that extended dotted line is your step function right?

28. dan815

but now all 3 functions have a way to be reprsented from t=0 time

29. anonymous

let me grasp this on a piece of paper

30. dan815

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31. dan815

okay wait maybe this is just silly lol

32. dan815

i wonder if we can just simply break out integral like this

33. anonymous

nah i think you're on the right track

34. dan815

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35. dan815

like can we just break the integrall ike this for a piecewise function

36. anonymous

what i don't get is why you sorta did the reciprocial on the bottom of the axis

37. dan815

yep u can do this just break it up and solve

38. anonymous

but why did you draw the function with other parts, thats what i don't get...

39. dan815

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40. dan815

forget that that is a long annoying way this is better

41. anonymous

lel haha

42. anonymous

so just worry about the orginal function?

43. dan815

yes just break it up piecewise like that

44. dan815

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45. anonymous

yep

46. anonymous

so we know the laplace of the first integral is 1/s^2 and the laplace of the second is 1/s

47. anonymous

A/s

48. dan815

yep just do these integrals like like normal style

49. anonymous

we would have to use by parts method on the first one tho?

50. anonymous

$\int\limits_{0}^{30}te ^{-st}dt=\ \frac{ -1 }{ s }(30e ^{-30s})-\frac{ 1 }{ s^2 }(e ^{-30s}-1)$

51. dan815

yep go on

52. anonymous

$\int\limits_{30}^{\infty}e ^{-st}=\frac{ 1 }{ s }e ^{-30s}$

53. anonymous

as lim t>infty of e^(-st)=0

54. anonymous

so grouping...

55. anonymous

so would that be it?

56. anonymous

where A=55 and k=(75-20)/(30-0)

57. dan815

lol maybe the other way was faster since we know the laplace transforms of the 3 functiosn right away hehe

58. dan815

but whatever good to know both ways

59. anonymous

whats the other way?

60. dan815

you can do it the other way and compare the answer u are getting with this integration it should be equal

61. anonymous

i was confused what you did at first..how would u set the equations up in the first way

62. anonymous

so how would you go about doing the other way?

63. dan815

okay so first identify these 3 graphs

64. dan815

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65. dan815

the first 2 are pretty straight forward can you figure out the 3rd one

66. dan815

laplace of the 3rd graph

67. anonymous

68. anonymous

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

69. anonymous

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

anyone want to enlighten me?

71. anonymous

sorry, the net just cut out

72. dan815

yep but now for the 3rd one we can use that step function trick

73. dan815

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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75. dan815

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76. dan815

so its just this one constant we tacked onto the function

77. anonymous

thats the heavside function right?

78. dan815

yes

79. dan815

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80. anonymous

so b would be the shift of 55 downards?

81. dan815

75 i think because u gotta get rid of the stuff after that http://prntscr.com/8jtl4f

82. anonymous

we aren't starting from the origin tho?

83. dan815

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

84. dan815

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85. anonymous

right

86. dan815

right so its heave side to this function

87. dan815

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88. dan815

just math it lol u get the theory now

89. anonymous

yeah now i see, so T(s) is just the sum of those 3 identities?

90. dan815

all u gotta think is what kind of functions can i add that i can take the laplace of easily

91. dan815

yeah

92. dan815

think of other ways to get starting functions

93. dan815

do this one

94. dan815

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95. dan815

these 3

96. dan815

the sum of (1) + (2) + (3) = is the function you want

97. dan815

you can see that right

98. anonymous

:/ not really,

99. dan815

see what happens when you add graph 1 and 2

100. anonymous

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101. dan815

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102. dan815

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103. dan815

all the values after 75 keep cancelling out

104. dan815

at every coordinate on x

105. dan815

or t

106. dan815

thats what it means to add functions right

107. dan815

at every t value we add both of their y values

108. dan815

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

109. anonymous

makes much more sense!!!

110. anonymous

ok so those two end up like that ok so we have

111. dan815

okay yes thats what ive been doing

112. dan815

now we add that constant function heavsided to get that constant line

113. dan815

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114. anonymous

yeeeeeeeeeee

115. anonymous

i see it now

116. anonymous

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

yeah

118. dan815

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

119. anonymous

so T(t) = t + u(t-30) -75 + 55?

120. anonymous

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

121. dan815

http://www.efunda.com/math/laplace_transform/rules.cfm list of some basic laplace properties

122. dan815

i gotta watch a lecture for my class right now, ill be back, work thru it

123. anonymous

np, i try get my head around it

124. anonymous

@dan815

125. anonymous

you finished?

126. anonymous

@ganeshie8 could you help me with this?