anonymous
  • anonymous
laplace transforms-process control
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
dan815
  • dan815
if you're ever in doubt just write out the laplace transform integral
dan815
  • dan815
|dw:1443058593494:dw|

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anonymous
  • anonymous
yep but f(t) is like two types of funtions... a ramp function which results in a new steady state value..
dan815
  • dan815
thats okay laplace transform is a linear operator so we can find laplace of each part and add it up
anonymous
  • anonymous
so you propose finding the first function i.e the ramp function?
dan815
  • dan815
yes
anonymous
  • anonymous
well a ramp function usually has a laplace of 1/s^2
dan815
  • dan815
|dw:1443058802969:dw|
anonymous
  • anonymous
yep. correct
anonymous
  • anonymous
i think we usually define it as A for the step change etc but yeah k is fine
dan815
  • dan815
okay wait we gotta do something else here, to make sure stuff after 30 mins is 0
anonymous
  • anonymous
want to find the slope or leave it in variable notation?
anonymous
  • anonymous
yep
anonymous
  • anonymous
thats what i'm confused about
dan815
  • dan815
|dw:1443059031652:dw|
dan815
  • dan815
okay you will find lapalce of these 3 functions
dan815
  • dan815
one is that first slope
anonymous
  • anonymous
i don't get what you did on the bottom of the axis
dan815
  • dan815
the 2nd is a slope line starting from down there with that U_c(t) shift
dan815
  • dan815
the one on the bottom is
dan815
  • 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)}
dan815
  • dan815
you will add this so that the part after 30 mins gets negated
dan815
  • dan815
then you will simple add the constant function A
dan815
  • dan815
|dw:1443059246798:dw|
dan815
  • dan815
basically what i did was rewrite this as 3 other functions which sum to what we want
anonymous
  • anonymous
that extended dotted line is your step function right?
dan815
  • dan815
but now all 3 functions have a way to be reprsented from t=0 time
anonymous
  • anonymous
let me grasp this on a piece of paper
dan815
  • dan815
|dw:1443059322182:dw|
dan815
  • dan815
okay wait maybe this is just silly lol
dan815
  • dan815
i wonder if we can just simply break out integral like this
anonymous
  • anonymous
nah i think you're on the right track
dan815
  • dan815
|dw:1443059492791:dw|
dan815
  • dan815
like can we just break the integrall ike this for a piecewise function
anonymous
  • anonymous
what i don't get is why you sorta did the reciprocial on the bottom of the axis
dan815
  • dan815
yep u can do this just break it up and solve
anonymous
  • anonymous
but why did you draw the function with other parts, thats what i don't get...
dan815
  • dan815
|dw:1443059801163:dw|
dan815
  • dan815
forget that that is a long annoying way this is better
anonymous
  • anonymous
lel haha
anonymous
  • anonymous
so just worry about the orginal function?
dan815
  • dan815
yes just break it up piecewise like that
dan815
  • dan815
|dw:1443059991702:dw|
anonymous
  • anonymous
yep
anonymous
  • anonymous
so we know the laplace of the first integral is 1/s^2 and the laplace of the second is 1/s
anonymous
  • anonymous
A/s
dan815
  • dan815
yep just do these integrals like like normal style
anonymous
  • anonymous
we would have to use by parts method on the first one tho?
anonymous
  • anonymous
\[\int\limits_{0}^{30}te ^{-st}dt=\ \frac{ -1 }{ s }(30e ^{-30s})-\frac{ 1 }{ s^2 }(e ^{-30s}-1)\]
dan815
  • dan815
yep go on
anonymous
  • anonymous
\[\int\limits_{30}^{\infty}e ^{-st}=\frac{ 1 }{ s }e ^{-30s}\]
anonymous
  • anonymous
as lim t>infty of e^(-st)=0
anonymous
  • anonymous
so grouping...
anonymous
  • anonymous
so would that be it?
anonymous
  • anonymous
where A=55 and k=(75-20)/(30-0)
dan815
  • dan815
lol maybe the other way was faster since we know the laplace transforms of the 3 functiosn right away hehe
dan815
  • dan815
but whatever good to know both ways
anonymous
  • anonymous
whats the other way?
dan815
  • dan815
you can do it the other way and compare the answer u are getting with this integration it should be equal
anonymous
  • anonymous
i was confused what you did at first..how would u set the equations up in the first way
anonymous
  • anonymous
so how would you go about doing the other way?
dan815
  • dan815
okay so first identify these 3 graphs
dan815
  • dan815
|dw:1443062354574:dw|
dan815
  • dan815
the first 2 are pretty straight forward can you figure out the 3rd one
dan815
  • dan815
laplace of the 3rd graph
anonymous
  • anonymous
sorry i had lunch
anonymous
  • anonymous
isn't the third graph simply the first one just in diff order...
anonymous
  • 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
anonymous
  • anonymous
anyone want to enlighten me?
anonymous
  • anonymous
sorry, the net just cut out
dan815
  • dan815
yep but now for the 3rd one we can use that step function trick
dan815
  • dan815
the 3rd graph is very similar to some step function line graph that is shifted down on the y axis now
dan815
  • dan815
|dw:1443065288309:dw|
dan815
  • dan815
|dw:1443065315083:dw|
dan815
  • dan815
so its just this one constant we tacked onto the function
anonymous
  • anonymous
thats the heavside function right?
dan815
  • dan815
yes
dan815
  • dan815
|dw:1443065438565:dw|
anonymous
  • anonymous
so b would be the shift of 55 downards?
dan815
  • dan815
75 i think because u gotta get rid of the stuff after that http://prntscr.com/8jtl4f
anonymous
  • anonymous
we aren't starting from the origin tho?
dan815
  • dan815
yeah its the heaveside aplied to a line shifted right and down
dan815
  • dan815
|dw:1443065665540:dw|
anonymous
  • anonymous
right
dan815
  • dan815
right so its heave side to this function
dan815
  • dan815
|dw:1443065735982:dw|
dan815
  • dan815
just math it lol u get the theory now
anonymous
  • anonymous
yeah now i see, so T(s) is just the sum of those 3 identities?
dan815
  • dan815
all u gotta think is what kind of functions can i add that i can take the laplace of easily
dan815
  • dan815
yeah
dan815
  • dan815
think of other ways to get starting functions
dan815
  • dan815
do this one
dan815
  • dan815
|dw:1443065897265:dw|
dan815
  • dan815
these 3
dan815
  • dan815
the sum of (1) + (2) + (3) = is the function you want
dan815
  • dan815
you can see that right
anonymous
  • anonymous
:/ not really,
dan815
  • dan815
see what happens when you add graph 1 and 2
anonymous
  • anonymous
|dw:1443066102562:dw|
dan815
  • dan815
|dw:1443066092625:dw|
dan815
  • dan815
|dw:1443066187929:dw|
dan815
  • dan815
all the values after 75 keep cancelling out
dan815
  • dan815
at every coordinate on x
dan815
  • dan815
or t
dan815
  • dan815
thats what it means to add functions right
dan815
  • dan815
at every t value we add both of their y values
dan815
  • dan815
if they have the same slope and one is starting exactly negative of it, then u have to keep cancelling
anonymous
  • anonymous
makes much more sense!!!
anonymous
  • anonymous
ok so those two end up like that ok so we have
dan815
  • dan815
okay yes thats what ive been doing
dan815
  • dan815
now we add that constant function heavsided to get that constant line
dan815
  • dan815
|dw:1443066363555:dw|
anonymous
  • anonymous
yeeeeeeeeeee
anonymous
  • anonymous
i see it now
anonymous
  • anonymous
so graph 2 is just a ramp function shifted 30units of t to the right and shifted down 75 units |dw:1443066491364:dw|
dan815
  • dan815
yeah
dan815
  • dan815
finding the laplace transform of these functions is much simpler, now u can add them to get the desired effect
anonymous
  • anonymous
so T(t) = t + u(t-30) -75 + 55?
anonymous
  • anonymous
cause the first has slope t, second has that shift of heavside function and the thirs is just a step function?
dan815
  • dan815
http://www.efunda.com/math/laplace_transform/rules.cfm list of some basic laplace properties
dan815
  • dan815
i gotta watch a lecture for my class right now, ill be back, work thru it
anonymous
  • anonymous
np, i try get my head around it
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
you finished?
anonymous
  • anonymous
@ganeshie8 could you help me with this?

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