laplace transforms-process control

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laplace transforms-process control

Mathematics
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if you're ever in doubt just write out the laplace transform integral
|dw:1443058593494:dw|

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

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