laplace transforms-process control

- anonymous

laplace transforms-process control

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- anonymous

##### 1 Attachment

- dan815

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

- dan815

|dw:1443058593494:dw|

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## More answers

- anonymous

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

- dan815

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

- anonymous

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

- dan815

yes

- anonymous

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

- dan815

|dw:1443058802969:dw|

- anonymous

yep. correct

- anonymous

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

- dan815

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

- anonymous

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

- anonymous

yep

- anonymous

thats what i'm confused about

- dan815

|dw:1443059031652:dw|

- dan815

okay you will find lapalce of these 3 functions

- dan815

one is that first slope

- anonymous

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

- dan815

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

- dan815

the one on the bottom is

- 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

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

- dan815

then you will simple add the constant function A

- dan815

|dw:1443059246798:dw|

- dan815

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

- anonymous

that extended dotted line is your step function right?

- dan815

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

- anonymous

let me grasp this on a piece of paper

- dan815

|dw:1443059322182:dw|

- dan815

okay wait maybe this is just silly lol

- dan815

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

- anonymous

nah i think you're on the right track

- dan815

|dw:1443059492791:dw|

- dan815

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

- anonymous

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

- dan815

yep u can do this just break it up and solve

- anonymous

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

- dan815

|dw:1443059801163:dw|

- dan815

forget that that is a long annoying way this is better

- anonymous

lel haha

- anonymous

so just worry about the orginal function?

- dan815

yes just break it up piecewise like that

- dan815

|dw:1443059991702:dw|

- anonymous

yep

- anonymous

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

- anonymous

A/s

- dan815

yep just do these integrals like like normal style

- anonymous

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

- anonymous

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

- dan815

yep go on

- anonymous

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

- anonymous

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

- anonymous

so grouping...

- anonymous

so would that be it?

- anonymous

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

- dan815

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

- dan815

but whatever good to know both ways

- anonymous

whats the other way?

- dan815

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

- anonymous

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

- anonymous

so how would you go about doing the other way?

- dan815

okay so first identify these 3 graphs

- dan815

|dw:1443062354574:dw|

- dan815

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

- dan815

laplace of the 3rd graph

- anonymous

sorry i had lunch

- anonymous

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

- 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

anyone want to enlighten me?

- anonymous

sorry, the net just cut out

- dan815

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

- dan815

the 3rd graph is very similar to some step function line graph that is shifted down on the y axis now

- dan815

|dw:1443065288309:dw|

- dan815

|dw:1443065315083:dw|

- dan815

so its just this one constant we tacked onto the function

- anonymous

thats the heavside function right?

- dan815

yes

- dan815

|dw:1443065438565:dw|

- anonymous

so b would be the shift of 55 downards?

- dan815

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

- anonymous

we aren't starting from the origin tho?

- dan815

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

- dan815

|dw:1443065665540:dw|

- anonymous

right

- dan815

right so its heave side to this function

- dan815

|dw:1443065735982:dw|

- dan815

just math it lol u get the theory now

- anonymous

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

- dan815

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

- dan815

yeah

- dan815

think of other ways to get starting functions

- dan815

do this one

- dan815

|dw:1443065897265:dw|

- dan815

these 3

- dan815

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

- dan815

you can see that right

- anonymous

:/ not really,

- dan815

see what happens when you add graph 1 and 2

- anonymous

|dw:1443066102562:dw|

- dan815

|dw:1443066092625:dw|

- dan815

|dw:1443066187929:dw|

- dan815

all the values after 75 keep cancelling out

- dan815

at every coordinate on x

- dan815

or t

- dan815

thats what it means to add functions right

- dan815

at every t value we add both of their y values

- dan815

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

- anonymous

makes much more sense!!!

- anonymous

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

- dan815

okay yes thats what ive been doing

- dan815

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

- dan815

|dw:1443066363555:dw|

- anonymous

yeeeeeeeeeee

- anonymous

i see it now

- 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

yeah

- dan815

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

- anonymous

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

- anonymous

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

- dan815

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

- dan815

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

- anonymous

np, i try get my head around it

- anonymous

@dan815

- anonymous

you finished?

- anonymous

@ganeshie8 could you help me with this?

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