Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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

|dw:1443058593494:dw|

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)

but whatever good to know both ways

whats the other way?

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...

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

|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

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

yeah

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

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?