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|dw:1443667583385:dw|

just to be clear, we can represent this as X(t)=u(t)+2u(t-1)

you can rewrite that as a summation of some other step functions

yes

SIK

uhh lemem see

getting the hang of this

actual problem is this sox

|dw:1443667664012:dw|

yeah what u want to do is actually u(t) + u(t-1)

|dw:1443667692857:dw|

X(t)=11u(t)+12u(t-1)

the graph is a bit confusing u want it 10 until 0?

10

what u did will result in 23 at your highest step

okay lemme get this right though u want 10 before 0?

u know for laplace transform it cuts off the part before 0

because if u look at the power series
summation n from 0 to inf x^n
this only converges when x<1

|X|<1

because another example, where

okay okay lemme make sure about 1 more thing

upto where do u want the 10 on the negative side

forever?

|dw:1443668012704:dw|

they have function X(t)=90(u(t)-u(t-0.1))

so pretty much i just applied it to get X(t)=11u(t)+12u(t-1) but need confirmation

whats the question again

|dw:1443668159355:dw|

11u(t)+12u(t-1) that wont give u that

i just need to find this function as step function

so i can solve this silly problem

lol okay ill show u best way to do this

just draw some step functions and see how to add them

neeeed

how about 10(u(t)+2u(t-1))?

question is this btw for anyones interest

|dw:1443668206318:dw|

10+u(t)+u(t-1)?

ya that works

say we have this drawing again

|dw:1443668443342:dw|

yep

they said the step function was x(t)=90(u(t)-u(t-0.1))

that doesnt work

thats the answer....

they are missing a 10

plug in a value for t lets say t=2

it legit has it in a book haha , omg I'm confused

maybe they just dont want to cover the 10, what they have there is

|dw:1443668705765:dw|

perhaps they just shift it down so its just normalises it

which really would affect the problem aye

wouldn't*

|dw:1443668833633:dw|

would be X(s)=10u(t)+11u(t-1)

X(s)=10*U(t)+u(t-1)

why isn't it 11u(t-1)? aren't we increasing an addition m^3 after t=1min

ohhhhhhh

pellet

lolq

yeah cause your steady state value is still at 10

yes

and u just add a step function of magnitude 1 extra YRESSSSDQIDLFYHQBLERBFHIRL

DAN u da man

i beeen telling u this lool

shhhhhhhhh

hahahahahaa

that just clicked aye

nigsss

lol!

i think laplace transforms might be a waste of time

who knows

its to introduce u to fourier transforms

ye but eventually DE become too hard with laplace, so u gotta use fourier transforms for PDES

i better get back to work T_T

yea np, peace