unit step function. #dan815

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unit step function. #dan815

Mathematics
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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

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Other answers:

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
has constant feed rate of 10m^3/min (theoretically) but at time t=0, a disturbance pushes 1m^3 of water into the tank and an extra 2m^3 of water at t=1min (So an additional 1m^3 of water per min)
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
yea, they draw these graphs just to show use the theoretical flow regime. yes it cuts off at 0 for laplacing, but we have questions that show graphs like these.
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
3weeks of lapalce transforms and its a subject inside this course, i rekn I'm doing well to grasp a lot of things haha
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
lel, we use lapalce transforms in process control so its easier to solve differential equations (approximate it using linearisation)
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
yeah tru dat but that was for eng math. i guess we aren't looking at too hard of ODE's but i guess they teach us this smack and we have to learn their way of calculating there questions
yea np, peace

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