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how do you write this as function? \[X(t)=10+\delta(t)+\delta(t-1)\]
Im not sure. I'll show you the question
i don't think its saying that the 1ft^3 of water still happen from t=0min to t=1 mins, otherwise it would be a step function (not an impulse function). Im not entirely sure
because in the end, after the last impulse, you want the inlet to be running at 10ft^3 of water (Since that is its steady state value; the impulses come from disturbances)
an impulse has infinite height right
yes, but i guess we say that an impulse of 2 is greater than an impulse of 1, hypothetically speaking, despite both having infinite height
oh i see what ur indicating is like the area
i think this is fine
so you think my function is reasonable?
i think u shud have 10
since 10 is not really a response
but i dont really know, it seems like its based on convention
i think u shudnt* have 10
|dw:1444124866547:dw| shouldnt it be like this?? the impulse is the derivative of this graph
then ur function would be X(t) = 10 + integral (unit impulse) + integral 2*unitimpulse
but they're saying sketch "response" of level vs time
that is a function of step functions, we can't assume that the disturbance continues after t=0 to t=1
not level vs time right...
i have the level v time sorted once i find the solution to the initial flow rate. i can just apply laplace and first order principles. but this is the important bit
i think its this
the 10 isnt there anymoore in response thats initial condition
so its just \[X(t)=\delta(t)+\delta(t-1)\]
well shouldnt it be 2* impulse
also depedns on how u drew the level graph
i forgot to add that in all of my equations
u can also add an impulse of 10 at like t=-inf or something
|dw:1444125308190:dw| perhaps u can just scale it back by -1 and you have\[X(t)=10\delta(t)+11\delta(t-1)\]
okay thats fine
its still only 2 for the 2nd one
and 11 for hte first one
i take it u wannna saw theres 10 in the tank and 1 added same time?
im not following, why is it 2, when we are inputting an impulse of a factor of 1 greater than the first impulse?
idont know why we need laplace transform and stuff since it jst ask to draw the level of the tank .. and from my graph we can easily find the levels at t=.5, 1, 1.5
@BAdhi we are continually removing water from the system
just show me your step function lo... itll be so much less confusing
so the height will change after the impulse
just do it depending on ur step function
but there is no step function...
lel im confused
i thought u drew that already
well its kinds ambiguous to me cause you aren't adding water from the first disturbance from t=0 to t=1, only instantaneously..
like all it is, is just a spike to the inflow, and thats it. it isn't flowing 11ft^3 of water from t=0 to t=1. only at an instantaneous period
any thoughts? i just need to find the function of inflow and i will be set. i just need to find what Q(t)=??
Q(t) being the inlet flow into the tank
again that 10 cannot be there
having a continuous 10 means ur level was rising constant at a rate of 10 m^3 over some time increment
then i don't know what to do.
because, we aren't really adding step functions, because it would have told us that from t=0 to t=1, 1ft^3 is being added to the tank, (that is a step function) but here it doesn't say.
so this would be an instantaneous disturbance
does anyone understand my thought process?
which topic is this question related to? :)
if anyone has a subscription to chegg then they have the answers, but yolo, I'm clean dry out of ideas