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sorry that should be db/dt

not dy/dt

hey what does linearise mean exactly?

does that mean to solve the differential equation or something else?

|dw:1438924011903:dw|

so we try to fit that curve with linear tangents

where the point we know as a basis is the steady state condition.

http://facstaff.cbu.edu/rprice/lectures/lineariz.html

its along these lines..

this is mearly the fundementals for laplace transforms

I was trying to find something easy to follow online but I can't find anything.

haha thanks for helping anyway

we are doing this for process control but i think ive nutted it out

to linearise ba^2 we need to partial differentiate each term at steady state

\[z=ba^2 \\ z_b=a^2 \\ z_a=2ab ?\]

so we get|dw:1438925115302:dw|

so thats a linear approximation of ab^2

so you apply that with all the other non linear terms

hmm...weird
I see where you got the last two terms

so hey would sin(ca) be \[
\approx \sin( c a_{ss})+c \cos(ca)a'\]

though that isn't linear :p

yeah i havent figured the sin term yet haha

might have to use an iterative procedure for that...hmm

this is interesting. its only the sin term which is weird to linearise

suggestions?