Calculate 3rd order Bessel filter 3db point

[RoveyDoveyGrovey]

Yep, im officially stumped, I keep ending up having to cube root a complex number when trying to solve for x1, my calculator no likey, neither do I for that matter. Maybe many simulations is the easiest way to go, although I have an inkling my lecturer would like the working. Well i've learnt one thing, repeatedly typing 10 digit numbers into my calculator is exceedingly tedious. I think i'll come back to this tomorrow.

 

[MrAl]

Hello again,

If you square 5.483698704*10^-51 you should get a reasonable number but you'll need a more advanced calculator.
Wow, you're doing this by using a hand calculator? That's not too good of an idea because it is too easy to make a mistake. This is something that really should be done in a program or on a programmable calculator.

BTW, how did you get 5.483698704*10^-51 for 'a' anyway?
Maybe you should give me your component values and we'll compare notes.

I just noticed that you posted a circuit that was made from a first order filter and a second order filter. This isnt the same as the other link in this thread. We did the math for a pure 3rd order SK LP filter, but if you are using that first+second order filter then the math might be simpler. Let me know.

 

[MikeMl]

When you get through playing with the equations, you can use this to check your answer

 

[RoveyDoveyGrovey]

HAHAHAHA, i've just found an infinitely easier way of working this all out, basically you use the normalised polynomials, these are numbers that can be found on many sites on the net and even this book i've been looking at while wondering 'what the hell do these numbers mean?'

anyway, you take the two filters in isolation

so, 1st order filter,

fc = 1 / (2xpixR1xC1)

fc = 1 / (2xpix12150x0.0033uF) = 3969.446Hz this is the cut-off for the 1st order part of the filter

Next you use the normalised number,

f0 = f02 x fc

where,
f0 = the cut-off freq that you've just calculated
f02 = the normalised number from the table
fc = the cut-off freq of the whole 3rd order filter

so f02 = 1.327 in this case

so fc = 3969.446/1.327 = 2991.29Hz which is pretty much exactly what PSpice predicts

Do it again with the 2nd order part of the filter,

fc = 1 / (2xpixsqrt(R2xR3xC2xC3))

fc = 1 / (2xpixsqrt(5860x10200x0.0033uFx0.0068uF)) = 4345.7Hz

Stick that into the f0 equation, f01 in this case equals 1.453

fc = 4345.7/1.453 = 2990.8Hz

Hey presto, no need to destroy your brain, you can go back to thinking laplace is where santa lives.

MrAI you have been a legend I really appreciate the help you gave me, hopefully this might help you at some point.

 

[RoveyDoveyGrovey]

Here's some of the tables for anyone who's interested,

 

[MrAl]

Hello again, Sounds good and interesting too, i'll have to take a better look once i get past a few computer problems i've been having lately.

 

[RoveyDoveyGrovey]

i think the normalised number may have something to do with all those horrible equations, it's just that some kind fellow has done all the hard work and all you need to do is look up the tables. anyway i've realised my dream of having a couple of simple equations to stick in excel and i've managed to calculate nearly 50 different combinations using different component tolerances and all of them closely match the simulations i did in pspice, so whatever the explaination, it works.