I think I' am starting to understand but what do I do with j since I don't have complex numbers? Secondly what does s represent and thirdly how was the transfer function derived(Is it standard?)?
OK, here are my explanations:
Starting with the equation as given in post#7 (for the second circuit):
Vout/Vin=(R2||Xc)/(R1 +R2||Xc)=N(s)/D(s) with N(s)=1/(1/R2+sC) and D(s)=R1+1/(1/R2+sC).
It is useful to rewrite this equation (by multiplying N and D by suitable terms) to get the so called "normal form" with s-polynoms in both numerator and denominator - the latter containing "1" as an absolute value.
This leads to
H(s)=Vout/Vin=[R2/(R1+R2)]/(1+s*RpC) with Rp=R1||R2
Now, if you compare this expression with the general first order transfer function (as given in post#2)
H(s)=Ao/(1+sT)
the actual time constant is T=RpC=1/wo ; Thus: wo=1/RpC (wo=cut-off frequency equivalent to the desired half-power frequency).
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Additional explanation: In this case, "s" is nothing else than an abbreviation for jw. That means: s=jw.
For s=jwo=j*(1/T) the denominator is D(s=jwo)=(1+j) and the magnitude is sqrt(1+1)=sqrt(2)=1.414.
That means: For s=jwo the numerator N(s) is multiplied by 1/1.414=0.707 which is eqivalent to a 3dB loss.
This justifies the definition of the cut-off frequency.