Oki I find the maximum amplitude
[LATEX]Amax = 1 + \frac{(C2 R1)}{(C1 R1 + C2 R2)}[/LATEX]
I find Amax by substituting S = jωo
But can you describe in more detail the way how you find 3dB frequencies.
Hi again,
Starting with the original transfer function:
Hs=(s^2*C1*C2*R1*R2+s*C2*R2+s*C2*R1+s*C1*R1+1)/((s*C1*R1+1)*(s*C2*R2+1))
in the numerator we subst s=j*w and we get:
HjwN=j^2*w^2*C1*C2*R1*R2+j*w*C2*R2+j*w*C2*R1+j*w*C1*R1+1
and after simplifying we get:
HjwN=-w^2*C1*C2*R1*R2+j*w*C2*R2+j*w*C2*R1+j*w*C1*R1+1
so the real part in the numerator is:
RPN=1-w^2*C1*C2*R1*R2
and the imag part in the numerator is:
IPN=w*C2*R2+w*C2*R1+w*C1*R1
and so the amplitude of the numerator is:
AmplN(w)=sqrt(RPN^2+IPN^2)=sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C2*R1+w*C1*R1)^2)
The denominator is:
(s*C1*R1+1)*(s*C2*R2+1)
and after subst s=j*w and finding the amplitude the same way we did the numerator we get:
AmplD(w)=sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C1*R1)^2)
Now we have an expression for the amplitude:
Ampl(w)=AmplN(w)/AmplD(w)
Now here we would subst w=w0 and calculate the amplitude at the center frequency to get the
max amplitude, but since you already have that we take that amplitude:
1+C2*R1/(C1*R1+C2*R2)
and divide that by the square root of 2 thus:
(1+C2*R1/(C1*R1+C2*R2))/sqrt(2)
and then set that equal to the amplitude we computed above AmplN(w)/AmplD(w):
(1+C2*R1/(C1*R1+C2*R2))/sqrt(2)=AmplN(w)/AmplD(w)
Written out this looks like this:
((C2*R1)/(C2*R2+C1*R1)+1)/sqrt(2)=sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C2*R1+w*C1*R1)^2)/sqrt((1-w^2*C1*C2*R1*R2)^2+(w*C2*R2+w*C1*R1)^2)
Now we simplify that expression and at this point it gets much simpler to substitute the actual values for the
variables C1, C2, R1, and R2. We could work symbolicially, but it would take a lot more work.
Doing that, we get a much more simple equation:
3.8457892807561354e-18*w^4-5.4217779933648401e-8*w^2+1.495274102079395=0
We then solve this for w, and reject negative values, leaving us with two positive values for w.
We then divide by 2*pi and get the two -3db frequencies F1 and F2. One is below F0 and one is above F0.
Of course sometimes we dont know the center frequency w0 yet, so we would have to solve for that first. To do that, we would compute the Amplitude as we did above and take the first derivative with respect to w and then set that equal to zero:
d(Ampl(w))/dw=0
and solve this for w and test each solution for validity in the amplitude equation, and that would give us w0. Again, doing this numerically is a *lot* easier than symbolically.