We use the table on page p.600 of the textbook to get the network
function.
The first circuit is what we are looking for, thus we get:
where p is given by 1/C2R2
and z is 1/C1R1
and k = R2/R1
thus we get:
For the Bode, we need the gain H and then phase shift.
Let’s call ω1 = 1/C1R1 and ω2 = 1/C2R2
To obtain the asymptotic Bode plot, we approximate (without the
constant k):
the numerator to 1 when ω < ω1 and to ω1 = 1/C1R1
when ω > ω1
the denominator to 1 when ω < ω2 and to ω2 = 1/C2R2
when ω > ω2
Thus we get H approximated to:
R2/R1 when ω < ω1
R2C1ω when ω1 < ω < ω2
C1/C2 when ω2 < ω
Thus for the logarithmic gain we have the following approximation:
20log10(R2/R1)
when
ω < ω1
20log10H = [20log10(R2/R1)
- 20log10(ω1)] + 20log10 ω when ω1 < ω <
ω2
[20log10(R2/R1)
- 20log10(ω1)] + 20log10ω2 when ω2 < ω
The gain will be first a horizontal straight line up to ω1 and
then will have a 20 dB/dec slope, up to ω2 where it is a horizontal
straight line again.
The phase angle of H is
The Bode diagram will be like the one on p. 593 of the textbook.