We have R1 in parallel with C, which gives an impedance of:
We then use the voltage divider to find v0 in terms of Vs
We the numeric values, we have:
So for the gain we have
Let’s call ω1 = 1/10-3 = 1000 and ω2 = 1/5.10-3
= 200
To obtain the asymptotic Bode plot, we approximate (without the
constant k = 5):
the numerator to 1 when ω < ω1 and to ω1 = 1000 when
ω > ω1
the denominator to 1 when ω < ω2 and to ω2 = 200 when
ω > ω2
Thus for the logarithmic gain we have the following approximation:
20log10(5)
when
ω < 200
20log10H = [20log10(5) - 20log10(ω2)] + 20log10 ω when
ω2 < ω < ω1
[20log10(5)
- 20log10(ω2)] + 20log10ω1 when ω1 < ω
The gain will be first a horizontal straight line at the value 20log10(5),
up to ω2 and then will have a 20 dB/dec slope, up to ω2 where it is a
horizontal straight line again. The last horizontal line is of value of [20log10(5)
- 20log10(ω2)] + 20log10ω1
The phase angle of H is
The Bode diagram will be somewhat like the one on p. 593 of the
textbook.