Problem 13.4-4

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.