Problem 13.4-10

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.