Problem 13.4-15

Let’s call ω1 = 200 and ω2 = 500.

Then we get:

                         14                   when ω <ω1

20 log (T(ω)) = A – 20 log ω    when ω1 < ω <ω2

                         B                   when ω > ω2

To solve for A and B we have

14 = A – 20 log (200) thus A = 14 + 20 log ω1

and B = 14 + 20 log ω1 – 20 log ω2

We set 14 = 20 log k

Thus

                         20 log k                                                when ω <ω1

20 log (T(ω)) =  20 log k + 20 log ω1 – 20 log ω            when ω1 < ω <ω2

                         20 log k + 20 log ω1 – 20 log ω2          when ω > ω2

So we the transfer function:

Which is the transfer function of any circuit like the 1^st one on page 600 of the textbook. Any circuit like this for which the components would be chosen in order to get

14 = 20 log k

ω1 = p = 200 = 1/R₂C₂

ω2 = z = 500 = 1/R₁C₁