Lets 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 1st
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