5.7 Frequency Response of Circuits cont.
M(3) = {1/[1 + (3/3)2]½} = 1/Ö2 and q(w) = -45° which is the transfer function’s half-power point
Why the moniker half-power point?
Suppose VS = VS/0° V. for all frequencies (i.e., all w)
Then the maximum output voltage Vmax.= VS occurs at w = 0 since V= M(w)VS and Mmax.= M(0) = 1
Impressing this voltage across a load resistor yields the maximum output power Pmax.= VS2/RL
The output voltage at w = 3 Rad./Sec. is M(3)VS = VS /Ö2 and the corresponding power is (VS /Ö2)2/RL = [VS2/RL]/2 = Pmax./2 (half of Pmax.)