- 1 Series resonance 131
the resistor is also a maximum. Now if the maximum power is PM = IM 2R at f0
then at fl (or f2) the power will be
P~ = P2 = (0.7071M) 2R = 0.5 IM2R = 0.5 PM
For this reason f~ and f2 are called the half-power frequencies.The dB notation
The logarithm to the base ten of the ratio of two powers P~ and P2 is called the
bel. Because this is rather large it is more usual to express power ratios in
decibels (dB). This means multiplying the logarithm of the ratio by 10. By
definition, a power P2 is 10 logao (P2/P1) dB above the power P1.
Since P = V2/R, we see that a voltage V2 is 10 loglo (V2/V1) 2 dB above a
voltage V1 i.e. V2 is 20 loglo (1/2/171) dB above V1.
For the half-power frequencies f~ and f2, the powers (PI and P2) are equal
to 0.5PM. Therefore P1 is IOIogIo(P~/PM) dB above PM, i.e. Px is
10 log10 (0.5 PM/PM) dB above PM"
Now 10 lOgl0 (0.5) - -3 so that P~ is in fact -3 dB above PM which means
that it is 3 dB below PM. It is said to be 3 dB down on PM. For this reason the
frequencies fl and fa are also referred to as the -3 dB points.Example 6.7
The maximum current in a bandpass filter circuit is 25 mA. Calculate the
current at the lower and upper cut-off frequencies.Solution
Let the current at the lower cut-off frequency be/1 and that at the upper cut-off
frequency be 12. Then 11 - 12 - 0.707 IM -- 0.707 • 25 -- 17.68 mA.Example 6.8
A bandpass filter circuit has a lower cut-off frequency (f~)= 12 kHz and an
upper cut-off frequency (f2) - 18 kHz. Calculate the bandwidth of this circuit.Solution
Using Equation (6.6) the bandwidth is B - f2 - fl = (18 - 12) kHz = 6 kHz.
Gain and phase diagrams
In the circuit of Fig. 6.12 the ratio Vo/V i is called the gain ratio and is denoted
by H. Since go = IR and Vi- I(R + j~oL + 1//jcoC) then
H = IR/I(R + jo~L + 1/jwC) (6.7)