492 Chapter 8. Signal Processing
Let us now substitutef=1/ 2 πRCin this relation. This gives
2log(
Vout
Vin)
=− 3 dB. (8.4.3)Hence, according to our definition, the frequency
fcutof f=1
2 πRC, (8.4.4)
is the cutoff frequency for this simple RC filter. This frequency is sometimes also
referred to as thebreakpointsince above this value the attenuation of the circuit
becomes too large. Quantitatively speaking, above the breakpoint the response of
the circuit varies by 6dBper octave of frequency increase (see Example below). On
the other hand at frequencies much lower than this cutoff frequency the attenuation
becomes negligibly small. This can also be seen from equation 8.4.2 where we can
neglect the second term in the denominator on the right hand side forf>>fcutof f.
This gives
2log(
Vout
Vin)
≈ 2log(1) = 0⇒Vout ≈ Vin,which simply means that at frequencies much lower than the cutoff value the output
voltage (or power) is nearly equal to the input voltage (or power).
Example:
Quantify the change in the response of a simple RC low pass filter with
change in frequency above the breakpoint offcutof f=1/ 2 πRC.Solution:
We start with equation 8.4.2 and note that at frequency higher than the cutoff
frequency the first term in the denominator on the right hand side can be
neglected.2log(
Vout
Vin)
=2log[
1
√
1+(2πfRC)^2]
≈ 2log[
1
√
(2πfRC)^2]
= −2log(2πfRC)
∝−2log(f)The above proportionality means that an 8 fold increase in frequency would
correspond to a 6dBchange in the response. Hence we can say that the
response of a low pass RC filter changes by 6dBper octave of frequency
increase.The simple RC low pass filter we just discussed is seldom used in modern radiation
detection systems, which are equipped with more complicated active circuits.