A generalized treatment for the mathematical relationships that exist between Q and bandwidth expressed in octaves for bandpass filters.
Aug 1, 1999 12:00 PM, Dennis A. Bohn
Analog and digital audio designers daily confront every imaginable aspect of active and passive filters. Most often, these are bandpass filters partially characterized by a quality factor, Q, and a bandwidth, BW. It seems that there have been enough books published on active and passive filter design to fill a modest office library, but there are certain aspects of the relationship between bandwidth as expressed in octaves versus Q that are poorly documented - if at all. The purpose of this article is to derive the mathematical relationships between these two variables and provide a useful look-up table for each, as well as a handy Microsoft Excel spreadsheet program, which may be downloaded from the Web at www.rane.com.
Definitions and review
A bandpass filter is characterized by three major parameters - center frequency, amplitude response (gain) and bandwidth. Center frequency is the frequency at which the amplitude is maximum. Gain is the maximum amplitude response occurring at the center frequency. Bandwidth (or passband) is the frequency range between the -3 dB points located on either side of the center frequency. Bandwidth is expressed in several ways - in frequency, as being so many Hertz wide; or in octaves, as being so many octaves (or fractional octave) wide, or in decades, as being so many decades (or fractional decade) wide. Far and away, the most common audio usage is to express bandwidth in octaves. It is here that the literature falls short in giving sufficient mathematical relationships to allow answers to be expressed easily in octaves.
When first designing an audio filter, normally the required bandwidth in octaves is known and the associated Q needs to be calculated. Once the filter has been designed, then Q is easily calculated by measuring the -3 dB frequency points, taking the difference and dividing that into the center frequency, and lastly, the bandwidth in octaves is then calculated.
A third situation arises in which only Q is known, and the bandwidth in octaves is desired. This calculation is not obvious, nor is it easy. The next section presents the necessary closed solutions for each of these calculations. For reference purposes, Figure 1 shows a bandpass filter with its associated parameters labeled for clarity and is used for derivation purposes.
The calculations
Given the -3 dB points, to find BW and Q follow the subsequent steps. If the 3 dB points are known, then calculating the BW in octaves is straightforward:
Let f subscript 2 = yf subscript 1
where y is any positive real number.
Next, define N as the number of octaves of BW. N octaves means that:
y = 2 superscript N
It then follows that:
f subscript 2 = 2 superscript N f subscript 1
(1)
Solving for N gives:
(2)
and, by definition, we get:
N= log y over log 2
(3)
If BW in octaves is known without knowledge of the actual -3 dB frequencies, and Q is to be calculated, then the following development leads to the required formula. In general, f subscript 0 is the geometric mean of the skirt frequencies, f subscript 1 and f subscript 2, therefore:
Q = f subscript 0 over f subscript 2 minus f subscript 1
>From (1) above, we get:
f subscript 0 = square root of f subscript 1 f subscript 2
or:
f subscript 0 = square root of f subscript 1 (2 superscript N f subscript 1)
>From (3) and (1), it follows that we get:
f subscript 0 = f subscript 1 square root of 2 superscript N
or:
Q = square root of 2 subscript N (f subscript 1) over 2 superscript N f subscript 1 minus f subscript 1
(4)
Table 1 shows several examples of equation (4) for BW's commonly used in audio design work.
If only Q is known, and the bandwidth in octaves is desired then equation (4) must be turned around and expressed in terms of Q:
Q = square root of 2 superscript N over 2 superscript N minus 1
or:
square root of 2 superscript N = Q(2 superscript N minus 1)
Because:
y = 2 superscript N
then:
y = Q(y - 1)
Squaring the equation gives:
y = Q superscript 2(y superscript 2 - 2y + 1)
which becomes:
Applying quadratic solution, we get:
y superscript 2 - y (2Q superscript 2 +1 over Q superscript 2) + 1 = 0
(5)
Be aware that the squaring operation introduces an extraneous root appearing as the minus square root term. The plus square root term gives the correct answer. The minus square root term gives the reciprocal answer. For example, the plus answer leads to y, while the minus answer yields 1/y. N (the number of octaves of BW) is now found from equation (2). Table 2 gives a handy look-up reference of bandwidth (octaves) versus Q.
y = 2Q superscript 2 +1 over 2Q superscript 2 plus or minus square root of (2Q superscript 2 +1 over Q superscript 2) squared over 4 minus 1
Available programs
Table 3 lists programs for HP-11C handheld calculators that solve equations (2), (4) and (5). For those with Internet access and PCs, a Microsoft Excel worksheet is available for downloading from www.rane.com. Look for it in the Library section.
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