# A generalized treatment for the mathematical relationships that existbetween 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 aspectof active and passive filters. Most often, these are bandpass filterspartially characterized by a quality factor, Q, and a bandwidth, BW. Itseems that there have been enough books published on active and passivefilter design to fill a modest office library, but there are certainaspects of the relationship between bandwidth as expressed in octavesversus Q that are poorly documented – if at all. The purpose of thisarticle is to derive the mathematical relationships between these twovariables and provide a useful look-up table for each, as well as a handyMicrosoft Excel spreadsheet program, which may be downloaded from the Webat www.rane.com.

Definitions and review

A bandpass filter is characterized by three major parameters – centerfrequency, amplitude response (gain) and bandwidth. Center frequency is thefrequency at which the amplitude is maximum. Gain is the maximum amplituderesponse occurring at the center frequency. Bandwidth (or passband) is thefrequency range between the -3 dB points located on either side of thecenter frequency. Bandwidth is expressed in several ways – in frequency, asbeing so many Hertz wide; or in octaves, as being so many octaves (orfractional octave) wide, or in decades, as being so many decades (orfractional decade) wide. Far and away, the most common audio usage is toexpress bandwidth in octaves. It is here that the literature falls short ingiving sufficient mathematical relationships to allow answers to beexpressed easily in octaves.

When first designing an audio filter, normally the required bandwidth inoctaves is known and the associated Q needs to be calculated. Once thefilter has been designed, then Q is easily calculated by measuring the -3dB frequency points, taking the difference and dividing that into thecenter frequency, and lastly, the bandwidth in octaves is then calculated.

A third situation arises in which only Q is known, and the bandwidth inoctaves is desired. This calculation is not obvious, nor is it easy. Thenext section presents the necessary closed solutions for each of thesecalculations. For reference purposes, Figure 1 shows a bandpass filter withits associated parameters labeled for clarity and is used for derivationpurposes.

The calculations

Given the -3 dB points, to find BW and Q follow the subsequent steps. Ifthe 3 dB points are known, then calculating the BW in octaves isstraightforward:

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 dBfrequencies, and Q is to be calculated, then the following developmentleads to the required formula. In general, f subscript 0 is the geometricmean 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 fsubscript 1 minus f subscript 1

(4)

Table 1 shows several examples of equation (4) for BW’s commonly used inaudio 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 rootappearing as the minus square root term. The plus square root term givesthe 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 2gives 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 Excelworksheet is available for downloading from www.rane.com. Look for it inthe Library section.