Parametric Equalizer "Q" Definitions and Bandwidth (page 2)

# Calculating the Bandwidth of the Parametric Equalizer

In his PDF about PEQ filters, RBJ discusses the definition of bandwidth. In traditional low-pass and bandpass filters, the -3 dB point is usually the reference for computing bandwidth. But since a PEQ filter might have a boost or cut less than 3 dB, such a definition would make no sense. To solve this problem, RBJ settles on the "half-gain bandwidth". In this definition, the upper and lower frequency points for computing bandwidth are found as the frequencies for which the boost (or cut) in dB is one-half the maximum boost (or cut) in dB. For example, for a PEQ with a maximum cut of 2 dB, the upper and lower bandwidth would be found as the two frequencies for which to cut is 1 dB.

RBJ computes the upper and lower half-gain points and expresses their ratio in octaves, deriving other useful relationships that I won't discuss here. When subtracted, his formulas for the half-gain points give the bandwidth in Hz. A casual inspection of these formulas gives the impression that the resulting formula for the bandwidth in Hz might be very messy. A recent discussion with John Mulcahy, the author of the Room EQ Wizard (REW) measurement software, brought the actual facts of the matter to light.

John explained to me that his generic PEQ type in REW used RBJ's definition of Q. In the description of the generic PEQ in the REW documentation, a very simple formula for the half-gain bandwidth appears as follows:

(11) | $${BW}_{\text{Hz}}={f}_{0}/{Q}_{b}$$ |

Where BW_{Hz} is the half-gain bandwidth in Hz, f_{0} is the center frequency in Hz, and Q_{b} is the Bristow-Johnson Q as defined in (6). Writing a program to test this equation using RBJ's formulas for the upper and lower half-gain frequencies showed equation (11) to be exact. The derivation of the bandwidth formula (11) follows.

RBJ begins his analysis by expressing the PEQ transfer function H(s) as follows:

(12) | $$H\left(s\right)=\frac{{s}^{2}+2K\alpha {\omega}_{0}s+{\omega}_{0}^{2}}{{s}^{2}+2\alpha {\omega}_{0}s+{\omega}_{0}^{2}}$$ |

This is equation (6) in RBJ's PDF file. Here he has used the damping factor α instead of Q, and his K (the gain at the center frequency) is equivalent to A_{0} of equation (1) herein. Equations (8) and (9) of RBJ's PDF file give the squares of the upper and lower half-gain frequencies respectively. I've taken his expressions, converted them to Hz^{2}, and expressed the upper and lower half-gain frequencies as f_{2} and f_{1} respectively. This gives:

(13) | $${f}_{2}^{2}={f}_{0}^{2}\left[1+2K{\alpha}^{2}+2\alpha \sqrt{K\left(K{\alpha}^{2}+1\right)}\right]$$ |

(14) | $${f}_{1}^{2}={f}_{0}^{2}\left[1+2K{\alpha}^{2}-2\alpha \sqrt{K\left(K{\alpha}^{2}+1\right)}\right]$$ |

To find the bandwidth in Hz, we must take the square root of the right side of (13) and subtract the square root of the right side of (14) from it. This gives an equation of the form:

(15) | $$B{W}_{\text{Hz}}={f}_{0}\left(\sqrt{a+b}-\sqrt{a-b}\right)$$ |

where

(16) | $$a=1+2K{\alpha}^{2}$$ |

and

(17) | $$b=2\alpha \sqrt{K\left(K{\alpha}^{2}+1\right)}=2\alpha \sqrt{K}\sqrt{K{\alpha}^{2}+1}$$ |

In hopes of getting some cancellation involving the a and b terms, we'll solve for the square of the bandwidth. Squaring both sides of (15) gives:

(18) | $$B{W}_{\text{Hz}}^{2}={2f}_{0}^{2}\left(a-\sqrt{{a}^{2}-{b}^{2}}\right)$$ |

Next, we need to solve for α in terms of K and Q_{b} (the Bristow-Johnson Q). Using (6) and (12), we get:

(19) | $$\alpha =\frac{1}{2\sqrt{K}{Q}_{b}}$$ |

Substituting (19) into (16), we obtain:

(20) | $$a=\frac{2{Q}_{b}^{2}+1}{2{Q}_{b}^{2}}$$ |

Evaluating the two terms on the rightmost side of (17), and using (19), we get:

(21) | $$2\alpha \sqrt{K}=1/{Q}_{b}$$ |

and

(22) | $$\sqrt{K{\alpha}^{2}+1}=\frac{1}{2{Q}_{b}}\sqrt{4{Q}_{b}^{2}+1}$$ |

Substituting (21) and (22) into the rightmost side of (17), we get:

(23) | $$b=\frac{1}{2{Q}_{b}^{2}}\sqrt{4{Q}_{b}^{2}+1}$$ |

Now we can substitute (20) and (23) into (18) to solve for the square of the bandwidth. Using (20) and (23), we obtain:

(24) | $${a}^{2}-{b}^{2}=1$$ |

Plugging (20) and (24) into (18), we finally get:

(25) | $$B{W}_{\text{Hz}}^{2}={\left(\frac{{f}_{0}}{{Q}_{b}}\right)}^{2}$$ |

This proves equation (11), namely:

(26) | $${BW}_{\text{Hz}}={f}_{0}/{Q}_{b}$$ |

The half-gain bandwidth of an audio PEQ filter is its center frequency divided by its Bristow-Johnson Q.