Designing De-emphasis Digital Filter for old CDs Part 3: Using Reference Equation

 There is a reference de-emphasis equation on this page (Thanks miguelito-san) : https://forums.stevehoffman.tv/threads/cd-dat-with-pre-emphasis-how-to-de-emphasize-correctly.88541/

$T_{\mathrm{dB}}(\mathit{f})=10{\log }_{10}\left(\left(\frac{1+\frac{1}{{(0.000015·2\mathit{\pi }\mathit{f})}^2}}{1+\frac{1}{{(0.00005·2\mathit{\pi }\mathit{f})}^2}}\right)\right)-10.4576\dots \dots (1)$
$\mathit{f}:\; frequency\; (Hz)$
$T_{\mathrm{dB}}(\mathit{f})\; :\; De-emphasis\; filter\; gain\; (dB)$

This decibel is root power quantity therefore actual gain magnitude value T(f) is:

$\mathit{f}:\; frequency\; (Hz)$
$T(\mathit{f})\; :\; De-emphasis\; filter\; gain\; magnitude\; value\; at\; frequency\mathit{f}$

Note 1: Math equations of this article uses MathML and it seems it is displayed correctly only on Firefox (as of November 2020).

Note 2: It seems 0.00005 and 0.000015 of equation 1 means 50 microseconds and 15 microseconds respectively and it is called 50/15 microsec emphasis.

On this article, equation (2) is used to create frequency sampling FIR digital filter.

Calculation steps are very similar to Part 2.

Example : Calculation of M=9 taps FIR filter coeffs using equation $T(\mathit{f})$

When sampling frequency==44100 Hz and Desired FIR filter taps M==9,

Frequency sampling index $k=0,1,2,\; ..,\frac{M-1}{2}=4$

$k\; =\; 0,\; 1,\; 2,\; 3\; and\; 4$

Frequency sampling angle frequency ${\omega }_{\mathrm{k}}=\frac{2\pi k}{M}:$

• $k\; =\; 0\; :\; {\omega }_0=\; 0,\; freq0\; =\; 0\; Hz$
  
• $k\; =\; 1\; :\; {\omega }_1=\; 2\pi /9,\; freq1\; =\; (44100/2\pi )(2\pi /9)\; =\; 44100\; /\; 9\; Hz\; =\; 4900\; Hz$
• $k\; =\; 2\; :\; {\omega }_2=\; 4\pi /9,\; freq2\; =\; (44100/2\pi )(4\pi /9)\; =\; 44100\; *\; 2\; /\; 9\; Hz\; =\; 9800\; Hz$
• $k\; =\; 3\; :\; {\omega }_3=\; 6\pi /9,\; freq3\; =\; (44100/2\pi )(6\pi /9)\; =\; 44100\; *\; 3\; /\; 9\; Hz\; =\; 14700\; Hz$
• $k\; =\; 4\; :\; {\omega }_4=\; 8\pi /9,\; freq4\; =\; (44100/2\pi )(8\pi /9)\; =\; 44100\; *\; 4\; /\; 9\; Hz\; =\; 19600\; Hz$

Get filter gain on those frequencies $Hr({\omega }_{\mathrm{k}})$ using Equation (2):

• $Hr($ {\omega }_0)\; =\; T(0)\; =\; 1$$
• $Hr($ {\omega }_1)\; =\; T(4900)\; =\; 0.60004356$$
• $Hr($ {\omega }_2)\; =\; T(9800)\; =\; 0.420524967$$
• $Hr($ {\omega }_3)\; =\; T(14700)\; =\; 0.361602897$$
• $Hr($ {\omega }_4)\; =\; T(19600)\; =\; 0.336724814$$

Then G(k) is calculated from $Hr({\omega }_{\mathrm{k}})$ by $G(k)\; ={\mathrm{(-1)}}^{\mathrm{k}}Hr($ {\omega }_{\mathrm{k}})\; \dots \dots (3)\; :$$

• $G(0)\; ={\mathrm{(-1)}}^{0}Hr($ {\omega }_0)\; =\; 1$$
• $G(1)\; ={\mathrm{(-1)}}^{1}Hr($ {\omega }_1)\; =\; -0.60004356$$
• $G(2)\; ={\mathrm{(-1)}}^{2}Hr($ {\omega }_2)\; =\; 0.420524967$$
• $G(3)\; ={\mathrm{(-1)}}^{3}Hr($ {\omega }_3)\; =\; -0.361602897$$
• $G(4)\; ={\mathrm{(-1)}}^{4}Hr($ {\omega }_4)\; =\; 0.336724814$$

FIR filter coefficients h(n) can be calculated by $h(n)\; =\frac{1}{M}\left\{G(0)+2\sum _{\mathrm{k=1}}^{\mathrm{(M-1)/2}}G(k)cos\frac{2\pi k(n+1/2)}{M}\right\}\dots (4)$

• $h(0)\; =\; 0.030212113446082472$
• $h(1)\; =\; 0.040656939222222181$
• $h(2)\; =\; 0.063594885887469199$
• $h(3)\; =\; 0.11899203499978166$
• $h(4)\; =\; 0.49308805288888891$

Finally, this FIR filter is symmetry (linear phase), therefore h(8) = h(0), h(7) = h(1), h(6) = h(2), h(5) = h(3).

• $h(0)\; =\; 0.030212113446082472$
• $h(1)\; =\; 0.040656939222222181$
• $h(2)\; =\; 0.063594885887469199$
• $h(3)\; =\; 0.11899203499978166$
• $h(4)\; =\; 0.49308805288888891$
• $h(5)\; =\; 0.11899203499978166$
• $h(6)\; =\; 0.063594885887469199$
• $h(7)\; =\; 0.040656939222222181$
• $h(8)\; =\; 0.030212113446082472$

We've got all the 9 FIR filter coefficients h(n).

Evaluating Frequency Response of FIR filter

Now one FIR filter is available for testing. FIR filter gain of arbitrary angular frequency ω can be calculated using the following equation:

$Gain(\omega )\; =\sum _{\mathrm{k=0}}^{\mathrm{M-1}}h(k)e^{\mathrm{-jk\omega }}\dots \dots (5)$

Real part and imaginary part of (5) can be calculated separately:

${\mathrm{Gain}}_{\mathrm{real}}(\omega )\; =\sum _{\mathrm{k=0}}^{\mathrm{M-1}}h(k)cos(-k\omega )\; \dots \dots (5r)$
${\mathrm{Gain}}_{\mathrm{imaginary}}(\omega )\; =\sum _{\mathrm{k=0}}^{\mathrm{M-1}}h(k)sin(-k\omega )\; \dots \dots (5i)$

And FIR filter Gain magnitude is calculated as follows:

Finally Gain in decibel is: ${\mathrm{Gain}}_{\mathrm{dB}}(\omega )\; =\; 20{\log }_{10}\left\{{\mathrm{Gain}}_{\mathrm{magnitude}}(\omega )\right\}\dots \dots (7)$

For our M=9 FIR filter, frequency response can be calculated using equation (7). And this time reference equation is available: desirable gain of any frequency can be calculated, it is possible to compare gain values at as many frequency points as you wish. I compared 10Hz to 22040Hz semitone step frequency points and created Fig.1.

Fig.1. 9 taps FIR de-emphasis filter frequency response.

Comparing to the original de-emphasis table, max error of this FIR filter is 0.878 dB on 7360Hz, It is poor, and the quality can be improved by increasing M.

Searching the best FIR filter tap number M

FIR Filter Taps M	Max Error (dB)
9	0.878
15	0.1704
17	0.115
19	0.0631
21	0.0522
23	0.0231
25	0.0272
27	0.00882

Table 1: FIR Filter taps and max error 

Fig.2 : Frequency Response of FIR Filter, taps=19  note: frequency axis is logarithmic.

Fig.3 : Frequency Response of FIR Filter, taps=27  note: frequency axis is logarithmic.

Fig.4 : Gain Error of FIR Filter, taps=27

De-emphasis FIR Filter Coefficients for 44.1kHz PCM data

19taps linear-phase FIR filter coefficients h(n) of max error = 0.0631 dB is as follows:

0.0022333652179533105,
0.0027676001211334594,
0.0042218013139663415,
0.0065438712761329912,
0.011312237381544295,
0.01877355043394098,
0.03398288954901494,
0.059120380386441337,
0.11593361915957012,
0.49022137032060437,
0.11593361915957012,
0.059120380386441337,
0.03398288954901494,
0.01877355043394098,
0.011312237381544295,
0.0065438712761329912,
0.0042218013139663415,
0.0027676001211334594,
0.0022333652179533105,

27taps linear-phase FIR filter coefficients h(n) of max error = 0.00882 dB is as follows:

0.00031102739649091732,
0.00036885685453166665,
0.00057659816229764602,
0.00082952248115418167,
0.001449970925925961,
0.0022387507393955598,
0.0039420948394406248,
0.0063363475292175873,
0.011215231698621361,
0.018685854350970088,
0.033951734838472802,
0.059076192885731141,
0.11592376177923121,
0.49018811103703697,
0.11592376177923121,
0.059076192885731141,
0.033951734838472802,
0.018685854350970088,
0.011215231698621361,
0.0063363475292175873,
0.0039420948394406248,
0.0022387507393955598,
0.001449970925925961,
0.00082952248115418167,
0.00057659816229764602,
0.00036885685453166665,
0.00031102739649091732,

Filtering sound files using sox

With sox, it is possible to use those coefficients to filter the 44.1kHz PCM sound files. The following examples inputs inFile.flac and apply 27taps De-emphasis FIR filter and output it as outFile.flac:

sox inFile.flac outFile.flac fir 0.00031102739649091732 0.00036885685453166665 0.00057659816229764602 0.00082952248115418167 0.001449970925925961 0.0022387507393955598 0.0039420948394406248 0.0063363475292175873 0.011215231698621361 0.018685854350970088 0.033951734838472802 0.059076192885731141 0.11592376177923121 0.49018811103703697 0.11592376177923121 0.059076192885731141 0.033951734838472802 0.018685854350970088 0.011215231698621361 0.0063363475292175873 0.0039420948394406248 0.0022387507393955598 0.001449970925925961 0.00082952248115418167 0.00057659816229764602 0.00036885685453166665 0.00031102739649091732

Next article: Creating Pre-emphasis Audio CD