1
$\begingroup$

Can a simple IIR biquad filter be used as an audio de-emphasis filter for broadcast FM radio? If so, how does one translate the filter requirements (50 uS time constant in the U.S.?) into parameters for one of the common (e.g. RBJ's) biquad cookbook algorithms, given typical audio sample rates? Is the formula for a low shelving filter appropriate?

$\endgroup$
  • $\begingroup$ it seems to me that the de-emphasis is a 1st-order LPF with the corner frequency at $\frac{1}{2 \pi \, 75 \, \mu S}$. all of the shelves and filters in the cookbook are 2nd order. you can experiment and see if a shelf will match up well enough. $\endgroup$ – robert bristow-johnson Oct 5 '16 at 6:29
1
$\begingroup$

You could also implement a 2nd order IIR filter which has response of 1st order filter by using some other TF but one used in RBJ's cookbook. Here's one way to do it using MZT (C++).

double Fs = 44100; // higher sample rate results better filter accuracy 
double T1 = 0.000075; // time constant
double p1 = -exp(-1.0/(Fs*T1));
double z1 = 1 + z1;

a0 = 1.0;
a1 = p1;
a2 = 0;

b0 = 1.0;
b1 = z1;
b2 = 0;

% swap between a1, b1 to select pre- or de-emphasis

Coefficients from above calculations adds +13.63dB gain so you need to normalize before use.

Same as Octave code:

% PACKAGES
pkg load control
pkg load signal

clear all;
clc;

fs = 44100;
samplingtime = 1/fs;

% analog prototype
A2 = [1];
B2 = [0.000075 1];
Ds = tf(B2, A2); % swap between A2 and B2 to select pre- or de-emphasis
Ds = Ds/dcgain(Ds);

% MZT
T1 = 0.000075; % 75us
z1 = -exp(-1.0/(fs*T1));
p1 = 1+z1;

a0 = 1.0;
a1 = p1;
a2 = 0;

b0 = 1.0;
b1 = z1;
b2 = 0;

% swap between a1, b1 to select pre- or de-emphasis

Bmzt = [b0 a1 b2]
Amzt = [a0 b1 a2]

DzMZT = tf(Amzt, Bmzt, samplingtime);
DzMZT = DzMZT/dcgain(DzMZT);

%% Plot
wmin = 2 * pi * 20.0; % 20Hz
wmax = 2 * pi * ((fs/2.0) - (fs/2 - 20000)); 20kHz

figure(1);
bode(Ds, 'b', DzMZT, 'c', {wmin, wmax});
legend('Analog prototype', 'MZT 2nd order','location', 'northwest');
grid on;

There are few other ways to calculate the coefficients as like what mentioned here.

If you don't want to code then here are (75us) IIR coefficients for most common Fs:

44100
den = [ 1 -0.2430506562509079 -0.3653116688328392 ]
num = [ 0.226236781592716 0.1524695626369773 0.01293133068655965 ]

48000
den = [ 1 -0.2625656192085276 -0.3736509425970024 ]
num = [ 0.2102915623850396 0.1415200473180778 0.01197182849135262 ]

88200
den = [ 1 -0.3702185951814965 -0.419953874650389 ]
num = [ 0.1217100727211314 0.08132561501764404 0.006791842429339071 ]

96000
den = [ 1 -0.3813262768034237 -0.424764645791489 ]
num = [ 0.1125121591800709 0.07513002079134688 0.006266897433669414 ]

176400
den = [ 1 -0.4405600251706213 -0.4505357896180226 ]
num = [ 0.06328932872397297 0.04212236025403318 0.003492496233349943 ]

192000
den = [ 1 -0.4464885169088165 -0.4531262255096845 ]
num = [ 0.05834721577148291 0.03882109525679095 0.003216946553225094 ]

352800
den = [ 1 -0.4775710488916642 -0.4667412351991018 ]
num = [ 0.03239176683609103 0.02151798489625147 0.001777964176891494 ]

384000
den = [ 1 -0.4806342452619782 -0.4680860610215655 ]
num = [ 0.02982988806358219 0.01981315821469544 0.00163664743817871 ]

Magnitude error less than 0.1 dB (20Hz-20kHz)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.