How to Design a Pre-emphasis filter knowing the gain, a zero and a pole?

Question

I am simulating an FM radio signal in order to use it as a waveform in passive radar; for this purpose I'm following an article: Analysis and Emulation of FM Radio Signals for Passive Radar

In a certain step, I need to design an emphasis filter for the rdiophonic signal, the filter has these characteristics:

• a gain of -5.4 dB ?(is it the gain at half of sampling frequency?)
• a zero imposed by international standards (can't find the zero)and a pole - to make it physically realizable - at 40 kHz. I know the response function of the Pre-emphasis filter

$$ H(j \omega) = 1 + \frac{j \omega}{\omega_0} $$

or in z domain:

$$H(z) = 1-B \cdot z $$

, can't understand where is the pole? Of course i will use a FIR filter ,but how to find the coefficient $B$.

thanks for all

Answer 1

Traditionally, these filters were implemented as analog filters, and they are different from the usual discrete-time first-order FIR pre-emphasis filter used in speech processing.

The transfer function of such an analog pre-emphasis filter is

$$H(s)=g\frac{1+s\tau_1}{1+s\tau_2},\qquad\tau_1\gt\tau_2,\tag{1}$$

where $g$ is the DC gain, and $\tau_1$ and $\tau_2$ are two time constants determining the locations of the zero and the pole:

$$s_0=-\frac{1}{\tau_1}\\s_{\infty}=-\frac{1}{\tau_2}\tag{2}$$

If the pole (or the zero) are given in terms of frequencies $f_i$, then the following relation holds:

$$\tau_i=\frac{1}{2\pi f_i},\qquad i\in\{1,2\}\tag{3}$$

So for a given frequency $f_i$ you can use $(3)$ to determine the corresponding time constant, from which you obtain the desired transfer function $(1)$.

If you want to implement that filter in discrete time, the most common option is to use the bilinear transform:

$$s=\frac{2}{T}\frac{z-1}{z+1}\tag{4}$$

where $T$ is the sampling period, i.e., the inverse of the sampling frequency. Note that $(2)$ is the bilinear transform without pre-warping, which means that the frequency axis will be warped, so neither the pole frequency nor the zero frequency of the analog filter will be exactly realized in the discrete-time domain. You can pre-warp the analog frequencies (or time constants) to make sure that the resulting discrete-time filter has the desired pole and zero frequencies. In order to realize a (pole or zero) frequency $f_i$ in the discrete-time domain, you need to use the following time constant in the analog filter transfer function $(1)$:

$$\tau_i=\frac{T}{2\tan\left(\frac{\pi f_i}{f_s}\right)}\tag{5}$$

where $f_s=1/T$ is the sampling frequency.

You can find more information on pre-warping in this answer.

Answer 2

I want to post the code of the pre-emphasis filter, a matlab transtlation od the code shared by Andy Walls. I think it works perfectly,well I think...

%This code is a succint translation to matlab of the code shared by Andy Walls
%pre-emphasis filter with a gain of -5.4dB, a pole at 40khZ and a constant
%time of 50 µs

%---------Specifications---------------------------------------------------

fs=48000;
to=50e-6;
w_cl=1/to;
f_cl=w_cl/(2*pi);
f_ch=40e3;
g_dc=10^(-5.4/20);

if f_ch >= fs/2 % In my case f_ch=40000 hZ
    f_ch=0.925*fs/2;
end 

w_ch=2*pi*f_ch;

%------use of bilinear transformation for disigning digital filter---------

%Convert digitial specifications into the analog domain, by prewarping
% digital frequency specifications into analog frequencies.

w_cla = (2*fs)*tan(w_cl/(2*fs));
w_cha = (2*fs)*tan(w_ch/(2*fs));

%---------------------Calculation Of poles and zeros for digital filter
%------using the bilinear transformation -----------------------------------

% H(z) = H(s)|s = (2/T)(1-z^-1)/(1+z^-1)
kl=-w_cla/(2*fs);
kh=-w_cha/(2*fs);
z1 = (1.0 + kl) / (1.0 - kl);
p1 = (1.0 + kh) / (1.0 - kh);
b0 = (1.0 - kl) / (1.0 - kh);

%Since  this filter  has a Dc gain of -5.4 dB gain 
% i have to correct with a gain g given as follows

g=(abs(1.0-p1)/(bo*abs(1-z1)))*g_dc;
b=[g*bo -g*bo*z1];
a=[1 -p1];
freqz(b,a);