Discrete Low Pass Filter

Hi people,

I'm trying to modell a discrete low pass filter first order with a discrete time integrator.

I want to filter Frequencies up to 5 Hz.

Filter equation is y(k)= (1-a)*y_(k-1) + a*y_(k) where a is (T_s/(T_f+T_s))

How do I have to chose the T_f and T_s to filter under 5Hz

Thank you

Answer 1

[Mathieu NOE]

hello

the a value will give you a cut off frequency of fc if you choose a = 2*pi*fc

[Bleron Kroni]

Hi,

you are saying a = (T_s/(T_f+T_s))

and I have to equalize 2*pi*fc = (T_s/(T_f+T_s))

I have to compansate a with T_f (filter )and T_s --> T_f > T_s for lowpass filter

[Mathieu NOE]

hello

if you know your sampling rate then T_s is fixed . Now I don't have your T_f definition, but I can show you how the cut off frequency , sampling frequency and a factor are linked together

try this :

fc = 5;     % cut off frequency (example)
Fs = 250;   % sampling rate ; suggestion, depends of signal attributes

wc = 2*pi*fc / Fs ;     % Fs = sampling rate, fc = cut off frequency
a = 1-exp(-wc);


% Filter equation is   y(k)= (1-a)*y(k-1) + a*e(k); 
% represented in H(z^-1) transfer function 
numd = [a 0];
dend = [1 a-1];

freq = logspace(-1,1,200);

[mod,phas] = dbode(numd,dend,1/Fs,2*pi*freq);

figure(1)
semilogx(freq,20*log10(mod));
xlabel('Freq (Hz)');
ylabel('Gain (dB)');

you will see on the graph that the -3 dB cut off point is at fc = 5 Hz

you can of course change Fs and fc to match your needs

all the best