How to filter a signal with a transfer function?

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Hi!
I have a discrete signal (a vector with many values) I want to filter using a low pass filter but I only have the tranfer function in the form. How can I do this since the common filter functions dont use transforms?

Accepted Answer

Star Strider
Star Strider on 26 May 2021
The easiest approach is to first let the Control System Toolbox solve it, then realise it as a discrete filter using the numerator and denominator vectors —
z = tf('z');
H = (1-z^-6)^2 / (1-z^-1)^2
H = z^14 - 2 z^8 + z^2 -------------------- z^14 - 2 z^13 + z^12 Sample time: unspecified Discrete-time transfer function.
Num = H.Numerator
Num = 1×1 cell array
{[1 0 0 0 0 0 -2 0 0 0 0 0 1 0 0]}
Den = H.Denominator
Den = 1×1 cell array
{[1 -2 1 0 0 0 0 0 0 0 0 0 0 0 0]}
figure
freqz(Num{:}, Den{:}, 2^16)
Use ‘Num{:}’ and ‘Den{:}’ with filtfilt to filter the signal.
Remember that it will be necessary to define a sampling interval, ‘Ts’ where ‘Ts=1/Fs’ where ‘Fs’ is the sampling frequency.
.
  2 Comments
Ana Gabriela Guedes
Ana Gabriela Guedes on 26 May 2021
Edited: Ana Gabriela Guedes on 27 May 2021
Thank you! Where sould I insert de sampling interval?
I'm getting the error that filtfilt arguments should me double matrices and I only have row vectors
Star Strider
Star Strider on 27 May 2021
No, use the filtfilt function to filter ‘x’.
Assuming that the sampling interval or sampling frequency is defined in the first tf call:
Fs = 1000; % Define Sampling Frequency
Ts = 1/Fs;
z = tf('z',Ts);
H = (1-z^-6)^2 / (1-z^-1)^2
H = z^14 - 2 z^8 + z^2 -------------------- z^14 - 2 z^13 + z^12 Sample time: 0.001 seconds Discrete-time transfer function.
Num = H.Numerator
Num = 1×1 cell array
{[1 0 0 0 0 0 -2 0 0 0 0 0 1 0 0]}
Den = H.Denominator
Den = 1×1 cell array
{[1 -2 1 0 0 0 0 0 0 0 0 0 0 0 0]}
figure
freqz(Num{:}, Den{:}, 2^16, Fs)
x = [-233....-667]
x_filtered = filtfilt(Num{:}, Den{:}, x)
and go from there.
.

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