How to filter a signal with a transfer function?

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Ana Gabriela Guedes
评论： Star Strider 2021-5-27
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?

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Star Strider
Star Strider 2021-5-26
Ran in:
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 个评论
Ana Gabriela Guedes
Ana Gabriela Guedes 2021-5-26
编辑：Ana Gabriela Guedes 2021-5-27
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 2021-5-27
Ran in:
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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