How to filter a signal with a transfer function?

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Ana Gabriela Guedes
Ana Gabriela Guedes 2021년 5월 26일
댓글: Star Strider 2021년 5월 27일
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 2021년 5월 26일
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.
.
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Ana Gabriela Guedes
Ana Gabriela Guedes 2021년 5월 26일
편집: Ana Gabriela Guedes 2021년 5월 27일
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 2021년 5월 27일
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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