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I have a high pass filter given by the time-domain relationship between input and output:

$x'(t) = x(t) - \alpha \times x(t-1)$.

I would like to get its cut-off frequency. How can I proceed?

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First you discretize you equation (note: I assumed that your $x(t-1)$ means a delay of one samples and not 1 second, but you can adapt easily if that's not the case) :

$$ x'[n] = x[n] - \alpha \times x[n-1] $$

Where $$ x[n] = x(n\times T_s)$$

Then you use the $\mathcal{Z}$ transform : $$X'[z] = X[z] - \alpha \times X[z]\times z^{-1}$$

And then you get the transfer function in the Z domain : $$ \frac{X'[z]}{X[z]} = 1 - \alpha \times z^{-1}$$

This is a causal FIR filter, called a 1-zero filter with its zero at $\alpha$. Your cutoff frequency will depend on your sample frequency.

Illustration :

frequency response

Code for the illustration :

figure,
hold on
for alpha=0.1:0.1:1
    h = freqz([1 0], [1 alpha]);
    f = 0:1/length(h):1-1/length(h);
    semilogy(f, 20*log(abs(h)));
    hold on
    grid on
end
legend('0.1', '0.2', '0.3', '0.4', '0.5', '0.6', '0.7', '0.8', '0.9', '1');
title('normalized frequency response')
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  • $\begingroup$ Why do you discretize a continuous time equation :-) ? $\endgroup$ – Fat32 Aug 26 '17 at 10:00
  • $\begingroup$ @Fat32 Well I assumed it's a digital filter and to find its transfer function I needed to factor the $x(t)$... That's how we learn it at school :) $\endgroup$ – Florent Aug 27 '17 at 22:26

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