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A discrete-time first-order high pass filter with unity gain at Nyquist and a zero at DC is described by the following difference equation: $$y[n]=\frac{1+\alpha}{2}\big(x[n]-x[n-1]\big)+\alpha y[n-1],\qquad -1<\alpha<1\tag{1}$$ Its transfer function is given by $$H(z)=\frac{1+\alpha}{2}\frac{1-z^{-1}}{1-\alpha z^{-1}}\tag{2}$$ Evaluating the squared ...


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Here are couple examples: % R is the resistance value (in ohms) % C is the capacitance value (in farrads) % fs is the digital sample rate (in Hz) % Constants RC = R * C; T = 1 / fs; % Analog Cutoff Fc w = 1 / (RC); % Prewarped coefficient for Bilinear transform A = 1 / (tan((w*T) / 2)); % using Bilinear transform of % % 1 ( 1 - z^-1 ...


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I've used this method in Octave: function [c] = Hz_at_3dB(b, a, fs, samples) [H,W] = freqz(b,a,samples); magresp = 20*log10(abs(H)); maxresp = max(magresp); [I,~] = find(magresp < maxresp-3.0103,3,'first'); c=(fs/2 * W(I(1)))/pi; EDIT: SECTION 8.4: STANDARD RESPONSES - https://www.analog.com/media/en/training-seminars/design-...


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In general there is no straightforward analytical solution. As you know, you need to solve $$\left|H(e^{j\omega_c})\right|=\frac{1}{\sqrt{2}}\tag{1}$$ for $\omega_c$, where it is assumed that the maximum filter gain equals $1$. For Butterworth filters, the specified cut-off frequency always equals the $3\textrm{ dB}$ frequency. This is not the case for other ...


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