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1

The problem with your example is that $\infty\cdot 0$ isn't necessarily equal to zero. The only way to judge what is happening in the limit $K\to\infty$ is to divide the original equation by $K$: $$\frac{D(s)}{K}+N(s)=0\tag{1}$$ Now it is obvious that for $K\to\infty$ the actual value of $D(s)$ is irrelevant, as long as it is finite. Consequently, the only ...

4

Or another way, visually, with the plot of the magnitude response of your filter shown below you see that it goes to zero at $\pi/2$. So what you could do is use the relationship between the angular frequency $\omega$ in [radians/sample] and the relative frequency $f$ in [cycles/sample] $$\omega = 2\pi f = 2\pi \left(\frac FF_s\right)$$ Then compute $F_s$ ...

3

The general approach would be to find the zeros of the filter's transfer function. If you do things right it will turn out that there are two complex conjugate zeros on the unit circle at $\pm j$, i.e., at half the Nyquist frequency. Now you need to choose the sampling frequency such that $80$ Hertz corresponds to half the Nyquist frequency.

0

One option could be to make a track with an impulse, and filter that with your filter in Audacity, and save the resulting impulse response to extract the filter kernel, and then apply that in Python.

1

You can use the manage/export functionality to convert your Audacity filter preset into a text file. The file contains frequency/value pairs plus some side info. It looks like Audacity builds an 8192 FFT grid and uses a B-spline to interpolate the frequency/value pairs over the entire grid. I assume that just do an inverse FFT plus some circular time ...

0

For the evaluation only at specific frequencies, you need to specify the frequency vector with at least two frequencies in it (see MATLAB's freqz). Below is the MATLAB code for the evaluation at the frequencies $\omega = 0, \pi/4, \pi/2, 3\pi/4, \text{and}\ \pi$. >> [h, w] = freqz([1, sqrt(2), 1], 1, [0 , pi/4, pi/2, 3*pi/4 pi]) h = 3.4142 + 0....

0

A simple moving average with a rectangular window will turn a step into a ramp, but no linear filter can give you a constant slope, because linearity implies that scaling the input (step) will scale the output (slope). Instead, what you want is a non-linear slew-rate limiter that steps towards the target at constant rate (eg. take the minimum of the "...

1

If the system described by the transfer function $H(s)$ is stable, you can obtain its frequency response by substituting $s=j\omega$, and use the relation that you found: $$S_Y(\omega)=S_X(\omega)\big|H(j\omega)\big|^2\tag{1}$$ where $S_X(\omega)$ and $S_Y(\omega)$ denote the power spectra of the system's input and its output, respectively.

2

\begin{align} Y(z) &= E(z) + \frac{1}{1-z^{-1}}(X(z) - z^{-1} Y(z)) \\ &= E(z) + \frac{1}{1-z^{-1}}X(z) - \frac{z^{-1}}{1-z^{-1}} Y(z) \\ \\ Y(z) + \frac{z^{-1}}{1-z^{-1}} Y(z) &= E(z) + \frac{1}{1-z^{-1}}X(z) \\ \left(1 + \frac{z^{-1}}{1-z^{-1}}\right) Y(z) &= E(z) + \frac{1}{1-z^{-1}}X(z) \\ \frac{1}{1-z^{-1}} Y(z) &= E(z) + \frac{1}...

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