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Also, an accelerometer has a very flat frequency response till its (high frequent) resonance point. But as the name gives away; the sensor provides you with acceleration data. If you want to know velocity, you need to integrate the signal once, and if you want to know displacement, you need to integrate the signal twice. Each integration step adds a pole at ...


$\frac{s^n + b_{n-1}s^{n-1}+\cdots+b_0}{s^n + a_{n-1}s^{n-1}+\cdots+a_0} = 1 + \frac{(b_{n-1}-a_{n-1})s^{n-1}+\cdots+b_0 - a_0}{s^n + a_{n-1}s^{n-1}+\cdots+a_0}$, and the inverse Laplace transform of 1 is $\delta(t)$. Is that enough to get on with?


Hi, a bit late to the game, but a solution for an Nth-order Low-pass Butterworth time-domain response written in C# is shown below. I hope it will (still) be of use. double ButterLowPass(double time, double Fmax, int order) { double angle, cosPart, expPart, h_of_t, t, scale; Complex Delta, RootN, Roots, Residue; // Best to use complex numbers, to work with ...


The function butter() computes the coefficients of a discrete-time ("digital") Butterworth filter, whereas the gain formula you used is valid for a continuous-time ("analog") Butterworth filter. According to the R documentation you can use butter() to compute an analog filter using the plane argument. The gain of a discrete-time ...


You probably meant to reverse them: The LPF should be at 1100 Hz, and the HPF should be at 100 Hz. Then you're keeping everything between 100 Hz and 1100 Hz, and throwing away lower and higher frequencies. will eliminates the low frequency samples Also remember filters don't eliminate everything in the stopband, they drop off with frequency, so some of ...


Your data have a non-zero DC component which is filtered out by the bandpass filter. So the filter output has approximately zero DC, which causes the corresponding vertical shift of the signal. If you want to retain the DC information in the signal you must use a lowpass filter instead of a bandpass filter.

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