# Tag Info

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This is a tricky problem. Roughly speaking, the root cause here is the original filter "destroys" information which cannot be recovered, so you have to make some trade offs The best way to design the inverse is to formulate the inversion as a least square error problem where the error criteria is used to dial in the trade offs for your specific ...

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Yes, the original filter has roots outside the unit circle: so when it's used inverted it will be unstable. One approach might be to form the minimum phase equivalent: minphase_filter = signal.minimum_phase(filter) that will guarantee roots inside the unit circle. The resulting impulse response is: this yields the filtered signal: and this allows s to be ...

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If you have a filter $\sum a_n z^n$, the deconvolution filter $1/ \sum a_n z^n$ attempts to cancel the earliest sample of the input by adding a scaled version of the filter, then proceed to the next sample (like a gaussian elimination process). If the first sample in the filter is small, the gain applied to the filter must be large, and that will cause the ...

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Image Processing Context In classic Image Processing the filters used are known. Hence being separable is a property of a given filter which is suitable to the task. In this context, separability only means we can have a more efficient way to apply the filter computationally while the end result is the same. So, in Image Processing, if you have a filter ...

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For any given problem definition, there's a filter that -- if you ignore execution time and hardware expense -- is "best"*. In general, that "best" filter isn't separable. Depending on the problem at hand, the degree to which the optimum degrades if you find the best separable filter will vary. So -- sometimes a non-separable filter ...

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If $A(t)$ is known, it can be zeroed in the synchrosqueezed representation - the remainder is then $B(t)$, recovered by inversion. $A(t)$ need not be known perfectly - just enough to indentify its time-frequency ridges.

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Of course it's valid for real coefficients because real number is a subset of complex number and the proof doesn't make any assumption whether the coefficient is complex or real. For a complex $d$ \begin{aligned} |A(z)|^2 &= A(z)A^*(z) = \frac{1-d^*z}{z-d} \frac{1-dz^*}{z^*-d^*}\\ &=\frac{1-(d^*z+dz^*)+|d|^2|z|^2}{|z|^2 - (d^*z+dz^*) + |d|^2} \end{... 2 Well, you cannot completely avoid images, you can only suppress them sufficiently. How much "sufficiently" is is completely up to your application! So, we can't tell you this. However, thanks to the fact that you're comfortably oversampling your 400 kHz, the first aliases would appear at 2.93 MHz (=lowest frequency in signal, i.e. -400 kHz, plus ... 0 I want to convert those numbers into poles/zeros That is impossible. These are just two points out of the frequency response, and you need the full frequency response for that. There's almost certainly an analog filter with many-poles/zeros involved here as anti-aliasing filter. You can't even with a perfect model of that filter get all these coefficients ... 0 Here is the trick-: I just found it out somewhere 1 What is Passband Ripple? Passband ripple \delta_1 is typically specified as the 0 to peak difference in the passband gain in the magnitude response of a filter. For a filter with unity gain (1), the ripple will oscillate between 1-\delta_1 and 1+\delta_1. Ripple causes some frequencies in the passband to be amplified and others to be attenuated. In ... 2 The function filtfilt filters the signal twice (forward and backward) in order to eliminate phase distortions. As a side effect, your signal is not filtered by the transfer function corresponding to the numerator and denominator polynomial coefficients supplied to the routine, but by its squared magnitude. You should use an ordinary filter routine, such as ... 1 The Audio EQ Cookbook has the general formulae for all-pass filters (APF). You can maybe cascade four of these APFs together and get 12° at 1 kHz and 360°+4° at 2 kHz. But it's a wild APF and will definitely have some "resonance" (of a sort) around 1.5 kHz. 1 This is a tricky problem. First you need to decide whether you need a "true" all pass filter or can do with an approximate one. The difference being|H_\text{trueAP}(\omega) = 1| \\ | |H_\text{apprAP}(\omega)| -1 | < \epsilon, \omega_1 < \omega < \omega_2  i.e. do you need the magnitude to be exactly one for all frequencies or do you ...

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