# Tag Info

10

If you consider poles of an integral transform domain to be important to the solution of differential equations: (as usual,) Euler did it first, 1753. One "importance" of poles is that they're part of a very useful representation for linear systems. They must've appeared when people started looking at functions as built from generating functions, so that'd ...

5

It is a common value because at $-3 \ \mathrm{dB}$ the power of the signal is reduced to half its value. I'll show a brief example to make it clearer. Suppose that you have a signal whose amplitude is $1 \ \mathrm{V}$. If we reduce its amplitude by $-3 \ \mathrm{dB}$, the new amplitude would be close to $0.707 \ \mathrm{V}$. The power of a signal is ...

5

This is called a binomial filter. Its transfer function can be written as $$H(z)=\frac{1}{2^M}\sum_{n=0}^M{M\choose n}z^{-n}\tag{1}$$ which gives $$h[n]=\frac{1}{2^M}{M\choose n},\qquad 0\le n\le M\tag{2}$$ for its impulse response. As mentioned in a comment, such a filter can be used to approximate a Gaussian filter. Also take a look at this related ...

4

It seems you mixed up CIC filters and recursive moving-average A recursive moving-average uses a comb-filter and an accumulator. $$H(z) = \frac{1}{N}\frac{1-z^{-N}}{1-z^{-1}}$$ The numerator corresponds to a comb filter (i.e a delay line of N elements and 1 adder). The denominator corresponds to an accumulator. The 1/N factor is to get a moving-average, ...

3

Zeroing out bins / attenuating them in discrete Fourier domain is universally a bad idea, due to the undesirable time-domain effects of that. Instead, use a audio processing program to apply an adjustable equalizer to the song of choice. Let's walk you through Free software: Get audacity; load your song in that, open "Effects"->"Equalizer" start with one ...

3

A plot of the normalized impulse responses, for the n = 2 through 10 Butterworth low pass filters, are given by H.J. Blinchikoff, A.I. Zverev, "Filtering in the Time and Frequency Domains", Wiley-Interscience, John Wiley & Sons, NY, ©1976, p. 113. This is shown below. They do not give the h(t) expressions in the book, at least where I have looked thus ...

3

There's no need to use numerical methods here. The most straightforward way to compute the output is to see that the filter's impulse response is given by $$h(t)=\sum_{k=1}^Nr_ke^{s_kt}u(t)=\sum_{k=1}^Nh_k(t)\tag{1}$$ where $N$ is the filter order, $u(t)$ is the unit step function, and $r_k$ are the coefficients of the partial fraction expansion of $H(s)$: ...

3

As an addition to Tendero's answer, it is mathematically simpler to define the bandedges as the -3.01 dB or "half-power" frequencies. The simplest example is a first-order low-pass filter: $$H(s) = \frac{1}{1 + \frac{s}{\omega_0}}$$ The frequency response is obtained by substituting: $s \leftarrow j \omega$ . Then $$H(j \omega) = \frac{1}{1 + j \frac{... 2 Design your desired magnitude response Make it minimum phase Cascade with any type of allpass filter to create your desired phase response 1 Yes, sorta kinda. Basically, if you can model your system accurately and the resulting model fits the Kalman filter paradigm, then yes. However, I think it's a mistake to give the filter position and velocity -- were I doing this, I'd just give it the wheel position and let the Kalman filter determine the velocity -- which it will do well to the extent ... 1 If you want to have a "numerical grasp" and you're not afraid of getting a little bit dirty, you can check the numbers with LTspice. I don't know how well you know to work with it, so I'll just explain it, feel free to ignore all the redundant info. Here you can download the archive, out of which you only need Filter.asy and filter.sub. Create a new ... 1 I assume that your experiment where you saw no effect was with signals that underwent a linear phase vs frequency process in their delay (constant group delay). Under many situations the phase of the channel in the frequency domain is non-linear, causing different frequency components to experience different delays (group delay variation). When you frequency ... 1 The input and the integrator output are shown below: So, you should consider two cases: t<0 and t>0. In the first case (t<0), you have$$\begin{align} y(t) &= \int_{-\infty}^t x(\tau)d\tau = -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}u(-\tau)d\tau \\ &= -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}d\tau = -\frac{1}{T}Te^{\...

1

Another source of the 3(dB) rule-of-thumb is that is how far two targets/signals/sources must be separated in order for the human eye to distinguish between the two. For example, if you have two rectangle functions, the edge of one must be past the -3(dB) edge of the other for them not to look like one single rectangle function.

1

First you need to find (calculate or look up) the transfer function of that circuit. It has the form $$H(s)=\frac{a}{s^2+bs+c}\tag{1}$$ where the constants $a$, $b$, and $c$ depend on the values of the resistors and capacitors. The $3$ dB cut-off frequency $\omega_c$ can be found by solving \big|H(j\omega_c)\big|^2=\frac12\big|H(0)\big|^2=\frac12\left(\...

1

A real signal with a phase shift does not technically exist since phase is $e^{j\phi}$, so requires an imaginary component except for phases that are modulo $\pi$. (The phase of a real signal is $0$ or $\pi$ radians only). You may be actually looking for a time delay. A constant time delay has a linear phase with frequency, and for the result to be real in ...

1

Normalizing the filterbanks by their widths is optional and totally up to you (similarly to the warping scale Mel/Bark). Depending on your application, you can start without normalization and see what results you are getting. Personally I prefer to keep it fixed and have one knob less for turning. There are more important parameters to tune, such as warping ...

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Can you refer to a reference? I think that I can help you.

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I think you can do a dft for a single frequency fine. Just sum over your window of a whole compression and rarefaction multiplied by the sine of the frequency your looking for, then do it for the cosine, then the amplitude is the hypotenuse if they were graphed on xy axis, and the phase is the angle.

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Here are coefficients for 5th and 6th order filters (fs=48kHz) I prepared using separate HPF (c2d)) and LPF (MIM): 4th order HPF, 1st order LPF: [0.588402730019084 -2.114578549207340 2.574286896638522 -0.919416694862366 -0.367726753456895 0.239032370868995] [1.0 -4.193437345479311 6.853084418397484 -5.397323870680988 2.009149234848459 -0.271472430592242] ...

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