# Tag Info

12

If you remove (for the time being) that leading factor $A$ as a constant gain factor: $$H(s)=\frac{s^2+\left(\frac{\sqrt{A}}{Q}\right)s + A}{As^2 + \left(\frac{\sqrt{A}}{Q}\right)s + 1}$$ what you get then is a symmetric, but otherwise general shelf that could be equally described as "LowShelf" or "HighShelf". In dB, the gain at the low ...

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"Zero-Mean" is the word that's commonly used to describe signals and signals with a zero average. "This is a zero-mean filter." If you really mean a filter that is specifically meant to cancel the DC component, a "DC blocker" is a name for that.

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The logical implications are the following: "non-recursive" $\Longrightarrow$ FIR IIR $\Longrightarrow$ "recursive" But the opposites are not necessarily true because a FIR system can be implemented recursively (transfer function poles can be cancelled by zeros). Of course, when referring to "recursive" or "non-recursive" we always talk ...

6

If a first-order IIR will do, modify that slightly, and you're done. So the usual first-order low-pass filter can be defined as $y_n = h(\theta_n)$ such that $y_n = y_{n-1} + a(\theta_n - y_{n-1})$. This works great for $\theta_n \in \mathbb{R}$. You want a low-pass filter that's defined on an interval that spans $360^\circ$. For reasons that will become ...

6

Given a set of values ${\left\{ {s}_{i} \right\}}_{i = 1}^{N}$, we're basically after: $$\arg \min_{x} \sum_{i = 1}^{N} \left| {s}_{i} - x \right|$$ One should notice that $\frac{\mathrm{d} \left | x \right | }{\mathrm{d} x} = \operatorname{sign} \left( x \right)$ (Being more rigorous would say it is a Sub Gradient of the non smooth ${L}_{1}$ Norm ...

6

Recently I have seen the paper Hui Yin, Yuanhao Gong, Guoping Qiu - Side Window Filtering. They suggest a really simple filtering framework for Edge Preserving Filter: Basically, what they suggest is filtering the image with a set of filter based on the Box Filter. This filter set is basically composed of 8 filters with different orientations and sub sets ...

5

You can equalize magnitude and phase simultaneously by defining a desired complex frequency response $$D(\omega)=M(\omega)e^{j\phi(\omega)}\tag{1}$$ with magnitude $M(\omega)$ and phase $\phi(\omega)$ chosen such that they compensate for the given magnitude and phase distortions. An FIR filter approximating $(1)$ can be designed by using the following error ...

5

We need to separate the concept of edge detection from the tools we use to apply the procedure. Edges are local property of the image. Being so local means we don't analyze the image in frequency domain but in spatial domain. Yet, a common step for edge detection is applying High Pass / Gradient Filter. Since those are Linear Shift Invariant operators we may ...

5

So for frequency hopping spread spectrum (FHSS),when the receiver is unaware of the transmitter's hopping pattern, does that mean the ADC sample rate needs to be higher than twice the entire hopping bandwidth? Yes. If I use a 2Msps (4x data BW) ADC it looks like I will still recover my data; sure it under-samples some of MHz modulated frequency but it does ...

5

If you use a parameterization with a (pole or zero) frequency and a Q-factor for numerator and denominator of a biquadratic function you get the following general second-order transfer function $$H(s)=G_{\infty}\frac{s^2+\frac{\omega_z}{Q_z}s+\omega_z^2}{s^2+\frac{\omega_p}{Q_p}s+\omega_p^2}\tag{1}$$ For a low shelving filter we want $H(\infty)=1$, i.e., $G_{... 5 This is a nice question. The math is actually pretty simple once you embrace the method I derived the Wiener Filter in - How Is the Formula for the Wiener Deconvolution Derived? So, here is the model: $$\boldsymbol{y}_{i} = \boldsymbol{h}_{i} \ast \boldsymbol{x} + \boldsymbol{w}, \; i = 1, 2, \ldots, n$$ Where$ \boldsymbol{w} $is an additive white ... 5 Image Processing Context In classic Image Processing the filters used are known. Hence being separable is a property of a given filter which is suitable to the task. In this context, separability only means we can have a more efficient way to apply the filter computationally while the end result is the same. So, in Image Processing, if you have a filter ... 4 This is three years later, but since I don't see the real answer posted here, I will post it. The correct answer is that if we are literally interpreting the original statement as a purely mathematical claim, taken at face value, then it is incorrect. There do exist causal filters, even minimum-phase ones with nice closed-form Fourier domain expressions, ... 4 Median filtering is non-linear, and pretty awesome about removing outliers. You just need to adjust the length of the filter based on the estimate of the frequency of the errored samples. 4 Or another way, visually, with the plot of the magnitude response of your filter shown below you see that it goes to zero at$\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$F_s$... 4 Their numbers work out right if you do the stated decimation before the stage, and you pay attention to the fact that decimations are cumulative. So stage one operates at 1/3 of the input rate, stage two operates at 1/6, stage three operates at 1/12, and stage four operates at 1/48. If you scale the adds and multiplies by those numbers, then their stated ... 4 Ripples in a spectrum are an indication of echoes in time. Specifically consider how the Fourier Transform of a sinewave is two impulses in frequency. Similarly, given the reciprocity of the Fourier Transform, a sinewave in frequency (in other words, ripples in the spectrum) is two impulses in the time domain. The Fourier Transform of the frequency response ... 4 The purpose of pulse shaping filters is not to overcome ISI as is implied in the OP's question. The only reason for using a pulse shaping filter is spectral efficiency, and in the process ISI can be introduced if not done properly. In order to limit bandwidth, pulse shaping must extend the time domain response for each pulse beyond a symbol boundary, but can ... 4 This must be an artifact of the implementation for the hilbert function (which to be clear is NOT the Hilbert Transform, but the analytic signal which consists of the signal plus it's Hilbert transform on the imaginary axis), in that any implementation of finite length will have a finite ripple. It doesn't appear the underlying filter using in the hilbert ... 4 Your result is correct, even if it seems a bit counterintuitive. Note that you don't compute the Fourier transform but the discrete Fourier transform (DFT). The system you want to invert has the impulse response $$g[n]=\frac12\delta[n-19]\tag{1}$$ Obviously, the inverse system must advance the incoming signal by$19$samples: $$g_{inv}[n]=2\delta[n+19]\tag{2}... 4 Is there an invertible low-pass filter No is there something particularly difficult about inverting a low-pass filter? Yes. Digital low pass filters (in the most common sense) have a zero at Nyquist which means that the inverse has infinite gain at Nyquist and is unstable. Seemingly having two identical coefficients in b as the first and only ... 4 First, make sure your problem is indeed faster in frequency domain. While asymptotically DFT based convolution should be faster in practice for small kernels it is not. One way to do it is to pad (See MATLAB's padarray()) the array and then use valid equivalent convolution as in Replicate MATLAB's conv2() in Frequency Domain. 4 Given a system with a known frequency response in the S-domain. Is there a way to find whether the system is linear and time invariant? If by "known frequency response in the s-domain" you mean a Laplace transfer function* as a ratio of polynomials in s -- yes. Laplace transform analysis on a system is not valid unless the system is linear and ... 4 the predominant stage of the speaker is the audio signal processing stage. That's a questionable statement. Transducers, enclosure design, physical management of diffraction and dispersion are arguably more important and can not be fixed with signal processing (analog or digital). Keep in mind that the speaker emits a complicated 3-dimensional sound field ... 4 A full convolution is Lx + Lh - 1 samples long. This can be truncated to two meaningful subsets, as documented in conv(). filter() is generally an iir filter, and will have an infinite output. Thus it is truncated at input length. If you like, you can read final state and insert initial state by adding parameters to i/o. Implementation of FIR and IIR ... 4 The frequency response of the corresponding filters must satisfy H^2(\omega)=H(\omega)\tag{1} This was already pointed out in Hilmar's answer, but the conclusion is not that the only option is H(\omega)=1 for all \omega. The correct conclusion from (1) is that the frequency response H(\omega) must equal either 0 or 1 for any frequency. One ... 4 From the transfer function alone it is generally impossible to say whether a system is causal or not. Only in combination with a given region of convergence (ROC), or, equivalently, with an assumption about stability, can we know for sure if a given system is causal or not. The given transfer function has a pole at s=1. There are two possible time domain ... 3 As you can see, the second measurement of 358 is an erroneous measurement. Why would that be erroneous? Phase is periodic with 2\pi or 360 degrees. That means 358 degrees is the same angle as -2 degrees. Your data looks perfectly fine if you look at it as 2, -2, 3, 5, 10, 18 , ... What you probably need is phase "unwrapping". If you see a ... 3 Matt L.’s exact answer in equation (3) is correct. But here is a simpler “exact” solution: 𝛼 = 2𝑦 / ( 𝑦 + 1 ), 𝑦 = tan( 𝜔𝑐 / 2 ) (11) This is computationally more robust. Eq (3) could lead to small errors depending on the resolution of your floating-point math. 3 A good way to deal with circular (directional) averages is to turn it into a vector average. To find the average angle \bar{\theta} of several angles \theta_n (in radians) then:$$ \bar{\theta} = \arg \left ( \sum_{n=0}^{N-1} e^{i \theta_n} \right )$$where$\arg\$ is the argument (angle) of the resulting complex sum. There's some more stuff about this ...

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