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A beamformer is basically a spatial filter. It can be passive, just like a temporal filter. Instead of samples separated by time, they are separated by space. A passive temporal filter can be a bandpass that is "aimed" or "steered" at a particular frequency. For passive spatial filters (i.e. beamformers), the filter can be steered towards a particular ...

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The LMS algorithm is based on the idea of gradient descent to search for the optimal (minimum error) condition, with a cost function equal to the mean squared error at the filter output. However, it doesn't actually calculate the gradient directly, as that would require knowing: $$E(\mathbf{x}[n]e^*[n])$$ where $\mathbf{x}[n]$ is the input vector and $e[... 6 Note that the inverse of an FIR system is IIR, and the same is true for the inverse of an IIR system, unless it is an all-pole system, the inverse of which would be FIR. So in most cases the ideal equalizer should have an infinitely long impulse response in order to perfectly invert the channel. In practice almost all adaptive equalizers are FIR filters ... 5 Your code reveals many misconception about what the CMA is supposed to achieve: your step size mu is much too small; note, however, that the optimal step size can only be found through experiment. the variabe noisedB appears to be the desired SNR of the received signal. An SNR of$0\,\text{dB}$as specified by you is very poor (the noise is as strong as the ... 4 In order to be able to choose an optimal value for the delay$\Delta$it's important to understand how the system works. The purpose of the delay is to decorrelate the desired signal$s(n)$and the signal component$s(n-\Delta)$at the input of the adaptive filter. This means that$\Delta$must be chosen such that the autocorrelation$R_{ss}(k)$of$s(n)$is ... 3 One way to do this is to look at modeling your signal: $$x[n] = x_h[n] + x_n[n]$$ where$x_h$is the hissing sound and$x_n$is the noise. If you can say that$x_n$is modeled as: $$x_n[n] = \sum_{k=K_1}^{K_2} a_k \sin(k\omega_0 n + \phi_k)$$ where$K_1$is the lowest harmonic of frequency$\omega_0$that makes it through your high pass filter,$K_2$is ... 2 I think this can be achieved through histogram stretching. Input pixel values (levels) are mapped using linear transform from$<\textbf{min}_{in},\textbf{max}_{in}>$to$<\textbf{min}_{out},\textbf{max}_{out}>. The intensity transform for input and output pixel values look like this: $$I_{out}=\frac{I_{in}-\textbf{min}_{in}}{\textbf{max}_{in}-\... 2 Your question has received quite few contributions, probably because of a lacking content. During a recent conference , I came across the PhD thesis: Détection en Environnement non Gaussien (Detection in a non-Gaussian environment). Since it is in French, I reproduce the abstract here: For a long time, radar echoes coming from the various returns of the ... 2 A simple exponential average has the following difference equation: y(n) = (1-alpha) * x(n) + alpha * y(n-1) To get a different rise and fall rate, you can dynamically choose alpha depending on whether x(n) is larger than y(n-1). For example, you could build have alpha = 0 when x(n) > y(n-1), and alpha = 0.99 otherwise. This could accomplish the "peak hold"... 2 The term smoothing length (also called observation interval) is simply the length of the equalizer, i.e. its number of filter taps. In the code the filter length is L+1, so L is the filter order. Similarly, the length of the channel refers to the length of the FIR filter simulating the dispersive channel. Again, the channel filter length is actually ChL+1. ... 2 This algorithm in general tries to solve an optimization problem each step, defined as,$$ \begin{aligned} & \underset{\textbf{u}}{\text{minimize}} & & \sum_{i\ =\ 0}^{N-1}\left[x^T\!(k+1+i)\, Q\, x(k+1+i) + u^T\!(k+i)\, R\, u(k+i)\right] \\ & \text{subject to} & & x(k+1+i) = f(x(k+i),\ u(k+i)) \qquad \forall\ i = 0, \ldots, N \\ &... 2 I think there is an error in your referenced independence assumption.w(k)$should be the update part$\Delta w(k)$i.e the$w(k+1)=w(k)+\mu \Delta w(k) =w(k)+\mu \frac{x(k)*e(k)}{c+x(k)^Hx(k)}$The error after convergence is uncorelated and zero mean. The two other assumptions are correct. If your filter is fast enough you may use it at ... 2 Recursive least squares (RLS) filters don't use gradient descent. As their name suggests, they use a least-squares fit to determine the optimum coefficients at each time step. Via clever formulation of the filter structure, one can use the calculations done from time step$n$to recursively calculate the updated coefficients for time step$n+1$without ... 2 The issue is possibly that the input signal you have chosen is not persistently exciting. This means that the signal doesn't "excite" enough modes of the filter in order to be able to accurately estimate its parameters. Another way to think about it is that it doesn't have enough energy in enough places in the spectrum: just at the frequency of the cosine, ... 2 That's a pretty broad question. Let's start with a noise cancelling head set, that's about the easiest device. A ANC headset has an internal microphone, that's placed as close to the ear canal as possible. It than runs a regular control loop (http://en.wikipedia.org/wiki/Control_theory): It calculates a speaker signal that tries to keep the signal at the ... 2 In its bare classical form the RLS algorithm, recursively (for every new iteration), solves the classical problem of least squares; by computing the optimal FIR transversal filter coefficients$w[n]$which minimizes the sum of error squares between the produced filter output$y[n]$and a desired response$d[n]$where the error sum is based on the entire data ... 1 No. OFDM isn't constant modulus (i.e. constant envelope) in time domain, if you look at it as one system. It's quite the opposite; it's known for its high PAPR (which you probably know!). This is the case, by the way, even if you use a CM modulation for the individual subcarriers. So, CMA's central requirement isn't fulfilled. Also, the whole point of ... 1 There is no hard rule regarding convergence speed of the block-LMS vs sample-by-sample LMS. It really depends on the scenario. On top of my head is the following two (stationary) scenarios: A very noisy scenario, where a single estimate of the gradient is not enough. In this case, the block-LMS has better gradient estimates and would usually result in ... 1 First, in science, a field is rarely closed, sometimes asleep only. Resistance to low-contrast, real-time, badly scanned, composite documents/writers or from aging medium seem to remain challenges, in similarity with other digital data: robustness, speed, poor acquisition, source separation or error concealment (the same topics, with generic names) are ... 1 To do system identification using a driving function, it is necessary that the driving function$x[n]$be broadbanded, meaning that the driving function has a Fourier Transform of non-zero value over a broad range of frequencies. The reason for this is that division by zero is a problem. Think of System Identification in terms of this most basic method: ... 1 Suppose that you are adapting$w$to minimize$\text{E}(y[n]-w[n]*u[n])^2$where $$y[n]=h[n]*u[n]+\nu[n]$$$y[n]$and$u[n]$are known and$\nu[n]$is an additive noise component. With a long enough FIR filter you can model any linear system$h$with any accuracy. You can opt for a shorter filter length in expense of limited modelling accuracy. A key ... 1 Prior to upsampling, you have a white signal meaning every single frequency in the Nyquist bandwidth from$-\pi$to$\pi$is represented. This is a requirement to obtain an impulse (because the Fourier transform of an impulse is a white spectrum). The reason that white signals are often used as inputs for purposes of system identification is that they excite ... 1 In all adaptive signal processing schemes, be it a Least Mean Squares (LMS), Recursive Least Squares (RLS) or a Kalman Filter, The fundamental concept is the update of some parameter: such as the vector filter coefficients of a Transversal Tapped Delay Line (an FIR indeed) structure of LMS or a state of a dynamical system as in the Kalman filter. The ... 1 I can add that LMS algorithm has a sample-based update. 1 In a very simple form: Gaussian-like noises are frequent, and derivatives of squared functions are easy to minimize. In very simple words: error, directly, cannot be used as a criterion. You can minimize a signed function to$-\infty$, which is not really useful here. As least, you need a cost function with a lower bound. Power (if by that you mean the ... 1 I think you are seeking for an algorithm but this may help. If you don't know how$ s $is distributed but you know its moments of first and second order, you can try with the linear estimator of the linear Minimum-mean Square Error (lmmse), with the following form, $$\hat{s}(y) = ay + b$$ with$$a = \frac{ \mathrm{C}_{xy} }{ \sigma_{y}^{2} } \\ b = m_{s} ... 1 As mentioned in the comments, the symbol$^H$denotes the conjugate transpose of a matrix or vector, which means that the vector/matrix is transposed and that all of its elements are conjugated. The bar over$y_k$means complex conjugate (note that$y_k\$ is a scalar, not a vector or matrix). When computing the gradient with respect to a complex variable in ...

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Typically adaptive algorithms are used for this type of DSP application. There are a lot of algorithms you can try. All of them are in the family of minimising algorithms / numerical optimalization algorithms (LMS, RLS ...) You can use an adaptive filter in a specific structure, that you tune continously with the algorithm. x(k) + n_1(k) ───────────────────...

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Try raising the input to the Nth power, then 1st- order low pass filtering, followed by the Nth root. If N is 2 then this is equivalent to rms with exponential forgetting factor. Bob

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CPU is cheap these days, try them all. Unless the data set is massively large 2-dim binary search should do it The problem is ill posed: for every given number of entries there will be many possible combination of Minimallänge and Pausendauer, so you need on extra constrain or assumption to get a unique answer A good way to visualize this would be the ...

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