# Tag Info

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Maximum likelihood (ML) estimator Here will be derived a maximum-likelihood estimator of the power of the clean signal, but it doesn't seem to be improving things in terms of root mean square error, for any SNR, compared to spectral power subtraction. Introduction Let's introduce the normalized clean amplitude $a$ and normalized noisy magnitude $m$ ...

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Common Approaches for Commercial Denoisers Commercial denoisers are different than what you'd see on most papers. While on papers the results are mostly using objective metrics (PSNR / SSIM) and are evaluated vs. Additive White Gaussian Noise (AWGN) with high level of noise real world images are mostly with moderate level of noise with Mixed Poisson ...

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Slope from all samples obtained To summarize the question's problem, you want to calculate the slope based on all samples obtained thus far, and as new samples are obtained, update the slope without going through all the samples again. On the page you cite is the equation for calculation of the slope $m_n$ that together with $b_n$ minimizes the sum of ...

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Update: I'm sorry to have to say that testing shows the following argument seems to break down under heavy noise. This is not what I expected, so I have definitely learned something new. My prior testing had all been in the high SNR range as my focus has been on finding exact solutions in the noiseless case. Olli, If your goal is to find the parameters ...

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I have several blog articles that solve this problem exactly in a theoretical sense, and quite accurately in an implementattion. 3. DFT Pure Tone Frequency Formulas Exact Frequency Formula for a Pure Real Tone in a DFT Two Bin Exact Frequency Formulas for a Pure Real Tone in a DFT Improved Three Bin Exact Frequency Formula for a Pure Real Tone in a DFT A ...

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I faced the same problem in the past. Perhaps there is a way without adding a delay but I haven't found it. You need to realize that your 3 first solutions (delay after vq, delay at the delta_freq and delay after the frequency) will yield the same result as omega_g is a constant and because your PI controller has fixed coefficients. Anyway, place the ...

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There are really great answers. I will try to give the Sequential Least Squares approach which generalizes to any Linear Model. Sequential Least Squares Model We're after solving the Linear Least Squares model: $$\arg \min_{\boldsymbol{\theta}} {\left\| H \boldsymbol{\theta} - \boldsymbol{x} \right\|}_{2}^{2}$$ Now imagine that we have new measurement ...

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$\operatorname{E}[|X(\omega)|^2]$ has the useful quality that if $S$ represents the clean signal and $N$ represents the noise, and $\operatorname{E}[S] = 0,$ $\operatorname{E}[N] = 0,$ and $S$ and $N$ are independent, then it follows from: $$X = S + N,$$ that: $$\operatorname{E}[|X(\omega)|^2] = \operatorname{E}[|S(\omega)|^2] + \operatorname{E}[|N(\omega)... 3 Basically I'd do something like:$$ \frac{ {\sigma}_{2}^{2} }{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2}} {r}_{1} + \frac{ {\sigma}_{1}^{2} }{ {\sigma}_{1}^{2} + {\sigma}_{2}^{2}} {r}_{2} $$This is the optimal weighing given knowledge of the Variance only (Well, linear). Basically it assumes the cross correlation is 0. Derivation There are many ways to derive ... 2 This is a really nice problem. Problem Formulation I will formulate it as following: Let  x \in \mathbb{R}^{n}  be a signal. Given  y \in \mathbb{R}^{n}  which is a noisy measurement of  x  such that  y = x + v  and  z  be a noisy measurement of the derivative of  x  such that  z = F x + w  where  F  is the finite differences operator. ... 2 I'm not sure what's you model is. Let's say it is something like:$$ y = H x + n $$Now, using the Least Squares model is optimal (In the MSE sense) when  n  is AWGN (It is the linear optimal estimator if the noise is white). So unless the noise in your model is colored, no gain by filtering the data before applying the Least Squares method. Now, what ... 2 Simply because that gives us a power spectral density, and that is usually more helpful than some amplitude density. But that's just a convention. We could just as well use \sqrt{E\left[\lvert X(\omega)\rvert^2\right]}. Your considerations with randomness are off. The whole point of the expectation operator E is to give you a non-random property of a ... 2 Scale-invariant minimum mean square error (MMSE) improper uniform prior estimators of transformed amplitude This answer presents a family scale-invariant estimators, parameterized by a single parameter which controls both the Bayesian prior distribution of amplitude and the transformation of amplitude to another scale. The estimators are minimum mean square ... 2 An interesting approximative solution of the maximum likelihood (ML) estimation problem is obtained by using the asymptotic formula$$I_0(x)\approx \frac{e^x}{\sqrt{2\pi x}},\qquad x\gg 1\tag{1}$$Using the notation and formulas from Olli's answer, the optimum ML estimate of the normalized clean signal amplitude satisfies$$\hat{a}=m\frac{I_1(2m\hat{a})}{...

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I understand your question like this You have new points $(x_i,y_i)$ coming in constantly, and would like to update the estimate of your slope $m$, analogously as you would with a running average (ie without computing the whole sums again for all the values). Suggestion Why don't you simply take the formula that is given in your link and split the terms ...

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I disagree with aspects of other the given answers. It depends on signal model that is assumed. If $x(t)=s(t)+n(t)$ such that the probability density $x(t)$ is $$p(x(t))=\frac{1}{\sqrt{2\pi}\sigma} e^{(x(t)-s(t))^2/(2\sigma^2)}$$ then $$E\{ x(t) \}= s(t) \quad \text{and} \quad \ne 0 \quad \text{even if the time average of} \; s(t)=0$$ and $$E\{ x(... 2 It's a spurious artifact due to not referencing your phase generation to the center of your data window, and not doing an fftshift before the fft to place your phase reference point at fft input index 0. e.g. For varying frequencies, either you have to know, generate, or want the phase at location T/2 of your continuous sinusoid. The phase result of an ... 2 A simple solution is to calculate the first coefficient of a DFT of appropriate length, using the summation formula instead of FFT. To get amplitude and phase, transform the result to polar coordinates. 2 There is a mistake int the conv function which you are using in your code. In ofdm, the channel must be convoluted with every symbol. the link provided in the above comments are ok, but you need to modify them according to your parameters. so replace the command of y = conv(x_ifft_p2s,h,'same'); by below command: for jj = 1 : n_ofdm_sym y(jj,:) =... 2 The clue is that without zero-padding, the circulary shifted sequence is just a shifted version of the periodic continuation of the original sequence. That's why without zero-padding the magnitudes of the DFTs of both sequences are identical. When you use zero-padding, the two sequences cannot be obtained from each other by pure shifting. So the DFTs must ... 2 A resistance isn’t a particularly dynamic state, should be an unknown constant, but you might have a bin of resistors where they might vary. Taking one “randomly” out the bin makes it a random variable. The bin of resistors will have a mean value, so perhaps that mean constitutes a state variable. Perhaps someone starts putting resistors in the bin from a ... 2 There are indeed many peak detection algorithms, and no clear consensus on which ones are "good" or "bad". But for what it's worth, your approach makes sense. Using median or other quantiles to detect sparse signals is common, e.g. the "median clipping" stage in Lasseck (2014), Large-scale identification of birds in audio recordings. In effect, you're ... 2 I'd do some small adjustments to your idea (You really nailed them). Assumptions The Signal Model - Signal + Additive White Gaussian Noise (AWGN) Probably we could generalize it more but this is beyond the scope of this question. The DFT of the signal contains Peaks with relatively small roll off This is important as we're almost saying the Signal is a ... 1 For simplicity I will show an approach Id' use on 1D signal (A row of real world image). You will be able to extend it and I will add few remarks on how you can even gain from having 2D data. The general idea is as sketched in Estimate the Discrete Fourier Series of a Signal with Missing Samples. The trick here is to exploit prior information. In our case ... 1 It depends on what you mean by SNR. It's a common joke in the DSP community to spell it out as "something to noise ratio", referring to the fact that there is no unique definition of SNR, so the term by itself means nothing. Define it yourself and use it appropriately. What's common is to define it as {\rm SNR} = \frac{P_{\rm s}}{P_{\rm n}} where P_{\rm ... 1 Does the calculation of the CRLB imply that only particular estimation algorithms can be used (e.g. estimating delay and Doppler from a complex ambiguity surface) which satisfy the CRLB inequality? All unbiased estimators satisfy the CRLB. You don't, however, have to use an unbiased estimator. Biased estimators might be lower in variance than the CRLB ... 1 As the name implies, a matched filter is found to be an exact replica of the signal of interest to be detected; just a reversed version as a result of an optimization. Hence your question reduces to whether the signal of interest is produced as a solution of some diferential equation. Even though it's true that there are signals as solutions to ... 1 Maximum Likelihood under the assumption of Additive White Gaussian Noise (AWGN) is always equivalent to finding the hypothesis with the minimum distance to given data. Since minimizing distance is equivalent (In the euclidean Space) of maximizing the correlation you can always build the idea of Match Filter for parameter estimation in the settings of ML ... 1 Now I wanted to show you how to get those minimum linear mean square estimator coefficients a and b for your given problem setup. The procedure is summarised from the book Statistical Digital Signal Processing_MonsonHayes. Given two random variables X and Y, we observe X and want to estimate Y using a linear estimator :$$ \hat{Y} = a\cdot X + ...

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So in your case doesn't the relation $x = n+y$ help ? I mean, assuming your derivation for the mean square estimtor is right, then to compute $E\{xn\}$ you would look for $E\{ (y+n)n\}$ and using properties of $x$ and $n$ you would get $$E\{xn\} = E\{(y+n)n\} = E\{yn\} + E\{n^2\} = 0.5 + 1 = 1.5$$

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