11

The normalization is basically a preconditioning to decrease condition number of the matrix $A$ (the larger the condition number, the nearer the matrix is to the singular matrix). The normalizing transform is also represented by a matrix in the case of homography estimation, and this happens to be usable as a good preconditioner matrix. The reason why is ...


9

My answer is for real scale $a$ and the fact that wavelet transform is usually defined in $L_2$ with norm $$||\Psi(\tau)|| = \int_\mathbb{R} \Psi(\tau)\Psi^*(\tau)\mathrm{d}\tau $$ So $$||\Psi_{a,t}(\tau)|| = \int_\mathbb{R} \frac{1}{|a|}\Psi(\frac{\tau-t}{a})\Psi^*(\frac{\tau-t}{a})\mathrm{d}\tau$$ Set $\tau' = \frac{\tau-t}{a} \implies d\tau' = d\tau / ...


9

Whether you scale the output of your DFT, forward or inverse, has nothing to do with convention or what is mathematically convenient. It has everything to do with the input to the DFT. Allow me to show some examples where scaling is either required or not required for both the forward and inverse transform. Must scale a forward transform by 1/N. To start ...


9

I can think of several reasons involving computational precision issues, but that probably would not do justice because mathematically we're defining it the same way no matter what, and mathematics knows no precision issues. Here's my take on it. Let's conceptually think about what DFT means in signal processing sense, not just purely as a transform. In ...


7

Actually, 3 different ways to put the scale factors are common in various and different FFT/IFFT implementations: 1.0 forward and 1.0/N back, 1.0/N forward and 1.0 back, and 1.0/sqrt(N) both forward and back. These 3 scaling variations all allow an IFFT(FFT(x)) round trip, using generic unscaled sin() and cos() trig functions for the twiddle factors, to ...


5

It depends on the normalization that you perform on the data. Note that for computing the Pearson correlation coefficient you subtract the means of the signals. This is normally not the case if you simply compute a mean squared error between the signals, unless mean removal is part of your normalization procedure. I assume you compute the Pearson correlation ...


5

FIR coefficients are $h[n]$, the same as the impulse response. there are $N$ non-zero taps. $$ h[n] = 0 \qquad \text{for } n<0, n \ge N $$ frequency response is $$\begin{align} H(e^{j \omega}) &= \sum\limits_{n=-\infty}^{+\infty} h[n] e^{-j \omega n} \\ &= \sum\limits_{n=0}^{N-1} h[n] e^{-j \omega n} \\ &= \...


4

To complete bjou's answer: A potential problem with min-max normalization is that it is very sensitive to outliers. For example, if your dataset contain 100 two-dimensional examples; and that the dataset looks like this: 1, 147 8, -252 3, 125 2, -605 ... 10000, -100 <- outlier 4, 200 min-max normalization will squash almost all values of the first ...


4

Both are reasonable approaches and it is foreseeable that either one could outperform the other empirically. The Euclidean distance assumes the data to be isotropically Gaussian, i.e. it will treat each feature equally. On the other hand, the Mahalanobis distance seeks to measure the correlation between variables and relaxes the assumption of the Euclidean ...


4

Assuming that $N$ is the length of your signal $s$, the normalized signal $s_n$ is given by: $$s_n = \dfrac{s}{\sqrt{\dfrac{\sum_{i=1}^{N}\left|s_i^2\right|}{N}}} $$ The denominator is nothing else than Root Mean Square value of your signal. Thus the code is doing a simple RMS normalization. You can think of it as a method of normalizing the average of ...


4

[EDIT] After a second read, the proposed normalization looks non standard. Suppose that $m\le x \le M$ ($m$ and $M$ denote the min and max). The scaling factor will be, depending on the situation: if $m\le M\le 0$: $-m$, if $m\le 0\le M$, and $|m|\ge M$: $-m$ if $m\le 0\le M$, and $|m|\le M$: $M$ if $0\le m\le M$: $M$ It turns out to be (if I do not err) ...


4

Question: Which parameter is suitable to indicate how "good" the measurement fits to the Kalman filter? To estimate a quality of association you can use likelihood function. The likelihood considers not only residual but also uncertainty and represented as scalar value: $$\mathcal{L} = \frac{1}{\sqrt{2\pi S}}\exp [-\frac{1}{2}\mathbf y^\mathsf T\mathbf S^...


3

from your posted waveform, i am assuming that this is a unipolar signal. that is $$ x[n] \ge 0 $$ in audio, it would be the same, except that we would be working on $|x[n]|$ instead. so first you want a sliding maximum of your signal, where the window length is $L$. $$ x_1[n] = \max_{0 \le i < L} \Big| x[n-i] \Big| $$ since your input signal $x[n]$ ...


3

Each pixel of the filter has a magnitude (intensity). The square function equivalently calculate the power density of the pixel. The normalization is to scale the filter power to 1. That's where I^2 at the denominator comes from.


3

The kind of filter you are looking for is a notch filter. Using filter design toolbox in Matlab you can get it as I'm showing you in the following picture: This has to be done in z-plane so there must be two poles at +i and −i since they cannot be included in region of convergence. Is my assumption correct? Yes! This is correct. Assuming you want to ...


3

Let's assume continuous time (rather than discrete time). If you do not process the windowed data at all, you would like the output (the sum of the windowed frames) to be equal to the original signal. Allowing scaling of the output by a constant scaling term, this is only possible if the sum of all of the time-shifted window functions is constant over time. ...


3

Again there is no wrong or right here. In the Alan Oppenheim's Discrete-Time Signal Processing book, the notation is as follows: when there are only continuous-time signals we use $\omega$ for radians per second frequency. when there are only discrete-time signals we use $\omega$ for radians per sample frequency when both types of signals are present, (as ...


2

You could normalize the average amplitude, i.e. YData = YData / mean(abs(YData)); Or you could normalize the signal power to one, i.e. YData = YData / sqrt(mean(abs(YData).^2)); If just the peaks are bothering you, you could use dynamic range compression, but that would introduce nonlinear distortions. As Phonon hinted, please tell us why you are not ...


2

The Wikipedia article states: "What makes the direct linear transformation problem distinct..is the fact that the left [X] and right [AY] sides of the defining equation [X = AY] can differ by an unknown multiplicative factor which is dependent on k" In the above X, A, Y are matrices. So to avoid having to estimate the factor, you simply normalise all the ...


2

There is no definite answer to that question since different workflows are applied to different tasks. For example in some applications you might want to normalize your input data samples or perform some kind of Automatic Gain Control (i.e. you work in fixed point and even for some low signal values you want your FFT to have good resolution). On the contrary ...


2

There are different conventions in scaling of the FFT, in MATLAB you need to scale it by $\sqrt{N}$, where $N$ is your number of samples. Saying it in matlabish: clc, clear all %% Create some sinusoidal signal with noise t = linspace(0, 6*pi, 1000)+randn(1,1000); s1 = sin(2*pi*t); % Calculate the norm of s1 s1_norm = norm(s1); display(sprintf('L2 norm of s1:...


2

Some of the channel effects can be indeed removed by doing the Cepstral Mean Subtraction/Normalization. Nevertheless that generally applies only to "convolutive" distortions that are constant. Any additive distortions, i.e. white noise, babble noise usually cannot be removed via CMS. But like you said this topic is handled via different methods to cope with ...


2

Wavelets play differents role in functional spaces, especially as unconditional bases (see What are unconditional bases and which wavelets have this property?). In $L_p$ spaces, if $|\psi|^p$ is integrable, the translation operator $t\to t-\tau$ preserves the norm, while the dilation $t\to t/a$ induces a scale factor of $|a|^{1/p}$, which can be corrected ...


2

here's an efficient sliding maximum algorithm that has cost that is $O(\log_2(L))$. below window_width is $L$. comes from Brookes: "Algorithms for Max and Min Filters with Improved Worst-Case Performance" IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS—II: ANALOG AND DIGITAL SIGNAL PROCESSING, VOL. 47, NO. 9, SEPTEMBER 2000 #define A_REALLY_LARGE_NUMBER 3....


2

After a few quick calculations, it seems to me that the trouble comes from poor notations for the root in your reference. If you read, in the final normalized matrix, $\sqrt{8/64}$ and $\sqrt{2/4}$ instead of $\sqrt{8}/64$ and $\sqrt{2}/4$ (along with the $\pm$ signs), then the final result is correct. The matrices $V_i$ are orthogonal. To normalize them, ...


1

First, shift: put the minimum to $0$, by compensating the actual minimum $m=−18.3667⋅10^5$ for every pixel: $p\to p - m$. Now your pixels are between $0$ and a new maximum $M = 9.3127 - m$. Finally you want the final image in $[0,255]$. The second operation is scale: multiply by something, so that $0$ remains at $0$, and $M$ is cast to $255$. So you have to ...


1

Search for radar plot to track association. There's a lot of algorithms on this subject. To your question: The residual itself will not give you information without its associated covariance matrix Try a chi-squared test on it. Putting a threshold on this scalar is called gating and it's a first step of plot to track association.


1

You need to understand that $f=f_s/4$ and "$4$ samples per cycle" are two ways of saying the same thing: $$\text{ # samples per cycle}=\frac{f_s\text{ samples per second }}{f \text{ cycles per second}}$$ The number of cycles per second is equal to the ratio $f_s/f$, where $f_s$ is the sampling frequency, and $f$ is the signal's frequency. So you just need ...


1

okay, since this is about scaling, i should be anal about definitions. discrete-time signal: $x[n]$ where $n$ are only integer values. DTFT: $$ X\left( e^{j\omega} \right) \triangleq \sum\limits_{n=-\infty}^{+\infty} x[n] \ e^{-j \omega n} $$ it is necessarily the case that $ X\left( e^{j(\omega + 2 \pi)} \right) = X\left( e^{j\omega} \right) $ for all ...


1

I think this is the equation you are using http://en.wikipedia.org/wiki/Normalization_%28image_processing%29 where b=newMin and a=newMax These values can also be found from your histogram. Usually you expand the new image to take up the full intensity range. The intensity range of your image is the X-axis of your histogram. If it goes from 0 to 1, then a=1,...


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