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I'll use the non-unitary Fourier transform (but this is not important, it's just a preference): $$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt\tag{1}$$ $$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega\tag{2}$$ where (1) is the Fourier transform, and (2) is the inverse Fourier transform. Now if you formally take the Fourier ...

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The error lies in the assumption that if $g(t)$ is the Hilbert transform of $f(t)$, then the Hilbert transform of $f(-t)$ must be $g(-t)$. This is not the case. Let $f^-(t)=f(-t)$. Then we have $$g(t)=\mathcal{H}\{f\}(t)=\frac{1}{\pi}\text{p.v.}\int_{-\infty}^{\infty}\frac{f(\tau)}{t-\tau}d\tau\tag{1}$$ and \begin{align}\mathcal{H}\{f^-\}(t)&=\frac{... 11 The most practical attempt that I am aware of is by Won and Berger (2005). They simultaneously recorded vocalizations at the mouth with a microphone and on the skull with a homemade vibrometer. They then estimated the relevant transfer functions with linear predictive coding and cepstral smoothing. 10 It is not impossible but it is not going to be a walk in the park too. What you would be trying to do is to add to the voice signal, those vibrations that are delivered to the ear via the bones and are not accessible to anyone else. But this is easier said than done in an accurate way. Sound propagation through a medium depends very much on its density. ... 9 I think it is kind'a similar to soft and hard thresholding using in wavelet de-noising. Have you come across this topic? pywt has already an in-built function for this purpose. Please take a closer look at this code and try to play with it: import pywt import matplotlib.pyplot as plt import numpy as np ts = [2, 56, 3, 22, 3, 4, 56, 7, 8, 9, 44, 23, 1, 4, 6,... 8 The closest orthogonal transform I know of that might meet your needs is the Slant Transform. It's based on sawtooth(ish) waves, but some of the basis functions do resemble triangle waves: (source: Applied Fourier transform) It was developed for image coding/compression, but it seems like a reasonable first approach for the analysis of long-term linear ... 8 Transliterations of Ukrainian names have different avatars in English (and in others languages as well). You can find Kravchuk polynomials, and other papers like On Krawtchouk Transforms or Krawtchouk polynomials and Krawtchouk matrices. You can find as well Kravchuk orthogonal polynomials. As they form an orthogonal basis of polynomials (as well as many ... 7 Correlation and convolution are basically the same operations. You can express the cross-correlation of two functions f(t) and g(t) by a convolution:R_{fg}(\tau)=f(\tau)\star g^*(-\tau)$$where \star denotes convolution, and * denotes complex conjugation. If you evaluate the cross-correlation at \tau=0 you get the inner product of f(t) and ... 7 The problem is not sufficiently specified, because the range of admissible values of n is missing. Here I make the assumption that we consider n>0. With this assumption we have$$X(z)=\sum_{n=1}^{\infty}x[n]z^{-n}=\sum_{n=1}^{\infty}\frac{z^{-n}}{n^2}\tag{1}$$And that's the point where we might get stuck, if we didn't have a list of mathematical ... 6 The fourier transform gives you very fine resolution in the frequency domain, but during the transformation, you loose all the information about when (for time signals) or where (for images) these frequencies occur in your input signal. The Gabor transform alleviates this problem by windowing the base functions of the fourier transform with a Gaussian ... 6 As mentioned in Batman's answer, the condition of the sequence being absolutely summable is only sufficient but not necessary. The Fourier transform can be extended to \ell_2 sequences, i.e. sequences for which$$\sum_{n=-\infty}^{\infty}|f[n]|^2<\infty$$is satisfied. A further generalization is possible if you allow distributions and their ... 6 This is achievable with two parallel all pass filters. The two all pass filters synthesize an odd ordered low pass filter whose pass band extends from -90º to +90º in the z-domain. (I will discuss this below).$$ G_{lowpass}(z) = \frac{A_0(z)+z^{-1}A_1(z)}{2} $$The low pass filter is then rotated by +90º so that its pass band extends from 0º to 180º, which ... 6 I would use a linear phase FIR Hilbert transformer, and use block processing, such as the overlap-add method. That means that you partition the input signal into contiguous non-overlapping blocks and compute the convolution of each block with your filter impulse response. The results are then overlapped and added. Overlap occurs because the result of the ... 6 It's all about structure. One early paper on this is A Unified Treatment of Discrete Fast Unitary Transforms, 1977: A set of recursive rules which generate unitary transforms with a fast algorithm (FUT) are presented. For each rule, simple relations give the number of elementary operations required by the fast algorithm. The common Fourier, Walsh-... 5 complex morlet was added Aug 10, 2007 ricker and cwt were added Sep 20, 2011 There's no indication that cwt is meant to be compatible with morlet. As cwt docstring says: Wavelet function, which should take 2 arguments. ... second is a width parameter, defining the size of the wavelet (e.g. standard deviation of a gaussian). The morlet function takes 4 ... 5 I have insufficient reputation to answer in the comments, so here goes: I believe Olli calculated his coefficients using some kind of genetic algorithm (I don't know the details). All I did was plot (from Olli's coefficients) the resulting pole/zero positions in the z-plane, and then take the logarithm to transform into the s-plane. Olli's poles and ... 5 Your first solution using the properties of the Fourier transform is correct. Your second solution is wrong, because you forgot to include the unit step function. Your function g(t) should be defined by$$g(t)=e^{-t}u(t)\tag{1}$$which gives for g(2t-1)$$g(2t-1)=e^{-(2t-1)}u(2t-1)=e^{-(2t-1)}u\left(t-\frac12\right)\tag{2}$$Consequently, the Fourier ... 5 Wavelet transforms can be more difficult to interpret than FFT at face value due to the various representations, nomenclature and output formats. I had to study more than 15 resources to get a good sense of the variety and which one is used by Pywavelets (which does not provide much theory or explanation in its documentation). In order to grasp the meaning ... 4 First order B-splines are triangles, and there exist algorithms to represent an arbitrary signal as a sum of B-splines. As mentioned, these splines do not form an orthobasis, but this is not necessarily a terrible thing. A good place to start is the the paper by Unser on efficient B-spline approximation. http://bigwww.epfl.ch/publications/unser9301.pdf 4 Because the left (top) edge of an image is unlikely to be a reflection of its right (bottom) edge, there are discontinuities all along the edges of an image when it is viewed as N-periodic. These discontinuities are represented in the frequency domain by high-frequency coefficients. From an image compression point of view, using the 2-D DFT as a transform ... 4 You are right that the (bilateral) Laplace transform can be interpreted as the Fourier transform of e^{-\sigma t}f(t). However, I think that the significance of the Laplace transform only becomes clear when s=\sigma+j\omega is viewed as a complex variable because then we can study the analytic properties of the system function. E.g., electrical networks ... 4 The plot is of$$\mid X\left(i\omega\right) \mid = \sqrt{\left(\frac{1}{a+j\omega}\right)\left(\frac{1}{a-j\omega}\right)} = \frac{1}{\sqrt{a^2 + \omega^2}}$$against \omega In particular \omega can be equal to -a. This checks out with Wolfram alpha 4 Strictly speaking you can't because without specifying the ROC, the inverse Laplace transform is generally not unique. However, in many contexts there is the implicit assumption of causality of the corresponding time function (i.e., x(t)=0 for t<0), which is equivalent to stating that the ROC is a right half-plane. 4 Looks like you need a general explanation of the discrete wavelet transform (DWT). DWT breaks a signal down into subbands distributed evenly in a logarithmic frequency scale, each subband sampled at a rate proportional to the frequencies in that band. The traditional Fourier transformation has no time domain resolution at all, or when done using many short ... 4 The generalisation of the concept of an analytic signal is not straight forward. I'm quite certain however that looking for such a generalisation with quarternions (or even octonions) will not turn out fruitful. Those generalise complex numbers primarily algebraically, attempting to preserve as much of the field structure as possible, and not so much as a ... 4 In contrast to Jason R's answer I claim that the Hilbert transform is a phase shift by -\pi/2 for real-valued signals. By definition, a phase shifter shifts the phase of a sinusoidal signal by some given phase \phi:$$x(t)=\cos(\omega_0t)\quad\Longrightarrow\quad y(t)=\cos(\omega_0t+\phi)\tag{1}$$Since$$\cos(\omega_0t)=\frac{e^{j\omega_0t}+e^{-j\...

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I give you some hints and then you can solve this homework. Your $x[n]$ has only $8$ nonzero values. Figuring out what happens to them (in an exhaustive way) is not difficult. Consider $(n-1)^2$ and change $n$ over $(-\infty, \infty)$. You can see that it is always non-negative. It means that we never can have $(n-1)^2=-1$. Hence the samples of $x[n]$ for $... 4 A single instantaneous phase estimate may or may not make any sense if there is more than one frequency peak in the signal's local spectrum. So, to get a better single frequency and phase estimate, you may want to first high-pass filter your signal to remove the spectrum of any D.C. bias or slow trend, and also band-pass filter the signal to remove any ... 4 Note that the antiderivative of a function is only defined up to a constant. Furthermore, note that if you integrate a periodic function, the result is not necessarily periodic. Let $$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}\tag{1}$$ Now we integrate$(1)$with an arbitrary lower integration limit$t_0$. I'll say more about that later. $$\begin{... 4 Your confusion comes from the fact that you use X(\cdot) for denoting both functions, the function of \omega and the function of f, but they are really two different functions, because$$X(\omega)=\hat{X}(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt,\quad\omega=2\pi f\tag{1}$$That's why in all the correspondences you mentioned you can just replace$...

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