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One of the definitive features of LTI systems is that they cannot generate any new frequencies which is not already present in their inputs. Please note that in this context a frequency refers to signals of the type $x(t)=e^{j\Omega_0 t}$ or $\cos(\Omega_0 t)$ which are of infinite duration, and are also referred to as eigenfunctions of LTI systems (...

9

You have two problems here. First, as hotpaw2 mentioned, double integration is sensitive to any non-zero bias in your accelerometer, which will certainly be present if you are using a MEMS accelerometer. Suppose you have a bias of $b$; then, your position at time $t$ is $$p(t) = \int_0^t \int_0^{t^\prime} b \ dt^{\prime\prime} \ dt^\prime = \frac{b}{2} t^... 9 The Gabor wavelet is basically the same thing. It's apparently another name for the Modified Morlet wavelet. Quoting from Wavelets and Signal Processing: [The Modified Morlet wavelet] does not satisfy the admissibility condition but is nonetheless commonly used. Sometimes this wavelet is called the "Gabor wavelet," but that term is improper because ... 9 You can make a simple algebraic argument, given the premise that you provided. If:$$ Y(\omega) = X(\omega) H(\omega) $$where X(\omega) is the spectrum of the input signal and H(\omega) is the frequency response of the system, then it's obvious that if there is some \omega in the input signal for which X(\omega) = 0, then Y(\omega) = 0 as well; ... 8 The term Doppler Shift is actually a bit of a misnomer. The frequencies are not actually shifted but they are scaled (see http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html for definition of shifting vs. scaling). It's a relative change not an absolute one. Both time and frequency domains are scaled: when the source is moving towards you, the ... 8 The Fourier transform operator \mathscr{F} is a linear one; i.e.,$$\mathscr{F}\{x(t)\}=X(f) ~,~ \mathscr{F}\{y(t)\}=Y(f) \implies \mathscr{F}\{\alpha x(t) + \beta y(t) \} = \alpha X(f) + \beta Y(f)$$And therefore the system that implements it will be linear. However time invariance, which is tested by the method$$\mathcal{T}\{x(t)\}=y(t) \implies \...

7

I believe that this "color graph" you are looking for is a spectrogram (although it looks to me more like a scalogram, but you did not mentioned wavelets). Let me give you an example in MATLAB of obtaining such plot: load handel nfft = 512; noverlap = 128; win = hamming(nfft); spectrogram(y, win, noverlap, nfft, Fs, 'yaxis') colormap('jet') So first line ...

7

For the source, go to end of the answer Suppose one day you got one note which has some thing written to it, say "Major frequency components are 10 Hz, 25Hz, 50 Hz and 100 Hz". Somehow, you understood that its time-series representation is a very important thing (may be master-piece work of a great musician, or some national security matter, anything). So ...

7

You're overlooking four things: The $\frac{1}{FFT\_size}$ normalization coefficient. Some FFT implementations have or do not have this factor. Check the definition of FFT as performed by matlab on the Mathworks site! Why are you looking at the real part only? The amplitude is conveyed by the modulus (magnitude) of the complex number. Here, the real part is ...

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

If you have a signal $$f[n]=\cos(\Omega_0n)$$ and you apply a time shift of $n_0$ you get $$f[n+n_0]=\cos(\Omega_0(n+n_0))=\cos(\Omega_0n+\Omega_0n_0)=\cos(\Omega_0n+\phi)$$ where $\phi=\Omega_0n_0$ is the phase shift. The other way around, if you have a phase shift of $\phi$, this is not always equivalent to a time shift of the original signal: $$g[n]=... 6 What a tricky question to overlook. Indeed I'm one of those who would immedieately press that Fourier transforms do lose time localization of the events as the comments stated. Yet it's certainly (mathematically and practically) true that any (transformable) signal waveform is exactly preserved under this reversible transform including all of its time ... 6 Fourier transforms generally yield complex spectral data. Under some technical conditions, they are bijections. From the Fourier transform, you can uniquely recover one single signal. However, when looking at spectra, the situation is different: signals can have the same amplitude spectrum and very different phases, as in the example below: Being bijections,... 6 If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:$$\begin{align*}\mathscr{F}\left\{x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)}\right\} &= \int_{-\infty}^\infty x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)} e^{-j2\pi f t}dt\\ \\ &= \int_{-\infty}^\infty x\left(\...

5

You can take images as 2D discrete signals. The "time" in 1D signals is actually two spatial dimensions in images (2D signals). You can measure "frequency" as well - imagine a line of white pixels with regular spacing. The spacing represents period $p$, and frequency is given by $1/p$. Hence the maximum frequency the discrete signal can contain is limited ...

5

Not at all. The Uncertainty Principle says that a function cannot be both limited in time and limited in frequency. More specifically, the product of the signal's widths in time and in frequency (i.e., its time extension $\Delta_t$ and its bandwidth $\Delta_f$) is bounded from below: $$\Delta_t\cdot\Delta_f\ge C\tag{1}$$ where the constant $C$ depends on ...

5

An FT (being invertible) does preserve all transient event time information, however this time locality information is usually preserved by distributed it as varying changes to the entire phase spectrum, which obscures (or is almost like encrypting) all the time locality information. Humans have to decrypt (inverse FT) the phase to make any sense out of ...

5

The result {4,1,2,3} is the circular convolution of {1,2,3,4} and {0,1,0,0} which you correctly get by taking the inverse DFT of the product of the DFTs of the two sequences. We can check this by doing the circular convolution the long way via matrix multiplication as follows: $x(n) = [1,2,3,4]$ $y(n) = [0,1,0,0]$ To solve for the circular convolution $x(... 4 complex morlet was added Aug 10, 2007 ricker and cwt were added Sep 20, 2011 There's no indication that cwt is meant to be compatible with morlet. As cwt docstring says: Wavelet function, which should take 2 arguments. ... second is a width parameter, defining the size of the wavelet (e.g. standard deviation of a gaussian). The morlet function takes 4 ... 4 For a continuous-time signal$x(t)$with Fourier Transform$X(f)$, the Fourier transform of$x(t-t_0)$is$e^{-j2\pi ft_0}X(f)$, that is, the value of$X(3)$, say, gets changed to$e^{-j2\pi 3t_0}X(3)$which has the same magnitude as$X(3)$but a different phase. For a discrete-time continuous-amplitude signal$x[n]$with DFT$X[k]$of length$N$, you have ... 4 It means that they are in phase---think of the signals as waves. Adding noncoherent signals results in cancellation, or fading. 4 Here is how you should do it. The two spikes in the FFT will have an amplitude of 1 each (sum those and you get the time domain amplitude of 2) clc close all clear all f = 1000; A = 2; Fs = 16000; t = 0:1/Fs:100/f; x=A*sin(2*pi*f*t); subplot(2,1 ,1) plot(x) % 100,000 = Fs % 10,000 = Length of the Signal % 100,000/10,000 = 100Hz <- First point in FFT =... 4 The coefficients of the Fourier series that you have computed are, in effect, the spectrum of the periodic signal consisting of the sum of signals$f(t)$delayed in time or advanced in time by integer multiples of$1$second. Mathematically, the Fourier transform of a periodic function has impulses in it with the impulse amplitudes being the Fourier series ... 4 In writing$h(\tau,t)$with$t$is time and$\tau$is delay, we are in the model that$\tau$varies "differently" from$t$. Or in other words, they are different notions in spite of the fact that they are both time unit. Similarly, in this equation $$y(t) = \int h(\tau, t) x(t - \tau) d\tau$$ The time index of$y(t)$varies "differently" from the time ... 4 A theorem, which I know as Weyl's, 1931, is: if$s(t)$and derived functions$ts(t), s'(t)$are in$L^2$with the related$\|\cdot\|$norm symbol then: $$\| s(t) \|^2 \leq 2\| ts(t) \| \| s'(t) \|$$ Equality is attained when$s(t)\$ is a modulated Gaussian/Gabor elementary function defined as: $$s'(t) / s(t) \propto t$$ or practically as: $$s(t) = ... 4 Wavelet transforms can be more difficult to interpret than FFT at face value due to the various representations, nomenclature and output formats. I had to study more than 15 resources to get a good sense of the variety and which one is used by Pywavelets (which does not provide much theory or explanation in its documentation). In order to grasp the meaning ... 4 The sinc is just fine an example as a signal with infinite support: Support being defined as the smallest interval in which the function has non-zero values, it's trivial to see that \text{sinc}(x)=\sin(x)/x (for all x but a single point) becomes arbitrarily small for large x, but never actually constantly \text{sinc}(x>\xi)=0 for any finite \... 4 What you are missing is that this is about a discrete-time system, because we're talking about poles and zeros in the complex z-plane and about poles inside or outside the unit circle. So there is no differential equation, but there is a difference equation:$$y[n]=\frac12y[n-1]+x[n]\tag{1}$$The corresponding impulse response is$$h[n]=\left(\frac12\...

4

With all due caution, no in both cases (title and body question). I'll start with the second one. Continuous wavelets use all dilations of the mother wavelet, which are not accessed with the STFT The STFT is complex in general, and the windowed sine is not. For the first one: I never tried it, and do not remember having seen it in use, and one should ...

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