17

One of the definitive features of LTI systems is that they cannot generate any new frequencies which are not already present in their inputs. One way to see why this is so, comes by observing the output's Fourier transform $Y(\omega) = H(\omega)X(\omega)$ $$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \longleftrightarrow Y(\omega)=X(\omega)H(\omega),$$ ...


14

Synchrosqueezing is a powerful reassignment method. To grasp its mechanisms, we dissect the (continuous) Wavelet Transform, and how its pitfalls can be remedied. Physical and statistical interpretations are provided. If unfamiliar with CWT, I recommend this tutorial. SSWT is implemented in MATLAB as wsst, and in Python, ssqueezepy. (-- All answer code) ...


10

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^...


10

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; ...


9

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 ...


9

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

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 ...


7

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(\...


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

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

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

An important theorem, known as Weyl's, 1931, is: if function $s(t)$ and related functions $ts(t)$, $s'(t)$ are in $L^2$ (square integrable) with the related $\|\cdot\|$ $L_2$ 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) \...


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

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

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

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

As far as my logic goes, I wont ge able to tell any of musical features just of an amplitude value in the time domain. That's wrong. The amplitude-over-time actually is the way the song is played by your sound card (we call that PCM). So, this is your song. What you really need to understand that your analog audio (i.e. the variations in air pressure over ...


4

A widnow $w[n]$ truncates and weights (tapers) an input signal $x[n]$, to produce $v[n] = x[n]. w[n]$., for subsequent spectral analysis of $x[n]$. A windows's effect on the input signal's true spectrum $X(e^{j\omega})$ is described by a convolution of $X(e^{j\omega})$ with $W(e^{j\omega})$ (window's Fourier transform); $$V(e^{j\omega}) = \frac{1}{2\pi} \...


4

For division there is no equivalent to the duality between multiplication and convolution. Note that from the existence of the Fourier transforms of $f_1(t)$ and $f_2(t)$, you cannot conclude anything about the existence of the Fourier transform of $f_3(t)=f_1(t)/f_2(t)$. So $f_3(t)$ might not even have a Fourier transform, and if it exists, it cannot be ...


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 ...


4

Let's assume you have 2 signals: vX and vY. So: clear(); numSamplesX = length(vX); numSamplesY = length(vY); numSamplesConv = numSamplesX + numSamplesY - 1; vTimeDomainConv = conv(vX, vY); vFrequencyDomainConv = ifft(fft(vX, numSamplesConv) .* fft(vY, numSamplesConv), 'symmetric'); max(abs(vTimeDomainConv - vFrequencyDomainConv)) %<! Should be < ...


4

Low-level intuition can be obtained by inspecting the phase transform, visually. Answer complements and is complemented by this one. (-- Answer code) We consider a pure sinusoidal tone; ideas extend naturally to more complex signals. Band of lines concentrated about a maximum in CWT, and a perfect line in ssqueezed right about f=8. Next, the phase ...


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