8

Creating the Frequency Vector The arrangement of the output of fft() depends on whether you use an odd or even number of points for your fft. I think this post nicely summarises how the frequencies are arranged. Have a look at it. Since you are using an even number of points, the Nyquist frequency, $F_N = F_s/2$, is present in the output of your fft, and is ...


6

Assuming you are referring to the standard definitions of odd and even symmetry, i.e. \begin{align} \text{even: }&& x[-n] &= x[n] \\ \text{odd: }&&x[-n] &= -x[n] \end{align} then it still holds that the DFT of an even sequence is real (i.e. the phase is 0 or $\pi$) and the DFT of an odd sequence is imaginary (i.e. the phase is $\pi/2$...


5

Well this goes to show that Fourier series is just approximation that gets more and more correct when you add more harmonics. Take a look at this: $\dfrac{4\sin\theta}{\pi}$ is just first harmonic. Harmonics are integer multiple of base frequency as you can see: $\sin3\theta$, $\sin5\theta$ etc. And the more of them you add to the first harmonic the more ...


5

If you did a continuous on off keying of a 10101010... pattern, then you would see sidebands as described since this is simply an up-conversion of the Fourier Transform of a 50% duty cycle square wave (moved to any carrier frequency). However if the data pattern for this case of a rectangular on-off keying was random, the resulting spectrum would be ...


4

The only signal, that really has just one frequency component, is an infinite sine signal. Limiting the signal duration in any way is bound to produce other frequency components, as time limiting can be thought of as multiplying with a rectangle window, which translates to convolution with an si-function in the frequency domain, thus introducing new ...


4

Know that each bin in the FFT is the value for the exponential $e^{j\omega t}$, not sine or cosine. So for the OP's case $5sin(2\pi 3 x-2)$ in terms of exponentials (see Euler's identity) this is: $$\frac{5}{2j}e^{j(6\pi x -2)} - \frac{5}{2j}e^{-j(6\pi x -2)}$$ Each is a DFT bin with that magnitude and phase given by the exponentials above, as long as ...


4

it seems like the bigger problem is that subsampling the signal would result in aliasing frequencies No, subsampling in frequency domain corresponds to aliasing in time domain. So the idea here is to purposefully alias in time domain so that you get a sub-sampled FFT. That is exactly why 'Claim 3.7' of paper mentions $y_i=\sum_{j=0}^{n/B-1}x_{i+Bj}$. These ...


4

I will introduce some terminology and intuition that will be helpful when reading other references. It will be neither complete nor completely rigorous. The measures that we first encounter in real analysis assign sizes (non-negative real numbers) to measurable subsets of $\mathbb{R}$; Lebesgue measure is the measure that agrees with the intuition we build ...


4

Do you know how ADS is plotting the spectrum? Plotting the spectrum without doing some kind of normalization will give you a higher magnitude. Once you determine the size of the FFT, normalizing by the length will give you the same magnitude no mater what sampling rate you choose. For example let's take two rectangular signals, one sampled at 1 MHz and the ...


3

Consider a wheel rotating counter-clockwise at one revolution per second. Its frequency of rotation is 1 Hz. If it rotates clockwise, its frequency of rotation is -1 Hz. It's as simple as that.


3

(Ideal) square waves are often drawn in a misleading way, because the vertical lines don't actually represent a signal value. The square wave actually jumps instantanously between two values, creating a discontinuity. In other words, a function that is defined like this: $$f(t) = \begin{cases} 0\,\,\,t<0\\1\,\,\,t>=0\end{cases}$$ is discontinuous. ...


3

Ultimately if your noise is white then the matched filter would be the best approach (multiply by one frequency and sum-- if it is complex you would multiply by the complex conjugate- both a form of correlation). This is equivalent to computing a single bin in the DFT (assuming your frequency is on bin center, otherwise equivalently the operations in the DFT ...


3

There is no problem in your plot if by fft you mean DTFT (or Discrete Time Fourier transform) and it is true to say "the points on this dash line are the DTFT result of corresponding frequency": If your function is a cosine: $$ x[n] = \cos(\omega_0n) \Longrightarrow X(e^{j\omega}) = \pi\delta(\omega-\omega_0)+\pi\delta(\omega+\omega_0)\ ; \qquad|\...


3

It's whatever makes the most sense for the problem at hand. Which can get confusing, and the same signal may have multiple definitions of "bandwidth" -- sometimes even in the same document, when, for example, one is reading up on or designing a communications system. Useful Bandwidth A signal's user is interested in the practically useful ...


3

See the MATLAB documentation: s = spectrogram(x) returns the short-time Fourier transform of the input signal, x. Each column of s contains an estimate of the short-term, time-localized frequency content of x.. Namely each column of the matrix s is the result of an fft() on some samples of the input. So the plot you see is the magnitude of the columns of s. ...


3

It's almost a matter of philosophy, i.e., difficult to argue hard facts. On the one hand all the features you mention can be extracted from the raw signals. So in theory the network should be able to learn how to do that if they provide meaningful information for the task at hand. This is what part of the ML community is claiming: feature engineering is dead,...


3

Conventional AM and narrowband PM signals look quite similar: $$x_{AM}(t) = A \cos(\omega_c t) + m(t) \cos(\omega_c t) $$ $$x_{NBPM}(t) = A \cos(\omega_c t) - A k_p m(t) \sin(\omega_c t) $$ And based on this, their spectrum also look quite similar. However, there's an important distinction between them: AM modulation changes (modulates) the amplitude of the ...


3

They have very different phase spectra. For a dirac the phase of the spectrum is zero for white noise, it's uniformly distributed over $[0, 2 \pi]$ They are both constant over the whole frequency range That's only true for the spectrum of the Dirac. For white noise, it's more complicated. The FFT of a chunk of white noise does NOT have constant spectrum, ...


3

I guess this probably just a mistake in your analysis code. Quantization noise is white and for a noise signal it's uncorrelated to the original signal, so spectrum of the original noise doesn't matter. I did repeat your steps and saw exactly what I expected: The quantization noise is white and the spectrum of the quantized signal follows the original signal ...


3

It's convention, they're equivalent: $$ \exp{\left(j2 \pi \frac{N}{2}n/N \right)} = \exp{\left(j2\pi \frac{-N}{2}n/N\right)} \\ \Rightarrow e^{j\pi n} = e^{-j \pi n} \Rightarrow \cos(\pi n) = \cos(-\pi n)=(-1)^n,\ j\sin(\pi n) = j\sin(-\pi n) = 0 $$ MATLAB and Numpy go $[-N/2, ..., N/2-1]$, which is unfortunate for analytic representations (+ freqs only). ...


3

Ripples in a spectrum are an indication of echoes in time. Specifically consider how the Fourier Transform of a sinewave is two impulses in frequency. Similarly, given the reciprocity of the Fourier Transform, a sinewave in frequency (in other words, ripples in the spectrum) is two impulses in the time domain. The Fourier Transform of the frequency response ...


2

sampled at 25.6 Hz, 25.6 kHz is much more likely. Your signal contains a strong fundamental (probably the turbine itself) plus some "fuzzy" stuff on top which is the vibration signal the author is apparently after. You see this clearly in the spectrum: a strong peak at low frequency and a bit of fuzz between 4k and 8k Apply a high pass filter to get rid ...


2

A high-pass filtered signal has lost its low frequencies, so you can refocus the spectrum on a higher frequency range. When you rectify it, the first phenomenon is that the signal becomes positive. Hence, it is not zero-mean anymore, and thus recovers low-frequencies that were attenuated in the high-pass filtered signal. A second is the presence of higher ...


2

One technique that I've seen used in determining bearing faults is using the kurtosis of the vibration signal. You can track as a function of time what Wikipedia calls the sample excess kurtosis. This is the kurtosis that is different from the kurtosis you would see if the signal was Gaussian distributed. The sample excess kurtosis is defined as: $$ \frac{...


2

Solution in python using Modpoly, Imodpoly and Zhang fit algorithm. Python library BaselineRemoval has Modpoly, IModploy and Zhang fit algorithm which can return baseline corrected results when you input the original values as a python list or pandas series and specify the polynomial degree. Install the library as pip install BaselineRemoval. Below is an ...


2

Given $ \left\{ x \left[ n \right] \right\}_{n \in M} $ where $ M $ is the set of indices given for the samples of $ x \left[ n \right] $. The trivial solution (Which it would be great to have a faster more efficient solution is what I'm looking for) would be: $$ \arg \min_{y} \frac{1}{2} \left\| \hat{F}^{T} y - x \right\|_{2}^{2} $$ Where $ \hat{F} $ is ...


2

A sinusoid with a frequency that is between bins in the FFT frequency domain is circularly discontinuous in the time domain. So you can't use the same IFFT results back-to-back without the noise from this discontinuity between each IFFT window, as the end of one window will have a value too far from the beginning of the next window (unless your frequency is ...


2

What does it mean exactly to "exist" vs "just theoretical"? Do we for some reason think that $cos(\omega t)$ exists while $e^{j\omega t}$ does not? Both are equally mathematical constructions that describe our physical world. We somehow conclude that the latter as a complex quantity does not exist but the former as a real quantity does, ...


2

Negative frequencies exist both mathematically and logically and you could probably accomplish the logical demonstration yourself if you want but I'll try. The mathematical demonstration is much more straightforward. OK so the logical approach would be this. Consider the energy flow in a tank circuit in a problem you are analysing. When the energy flows from ...


2

There are two numbers that square to be $-1$. Pick either one of those two numbers and call that "$j$". Then the other one is "$-j$". Doesn't matter which one is picked. The difference between $+j$ and $-j$ is only an arbitrary choice. A convention. Now multiply that $+j$ and $-j$ by a single non-zero real number. Doesn't matter which sign but let's ...


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