New answers tagged

2

(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. ...


0

Moving average filter is a good smoothing filter in time domain but is a bad low pass filter in frequency domain due to slow roll off and bad stop band attenuation. It is important to look at the time and frequency response of the filters before deciding on them.


0

1) Is this a theoretically accepted method to smooth data in DFT domain or only applies in time domain ? Sort of. It certainly can be done this way. Doing it directly in the DFT (instead of the PSD) is risky since it's a complex number you can get a lot of cancellation if the phase is fluctuating a lot. You also need to decide how you manage the "edges": ...


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 ...


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 ...


2

In the line Inverse_GShifted=real(ifft(GShifted)); I have changed to Inverse_GShifted=real(fftshift(ifft(GShifted))); This is because the ifft output is from $0 \le n \le N-1$, but your actual time output is symmetric about $n=0$. So what you have to do is to center it around $n=0$ by doing fftshift() command. It will give the output of $-N/2 \le n \le ...


1

The magnitude response of both the sine and the cosine waves are the same . The difference is in the phase response. The extra j term in the definition of the FT of sine introduces the $\pi/2$ in the phase response and since there is a minus sign between the two delta functions of the FT of the sine wave ($j*pi*(\delta(\omega+\omega0)-\delta(\omega-\omega0))$...


1

The assumption you made is that sum of 2 complex exponentials with opposite phase is always $sin$. That is not true. $x(t) = 4 + 4 e^{j\omega_0 4t + \pi /2} + 4e^{-j\omega_0 4t -\pi /2} + 2e^{j\omega_0 3 t + \pi /2} + 2e^{-j\omega_0 3 t +\pi /2} = 4 + 8cos(4\omega_0 t+\pi /2) + 4cos(3\omega_0t -\pi/2)$.


1

The video could've stopped at the answer: $x(t)=4+ 2e^{j(3\omega t + \pi /2)} + 4e^{j(4\omega t - \pi /2)} + 2e^{-j(3\omega t + \pi /2)} + 4e^{-j(4\omega t - \pi /2)}$ And this answer can be read directly from the plots. The idea is that the $n^{th}$ term in the sum is equal to $|C_n|e^{j(n\omega t+\angle C_n)}$. You get this from the Fourier series ...


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 ...


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.


1

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, but I don't see a ...


-1

With respect to a known given fixed point in time (and space if you assume a time-space coordinate system), a negative frequency sine wave is simply the negative of a sine wave or sine() function basis vector that starts with a phase of exactly zero at that same exact point in time. The fixed point in absolute time can also be defined to be periodic with ...


0

How can it be that the amplitude is −0.5 Primarily sloppy notation in the book. The frequency domain representation of a signal uses complex numbers, i.e. you would need two real numbers to describe it. Either Amplitude and Phase or Real and Imaginary Part. The correct way to describe this would Amplitude is $0.5$ and the phase is $\pi$, or Real part is $...


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You are likely looking for the "Power Spectral Density" which is determined by taking the complex conjugate multiplication of the Fourier Transform. The complex conjugate process will give the expected real result and represents a power quantity (which has no phase). For the DFT this is often presented in normalized fashion such as dBFS (dB relative to full ...


0

@jithin 's answer gave me an idea that I think answers my question. I'm going to define a new window function: $$ W_{\Delta t}(t) = \theta\left(t-\frac{\Delta t}{2}\right)\theta\left(\frac{\Delta t}{2} - t\right) $$ This is the same as $W_{\Delta t}(t)$ in the post except for the missing factor of $\frac{1}{\sqrt{\Delta t}}$. The PSD is then defined as $$ ...


1

I am assuming $X(t)$ is a passband signal having finite support (non zero value for only a finite range of frequencies within spectrum analyzers range). In your equation (1) what you have missed is $f_{LO}$ to $f_{IF}$ conversion. So $W_{\Delta t}$ should be $W_{IF}(t) = e^{j2\pi (f_{LO}-f_{IF})t}W_{\Delta t}(t)$. When you convolve $X(t)$ and $W_{IF}(t)$, ...


0

I suggest monitoring the phase versus time directly instead of frequency. Frequency is the derivative of phase so the slope of the phase would indicate the frequency. Detrend the phase slope for the starting frequency and then the point in time where the phase starts to ramp up should be easier to detect. The window in which to detect this change will be ...


-1

The word spectrum does NOT mean "a range of frequencies". Stated differently, the word spectrum does not mean "a frequency range". The word spectrum means "a description (a portrayal) of how much energy exists in a signal at all the different frequencies within some frequency range". That description could be a two-column table with one column listing the ...


0

I think the term spectrum in signal processing came from 'colour spectra' (https://en.wikipedia.org/wiki/Spectral_color) where in distant objects were used to determine whether they were moving away or towards earth. Not really sure of it though. So if you see the wiki link, there are different frequencies (or wavelengths) assigned to each colour band. ...


1

If yes then how can this be done ? Yes, it can be done. But doing it directly in the frequency domain is awkward, cumbersome and slow. The most efficient way is to generate it directly in the time domain using a rotating phasor. If you need it in the STFT domain, just generate a properly aligned frame of data in the time domain, window and FFT it. That's ...


1

If this is about using the STFT, then it's about frames of audio (or whatever signal class) and, if it's about frames, it's about windows. Usually the windows we want are complementary, they add to 1. An example would be a Hann window. So now imagine your sinusoid of an arbitrary frequency (mid bin or between bins or something else) being multiplied by a ...


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 ...


0

However, since I have taken the modulus, it must be impossible to go from spec back to audio correct? You need to resort to some approximations like the Griffin-Lim algorithm or various vocoders (WaveNet,WaveGlow,etc.). So does that mean that librosa.istft does NOT convert a spectrogram to a wav file? It's a bit more complicated than that. An FFT ideally ...


1

Spectrogram contains magnitude values only, that explains why the values start at 0 instead of some negative value. And scipy seems to calculate it with 24bit accuracy, where $2^{24}\approx 16.7\cdot10^6$.


0

What is the link between them? Power should not be function of frequency (it is the integral on all frequencies of power spectral density) As far as a Spectrum Analyzer is concerned, to display the power spectrum in dBm, what it does it to take the FFT and compute square of magnitude $P_n = 20 \times log_{10}(|X(f_n)|^2*1000/100)$ with $f_n$ being the ...


0

In addition to the resolution bandwidth fact, i think power spectral density has something to do with stochastic signals. If your signal is predetermined, you have to work with power instead of power spectral density. One good example is the very sine wave you mentioned. The mean power for that signal in its main frequency can be defined using the integral ...


1

I'm not sure but there's something called the time-frequency resolution limit, basically the shorter your windowing interval the broader your spectrum is going to be. The frequency resolution in the low frequency area may be poor because of this. Things like wavlets attempt to resolve this problem but i dont know anything about them.


0

I think the question as is, is unanswerable, but here I am making an attempt of correct it with these comments as answer. Should I upsample before or after rectifying? For the algorithm, if you are finally assessing against a threshold in magnitude, you should not care about down|upsampling. Do not add that. Does this process look correct? Strictly ...


1

It is null-null when the signal is bandpass and zero-null when the signal is baseband. So you will always take the positive frequencies.


1

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

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 ...


1

(I am relatively new to this field of study as well, but here is my input which I hope is of help). Frequency analysis techniques should be chosen that reflect the characteristics of the system you use. Fourier analysis in this way assumes that the data is stationary for each sample interval over which the data is collected. As you only take the data in 10 ...


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