23

Sampling at a higher frequency will give you more effective number of bits (ENOB), up to the limits of the spurious free dynamic range of the Analog to Digital Converter (ADC) you are using (as well as other factors such as the analog input bandwidth of the ADC). However there are some important aspects to understand when doing this that I will detail ...


18

Your understanding is correct. Goertzel algorithms give almost exactly the same result as 1 bin of a DFT or FFT of the same length or number of samples (and where the FFT twiddle factors are generated by a trig recursion), when used for frequencies that are exactly integer periodic in the Goertzel length. But many forms of the Goertzel algorithm provide ...


8

It is important to define the time and frequency widths $\Delta_t$ and $\Delta_{\omega}$ of a signal before discussing any special forms of the uncertainty principle. There is no unique definition of these quantities. With appropriate definitions it can be shown that only the Gaussian signal satisfies the uncertainty principle with equality. Consider a ...


8

One word for that technique is superresolution. Robert Gawron has a blog post here and the Python implementation here. Usually, this technique relies on each image being slightly offset from the others. The only gain you'd get from not moving between shots would be to reduce the noise level.


5

If you have no prior knowledge about the approximate locations of the frequencies, the Chirp Z-transform is of no immediate use to you. The Chirp Z-transform functions like a magnifying glass, so you need to know where you want to look and the Chirp Z-transform will show you the details. I would suggest you use an FFT to get an idea where the frequencies are,...


5

If you sample at a higher sample rate, you need to analyze (e.g. feed to your CNN) a proportionately longer sample vector to get about the same frequency resolution (or other characteristics of any vibrations, etc.) Or if the input size of your CNN is limited, you can filter and downsample the data to the previous length (and thus lower sample rate) ...


4

Intuitively, if You move the sensor $ N $ steps each at the size of $ \frac{1}{N} $ of its resolution you can get $ \times N $ more resolution. It is like a polyphase representation of the signal. Using estimation methods, any movement which is not an (Event with zero probability) integer multiplication of the resolution of the sensor, namely, fractional ...


4

Indeed, that's the Heisenberg Uncertainty Principle - you can't have both very good frequency and time resolution. You always have to sacrifice something. In case of Short Time Fourier Transform it's straightforward, but for wavelets are being 'squeezed', which is changing their frequency resolution. Figure below describes more than a thousand words: EDIT: ...


4

Apply a sliding window function of 500 samples length to the signal, for example a Hamming window. Choose a suitable overlap between windows e.g. 75%. Calculate the spectrum of each window and then average over all windows. Then the frequency bins lie exactly at the natural numbers as you wanted. But beware: each bin does not only represent one exact ...


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) = ...


3

how does the SDR sample rate translate to sample rate and bit depth of the resulting audio signal? Extremely indirectly. I haven't got precise mathematical formulas for you, but here's an overview of the topic which should let you figure out where you want to dig deeper. First of all, FM is an analog modulation. This means that there is no inherent sample ...


3

I would recommend Array Signal Processing - Johnson, Dudgeon. This book covers classical spectral estimation, Minimum Variance Distortionless Response (MVDR), Linear Prediction, and subspace methods (e.g. MUSIC and ESPIRIT). It provides examples of the resolution capabilities between these methods. The are quite a number of good references in this book if ...


3

I cannot give you all the theory behind this (as it literally fills books), but it turns out that Heisenberg becomes an exact equality for precisely this family of signals: $$ s_{t_0,\omega_0,\sigma,\phi,\gamma}(t) = \exp\left(-\left(\frac{t-t_0}{\sigma}\right)^2 + i \left(\phi + \omega_0 (t-t_0) + \gamma (t-t_0)^2\right)\right) $$ where all parameters are ...


3

Gray-scale resolution and image contrast are different. Gray-scale resolution looks at how many gray scales are being used to present the image. Image contrast, on the other hand, looks at the difference between the maximum and minimum values of the gray scales used to present an image. For example, a low gray-scale resolution, low contrast image might ...


3

Since your offset is 3/5ths (plus some integer), the truncation fractions will be different, resulting in a different sum of quantization noise. You can get rid of this difference by using an integer offset (including/after scaling) before truncation or rounding. (e.g. not (2^N)/5) Also, if your signal is not exactly periodic in your FFT aperture, then ...


2

I was not able to access the article you were referring to, but I think you may find this one quite interesting. The authors presented their version of Goertzel algorithm which can be used to find amplitudes and phases at frequencies that are non-integer multiples of the fundamental frequency in the given signal. That means their algorithm does improve the ...


2

Firstly you didn't mention your sample rate and how your time vector is defined. Yes, this is a quantization problem. The maximum amplitude in Buf2 is 1.6. If you quantize that to integers you only end up with the values -2,-1,0,1,2 so you have massive amounts of quantization noise. You are only quantizing this two 2 or 3 bits. The correct way to turn this ...


2

Consider shifted and scaled versions of a mother wavelet $\psi(t)$: $$\psi_{a,b}(t)=\frac{1}{\sqrt{a}}\psi\left(\frac{t-b}{a}\right),\quad a>0,\;b\in\mathbb{R}\tag{1}$$ By the definition of the Fourier transform $$\Psi(\omega)=\int_{-\infty}^{\infty}\psi(t)e^{-i\omega t}dt$$ it can be shown that the Fourier transform of $\psi_{a,b}(t)$ is $$\Psi_{a,b}...


2

If you a-priori assume that $k \cdot f$ is a stationary harmonic of periodic waveform of frequency $f$, that implies that you don't have any separable time locality of that harmonic. But any tighter bandwidth of a frequency estimation might come from the a-priori assumptions regarding the stationarity of the harmonics, not from the data.


2

Yes, the center pixel and focal length in pixels will change, as described in the link above. However, if you learn distortion parameters (radial and tangential) then they shouldn't change as resolution changes because they operate on the projective image plane (before multiplying by camera matrix) instead of pixel coordinates (after multiplying by camera ...


2

The uncertainty principle sets up a theoretical bound for resolution, so it is never written as an equality. The equality relationships you are encountering are for for a specific analysis context and analysis implementation. In this case the context is signal analysis so time/frequency are the conjugate variables of interest, and the implementation is the ...


2

You can compute any single FFT result bin using a complex Goertzel algorithm or filter. Realize however that if you want to compute a number of bins on the order of log N or greater (such as 10 extra bins with 10 Hz bandwidth and spacing), it may be faster to do the full longer FFT and throw away all the extra FFT result bins. So just compute 2 FFTs of 2 ...


2

One way to do it is to solve a MAP problem of the up scaled video and using the High Resolution images as a prior. Try looking at the articles - Super Resolution MAP.


2

I've taken the original tiff and just set a low threshold for 1bit quantization (conversion to black and white): so we can clearly make out that these artifacts look very circular, and placed in specific patterns around the image borders. I think these are artifacts could stem from a resizing of the image with imperfect anti-aliasing, though I'm certainly ...


2

In scale space the square of width of blurring kernel considered as the scale (variance of Gaussian and not its Standard deviation because if you sequentially blur an image with 2 Gaussian kernel variance of effective blurring kernel will be sum of variance of each kernel) but in multi-resolution analysis the resolution defined by the scaling parameter.


2

The definitions behind the concepts of multiresolution or multiscale may overlap somehow, and are sometimes used interchangeably. Let me provide the following distinction: resolution encompasses spatial discretization, while scale relates to a more continuous framework. The real world can be considered as continuous, and a true image $I(x,y)$ would have ...


2

Most technical terms, such as these two, do not get their definitions purely from their etymology, but rather from the context of application, by experience and by tradition of acceptance. And for this case, your understanding of separability as the allowence (or ability) of someone to distinguish between two closest time of arrivals seems a synonym to the ...


1

I object to the Wiki author's use of the words "resolution" and "sensitivity." Those words are vague and the author does not define or explain them in any meaningful way. If I was forced to use the author's poor terminology I'd say, "A fine-resolution window is a window having a narrow main lobe. (And windows whose main lobe widths are narrow have high-level ...


1

I haven't seen this answer here, so I'll post it. You can resample the signal to a lower sample rate before doing the FFT, which reduces the size of the FFT. If you know that the only frequencies that you are interested in are from 0..1000Hz, you can safely resample the original signal to an FS of 2000Hz and then do the FFT. This decreases the required ...


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