# Tag Info

## Hot answers tagged resolution

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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 ...

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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 ...

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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 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 ... 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 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 My input signal is a dc signal (sensor output) and im getting kind of a headache to understand why oversampling can increase the resolution of dc signals. The ADC puts out integers. So, let x be the integer that would come out of the ADC (100.3, say, or -333.3). Now let y = \lfloor r \rfloor be the quantization operation, where you get straight ... 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 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. 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 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 ... 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) $$where all parameters are ... 2 If you a-priori assume that k \cdot f is a stationary harmonic of periodic waveform of frequency f, that implies that you don't have any separable time locality of that harmonic. But any tighter bandwidth of a frequency estimation might come from the a-priori assumptions regarding the stationarity of the harmonics, not from the data. 2 Yes, the center pixel and focal length in pixels will change, as described in the link above. However, if you learn distortion parameters (radial and tangential) then they shouldn't change as resolution changes because they operate on the projective image plane (before multiplying by camera matrix) instead of pixel coordinates (after multiplying by camera ... 2 The uncertainty principle sets up a theoretical bound for resolution, so it is never written as an equality. The equality relationships you are encountering are for for a specific analysis context and analysis implementation. In this case the context is signal analysis so time/frequency are the conjugate variables of interest, and the implementation is the ... 2 I was not able to access the article you were referring to, but I think you may find this one quite interesting. The authors presented their version of Goertzel algorithm which can be used to find amplitudes and phases at frequencies that are non-integer multiples of the fundamental frequency in the given signal. That means their algorithm does improve the ... 2 Consider shifted and scaled versions of a mother wavelet \psi(t):$$\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}...

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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 ...

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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 ...

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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 ...

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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.

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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 ...

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The resolution of a FFT can be described as FS/N , where FS is the sampling frequency and N is number of points. If you want more resolution in your case just take aquire more data before running FFT function.. 40k/8192 gives ~5Hz per bin.

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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 ...

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What may explain why your spectrum is noisy is that you are computing it using a single burst of data. You will have to smooth it by averaging successive spectrums applied on your measurements. The number of samples that shall be used for the averaging have to be tuned depending on the speed of variation of your phenomena. After having done this, the other ...

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What the OP shows is the difference between truncation and rounding which in terms of error simply introduces an offset. They generate the same quantization noise spectrally; the standard deviation of the quantization noise (and equivalently noise power) is equivalents since DC offset is ignored in those computations. If there was concern with the offset ...

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