# Tag Info

## Hot answers tagged interpolation

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Three reasons to increasing the sampling rate further are 1) To relax the requirements of the post D/A conversion filtering for image rejection. 2) Increase signal SNR by spreading quantization noise for a fixed number of DAC bits across a wider frequency range. 3) Minimize passband droop in the D/A reconstruction. Reason 1 is the most dominant one in my ...

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This visual phenomenon appears because the maximum frequency is close to the Nyquist frequency, or half the sampling frequency. Sampling begins to approach the limit of $2$ samples per period, and thus the linear interpolation performed by Matlab becomes highly inaccurate. However, samples are correctly located, as you can see from the code where an higher ...

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Yes, at least in the above case it is possible. Though it might not be computationally as cheap as other methods such as curve fitting methods. I do not think injecting NaN gonna help, instead let's look at the ignored data as dimension reduction feature of CS. Here, the measurement matrix is basically an identity matrix that some of its row (120:130) ...

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If you represent a second-order polynomial $s(x)$ with Lagrange polynomials $L_i(x)$ and interpolation points $\beta_i$, $i=0,1,2$, such that $$s(x)=s(\beta_0)L_0(x)+s(\beta_1)L_1(x)+s(\beta_2)L_2(x)\tag{1}$$ then for the equality in $(1)$ to be satisfied, the polynomials $L_i(x)$ must have zeros at $x=\beta_j$, $j\neq i$, and they must equal $1$ at $x=\... 3 How can we know the used sample frequency fs by looking at y(t)? You can't. The sampling theorem states that any sampling higher than twice the highest signal frequency allows for perfect reconstruction. In you case any sampling frequency higher than 4 would result in the same$y(t)$. 3 To my eye, the equations in the slides look like$f(x) = a_1 x^1 + a_0 x^0 = a_1 x + a_0$. So it's an exponent on the$x$, not an index. As long as you are interpolating linear, the only exponents that you encouter are 1 (linear) and 0 (constant). When you go to quadratic, you'll see$x^2$terms appearing. Now,$a_1$is the coefficient of the linear term (... 3 The sinc function actually represents an ideal (brickwall) lowpass filter that's used to complete the interpolation process after the data has been expanded (zero stuffed) properly. So let me outline the time domain approach here: Assume you have data samples$x[n]$of length$N$, and you want to upsample this by the integer factor of$L$, yielding a new ... 3 Zero-order hold will result in a piecewise-constant waveform. Linear interpolation will result in a piecewise-linear waveform. If you want a piecewise-quadratic or piecewise-cubic or higher order polynomial interpolation, it will not appear much different from the original bandlimited waveform. 2 I'm top-editing this since it answers the question directly. The sinc series is fundamentally a$C/x$, so you can extract as many absolutely convergent series out of it as you want, but what is left over is still only conditionally convergent. Also, you can rescale$x$and it is still a$C/x$series. Saying you have a summation to or from infinity is an ... 2 I try to think of it this way: In the time domain, when downsampling occurs, the signal gets compressed; while on upsampling, the signal gets stretched. Then, from the Fourier transform we know that time stretching means frequency compression, and vice-versa. It may not be a rigorous answer, but hope this helps! 2 Here again let me note that exact Nyquist frequency for a pure sine wave should be avoided. Shannon-Nyquist sampling theorem requires that there's no impulse at the Nyquist frequency as a consequence of bandlimitedness, the content at the exact Nyquist frequency is taken to be zero. Then the following code demonstrates the approximate simulation of an ideal ... 2 1-d MMSE derivative filter for uniform spectral prior With some simplifying assumptions about the signal distribution, a filter with least square frequency response error is the Bayesian minimum mean square error (MMSE) filter for a uniform prior of the frequency spectrum. For such 1-d filters, an oversampling factor$\beta \approx 2\ldots2.5$seems optimal,... 2 Sinc() interpolation looks nice on paper or in text books but in practice it's rarely a good solution. The main problem is that the sinc() impulse response is infinitely long and it's not causal. Not only does it have infinite length, but it also decays only very slowly with time, so you typically need a large number of samples to get a decent accuracy. ... 2 For simplicity I will show an approach Id' use on 1D signal (A row of real world image). You will be able to extend it and I will add few remarks on how you can even gain from having 2D data. The general idea is as sketched in Estimate the Discrete Fourier Series of a Signal with Missing Samples. The trick here is to exploit prior information. In our case ... 2 The FFT of a strictly real input results in a conjugate mirrored result. For a sinewave, both a positive and a negative frequency peak will appear. If the input is not integer periodic over the full aperture or is zero-padded, the two response peaks will be Sinc shaped, with lots of extended ripples in the frequency domain. The positive and negative ... 2 Linear and more generally polynomials are pretty common methods for interpolation or extrapolation, easy to implement, and simple in a causal setting (prediction). Parabolic interpolation (in the Fourier domain) or Savitzky-Golay filtering are practically useful examples, especially for peak-like signals. Polynomials however can be poor at modeling ... 2 For the sake of simplicity, I'll explain the 1-D case; the 2-D case is completely analogous. Let$x[n]$be a finite length sequence with$n\in[0,N-1]$. Its discrete Fourier transform (DFT) is $$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N} kn}\tag{1}$$ The sequence$x[n]$can be obtained from$X[k]$via the inverse DFT (IDFT):$$x[n]=\frac{1}{N}\sum_{k=0}^{... 2 Note: historically the Gibbs phenomenon was described over a continuous-time domain periodic square wave which was constructed from a finite number of its Fourier series coefficients, however, one can interchange the domains and observe the same behaviour on the frequency response of a finite length discrete-time filter with its impulse response samples ... 2 Sample playback The basic idea of sample playback in musical applications is to keep track of each voice's playback position, to form an output sample by reading the source sample data at the playback position, to add a possibly time-varying playback step to the playback position, and to repeat this in a program loop until we have accumulated enough output ... 2 I have some very short signals in the range of 8 to 16 samples. These represent a bandlimited signal, sampled at or slightly above the Nyquist rate. Nope. A signal can't be limited in time and in frequency at the same time. If it's very short, than chances are the bandwidth is a lot higher than you think it is and that you've already picked up some ... 1 The error source is the kernel function, which should be a multiplication of two 1D sincs (without rotational symmetry) instead of the Sombrero function, which is a sinc function with rotational symmetry. 1 Yes inserting zeros does insert new frequencies in the unique digital spectrum that extends from$0$to$2\pi$radians/sample or equivalently$\pm \pi$radians/sample corresponding to$\pm F_s/2$where$F_s$is the sampling rate. The easiest way to see this intuitively is to consider a DC signal represented by a stream of constants, such as:$x_1 = \begin{...

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If you are upsampling a class of bandlimited signals, such as specch, using polynomial interpolators, then you are introducing erros into your results. This is because polynomial interpolators will not have the necessary conditions for being an image free upsampler. The remaining spectral images are observed as errors in the interpoated signal. However, if ...

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These are the formulas you want. There are different formulas whether you have an even or odd number of samples in your source wave definition. $x[n]$ is your source and $y_m$ is your output. Your $N$ source samples are indexed by $n$ going from 0 to $N-1$. Your $M$ output samples are indexed by $m$ going from 0 to $M-1$. These formulas calculate ...

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I have since found this paper, which provides a nice framework for how to think about this problem.

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Butterworth low-pass filters are not going to work for this, as demonstrated in the following. You'd need to use other types of filters, for example linear-phase finite impulse response (FIR). The magnitude frequency response of an order $N$ discrete-time Butterworth low-pass filter can be approximated by that of an analog prototype (from Wikipedia, with a ...

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The "discrete sinc kernel" is also known as a Dirichlet Kernel. It looks like a you got a 2D (u and v) going there. I'm not that familiar with that, but we just did a pretty good number on the 1D case. This post explains the inverse DFT can be interpreted as "epicycle drawing": How to get Fourier coefficients to draw any shape using DFT? This post then ...

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A pure sine wave sampled at exactly the Nyquist frequency is simply an alternating sequence of $C(-1)^n$, where C can be anything from zero to the amplitude of the sine wave depending on where in the cycle the points are sampled. Thus the only plausible reconstruction (which can be done as an infinite series of sincs, see Convergence of periodic sinc ...

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I am using this link for getting the reference by Richard Balmer : There are no references 30 and 32

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If you need to discriminate between 60 and 65 Hz signals, the fact that you're sampling at 25 kHz means that you're basically doing real time signal processing: the Nyquist Frequency is far above the frequencies of interest. Thus, from classical Fourier theory, you're going to need on the order of 1.0/5 Hz = 0.2 seconds of data, no matter how you process it....

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