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In addition to @hotpaw2 explanation, a graphic. There are two analog square waves (red and green), with different lengths. They are depicted with a fine sampling, denoted by crosses. Their actual sampling is denoted by circles. The red one is shorter than the green one, as can be seen in the interval $]0.7\;0.8[$. Yet, the sample points are the same. Thus, ...

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Math and Physics are not the same thing. They can be very close but infinity is hard to realize. Going from analog to digital, a more accurate description is an average of the waveform over a small time window and the average is in turn quantized. Filters don’t have infinite stop bands but 80 dB is usually more than good enough. There is always a small ...

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Possibly intuitive answer, skipping the math: There is at least one problem with sampling a non-antialiased square wave at a sampling frequency not a strict multiple of the square wave's frequency. Because a fixed number of samples don't fit into 1 period of the square wave at these ratios, and a non anti-aliased square wave quantizes to just 2 levels, ...

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I'm still not quite sure I understand what you're trying to achieve, but here's an example I cooked up in scilab. The input is 33 points of a sine wave: fs = 100; // 10kHz sampling rate Tin = 1/fs; // Input sampling period N = 16; tin = [-N:N]; // Times of samples f0 = 5; x = sin(2*%pi*f0/fs*tin + 1.123791823) clf subplot(211) plot(tin,x) The sampling ...

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Question 1: The anti-aliasing filter before the ADC is exactly for the purpose of rejecting high frequencies, that will become lower frequencies (i.e. aliasing) after the ADC. The digital lowpass after the ADC cannot help here, as the aliasing has already happened. Consider this example: Your ADC has a sampling frequency of Fs=100kHz. Your input signal is ...

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So the point is, oversampling with a factor of $N$ requires your signal to be confined to at most $\frac1N$ of your Nyquist bandwidth. Otherwise, you're not oversampling. If the signal is six time oversampled, then there's no loss of information/aliasing when reducing the sample rate by six. If I would oversample by 2 and then downsample it by 2 without ...

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Technically, the most reactive filter is the all-pass filter with a gain of 1, this filter has no phase shift at all. But it is not a really useful filter. Here's what you need to take in account : 1 - IIR versur FIR An IIR filter will have less phase shift than a FIR filter for the same cut-off frequency 2 - Cut-off frequency For an IIR low-pass ...

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I think you make this needlessly complicated. Let's do a specific example: let's say we want to go from 48 kHz up by 4/3 to 64 kHz and your signal bandwidth is 20 kHz. In the first step, you would up-sample by a factor of 4 by inserting 3 zeros between each sample. Your new sample rate is 192 kHz. Your original spectrum is preserved and you get mirror ...

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First of all, if your phasor oscillator has a step discontinuity (as opposed to a ramp one) at $\phi\bmod 1 = 0$, then you should use the four-point, fourth order polyBLEP residual, not the fifth-order polyBLAMP one. Second, the easiest way to simplify the piecewise polynomials is to see that the residuals are the successive integrals of the piecewise ...

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Aliasing does not happen in the analog world. It only happens when you sample a signal. If you have an ADC with sampling frequency $f_s$, then you'll have aliasing if the input signal has frequency components larger than $f_N=f_s/2$. To avoid aliasing, it's common to low-pass filter the signal before it is input into the ADC. That way, you make sure that ...

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How to calculate the variance of the noise samples $n[j]$ in terms of $N_0$ and $B$, where $n[j]$=$n_f(jT_s)$ and $T_s$ is the sampling period? Do you know how to calculate the variance of the process $\{n_f(t) \colon -\infty < t < \infty\}$? No? Hint: it is the area under the power spectral density curve of $\{n_f(t)\colon -\infty < t < \... 1 Suppose if your sampling rate is fs, then the corner frequency of the Anti Aliasing filter would be fs/2 1 From your statement, i understand a discrete-time downsampling of a sequence$x[n]$already converted to digital, without aliasing, previously. So to prevent any aliasing on the downsampled sequence$y[n]$, you should first apply an anti-aliasing lowpass filter to the input sequence$x[n]$. The cutoff frequency of the lowpass filter is given by the ... 1 This MATLAB/OCTAVE code makes a causal implemtation the system$G(z)$you mention. Note that$G(z)\$ is non-causal, so you have to assume delays to make it causal. You can convert it to non-causal if you have the full data available before processing. k = 5; N = 2*k+1; K = (k+1)^(-2); b = K*[1, zeros(1,k) , -2, zeros(1,k), 1]; a = [1,-2,1]; [Hk,w]=freqz(b,...

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If you do offline processing then consider using sgolay filter in matlab or the filtfilt function which is supposed to give zero phase shift. I particularly like sgolayfilt with coefficient 3 and window size 11. If you want zero phase filtering in real time then predictive filter such as the kalman filter is a good solution. Use a simple velocity filter ...

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Even though @Ben has already given a wonderful answer I would like to add that there's a technique called zero-phase filtering which allows you to have no phase delay at all. However, as I don't really know what you mean by reactive filter (fast?), I must say that in order to implement this in a causal way you will have to work with a buffer of stored ...

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By allowing a little error, you can get away with a limited length filter, and it will introduce less error in the frequency domain than sampling the signal directly without filtering first. Here is an illustrative only example using finite impulse response (FIR) windowed sinc filtering, which is compared to sinc filtering and direct sampling of a ...

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If the signal breaks the Nyquist criterion we'll only get a part of the signal. This needs to be clarified: the frequencies above Nyquist will fold back into the Nyquist band, so it's not just that you lose part of the signal; it's that the signal is corrupted (usually) beyond hope of reconstruction. As an example, consider a 4 kHz real signal sampled at 4 ...

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The higher the sample rate, to easier (and thus cheaper) it is to make the analog low-pass filter required to prevent aliasing (to limit aliasing to below your required noise floor). Making a filter with a sharp transition band is more costly. Often a much higher sample rate is cheaper to implement than a slightly sharper low-pass filter.

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In Karel Fliegel, Modeling and Measurement of Image Sensor Characteristics, Radioengineering, vol. 13, no. 4, December 2004, he gives the optical transfer functions (OTFs) for different detector photosensitive area shapes. OTF is the Fourier transform of the spatial domain impulse response. The spatial domain impulse response, also called the point spread ...

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When you use interpolation to decrease the sample rate, you need to first low pass filter the signal to a bandwidth below the new FsNew/2. You can do this by using a wider Sinc kernel related to the new lower sample rate. e.g. intead of T in the denominator on the Sinc parameter in your interpolation kernel, use 1/FsNew.

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Oversampling of audio can be used to reduce aliasing artifacts that result from non-linear processing. Harmonics generated by the nonlinear process can be filtered from the oversampled signal by a lowpass filter. A graphics analog of this is supersampling. If Wikipedia is to trust, multi-sample antialiasing is a further optimization in which only the mixing ...

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No. The reason is that MSAA works in the visual domain. Optically, we have hardwired edge detectors. MSAA eliminates unintended edges in generated images. Acoustically, we do not have such edge detectors, and therefore there's no reason to remove them. Also, loudspeakers already filter out such edges mechanically.

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Another way to think about it is that aliasing always occurs during sampling. The entire spectrum always gets folded down and aliased with the signal of interest. Even if that spectrum content is zero. The closer you can get the unwanted spectrum to zero, the less aliasing noise you will have contaminating your signal band of interest. The best case ...

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