# Tag Info

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Continuing from my comments. The units equation in the comments usually has these variables: $$f_{Hz} = k \cdot \frac{F_s}{N}$$ Where $k$ is the bin index. This can be rearranged a little bit to: $$f_{Hz} = \frac{k}{N} \cdot F_s$$ This equation shows that the frequencies associated with the bin indexes increases linearly up the bin scale until it ...

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If you don’t know the sample spacing, you don’t know the sample rate. If you know that samples are uniformly spaced, you can do a DFT and get relative frequencies. You won’t be able to map one dft bin to a specific frequency (in Hertz) unless there is some additional side info (perhaps a known reference hidden within the data).

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There is nothing unordinary about it. Signals can have a period that is not an integer multiple of samples, and that is normal. For example, if you record a 440 Hz sine wave with a sampling rate of 48 kHz, it is not an integer multiple. It takes 11 full sine wave periods to hit an integer amount of 1200 samples.

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After upsampling with zero-insert, the higher frequency images need to be filtered out, which will serve the purpose of growing the zero insert values to the correct interpolated value in between samples. Once this is done, subsequent decimation operations can be properly done since every sample will then properly represent the signal. For a signal that ...

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Note that the cosine has a positive and a negative frequency component. Consequently, in the range $[0,f_s/2]$ you get a component at $-50\textrm{ Hz}+75\textrm{ Hz}=25\textrm{ Hz}$.

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Your x-axis is apparently varying between plots. And Matlab plots joining lines between discrete points are misleading. So, here is an other version, with more sampling times: The Nyquist-etc sampling theorem tells you (somehow) that if you don't have at least two samples per period, you are likely to loose most of the signal's information. Until $0.095$ ...

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You are sampling a sine function that should return zero, but because of finite precision argument into that sine function, the error tends to increase towards the right?

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You sample at times $nT_s$, so the values of the sampled signal are $$x_d[n]=x(nT_s)=\sin(10\pi nT_s)\big |_{T_s=0.1}=\sin(n\pi)=?$$ Can you take it from here?

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Yes; what you describe doesn't mathematically look any different than a single sampler at twice the rate. But they sensors need to sense the same, band-limited thing! Actually, that's how some very high-speed ADCs work. If you're not actually doing GS/s, getting a faster ADC would probably be easier than staggering multiple ADCs.

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How are you changing the number of samples ? If you are just truncating the 2048 samples $X$ to 1024 samples $X^t$, then the samples won't be complex conjugate anymore. First sample of the 2048 length sequence will be considered as DC and then the remaining should be conjugate symmetric around $N/2$. That means, $X[k] = X^*[2048-k], \forall k \in \{ 1, ... 1 Hint:$f(t)$is the expected output after passing a sampled function through the first order hold function with impulse response$h_{FOH}(t)$. The sampled function is$f(kh)$, with$k$being the sample index and$h$being the sample interval, with zeros elsewhere as a continuous time process. The output is the continuous-time convolution of$f(kh)$with$...

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...clock would vary for each device. Unless you can afford high end studio devices that can synchronise to a common clock, this is taken for granted. However, saying that, you also need to balance your requirements. If you are using a relatively good "audio interface" (i.e. not a random off the shelf sound card) then you can expect a pretty accurate clock. ...

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I'd say that this is not only "similar to a cross-domain equivalent to Nyquist's Sampling Theorem", but it simply is the sampling theorem. The sampling theorem does not specify the domains of the signals involved; it is rather a mathematical condition that a function of a continuous variable needs to satisfy such that it is perfectly represented by ...

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If you have a closed-form expression for the weighting function, you can use it directly. Here, the weight is linear, so you can replace coef_vec by xand be ok. fun = @(x)x.*exp(-x.^2).*log(x).^2; q = integral(fun,1,10); If not, an option is to define coef_vec on a larger interval, interpolate it with a suitable closed-form interpolating function, and use ...

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Because sampled signals Spectrum will have copies of original spectrum at multiples of $f=F_s$. Even if aliasing happened due to choice of $F_s$, the spectrum of sampled signal $x(n.T_s)$ will be periodic in frequency always with period $F_s$. And you can take any one such $k^{th}$ copy or period and integrate from $\frac{k-1}{2}F_s$ to $\frac{k+1}{2}F_s$. ...

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If I understood that right, you are trying to minimize the infinity norm over a discrete set of points. If that is the case you set over which you are trying to minimize is not convex and this is very much a NP hard problem, unless the set is very small, so that we can rigorously loop over it. Those are my thoughts on this.

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This elaborates on my comment and my linked FTIR example. Once the FTIR spectrum is obtained, the phase spectrum can be obtained via the Hilbert Transform of the spectrum and that can be performed as a convolution. The Hilbert Transform can be performed by various software packages. In Igor Pro (v. 6.3), the Help text on Hilbert Transforms (page 723 of the ...

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