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When you quantize a signal, you introduce and error which can be defined as $$q[n] = x_q[n]-x[n]$$ where $q[n]$ is the quantization error, $x[n]$ the original signal, and $x_q[n]$ of the quantized signal. The maximum quantization error is simply $max(\left | q \right |)$, the absolute maximum of this error function. Dx in this definition seems to be the ...

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The sampling theorem states that $f_\mathrm{S} \geq 2f_\mathrm{max}$, where $f_\mathrm{S}$ and $f_\mathrm{max}$ are the sampling and maximum signal freuqency, respectively. But there's an additional condition: The equal sign only holds if the signal spectrum does not contain a dirac impulse at $f_\mathrm{S}/2$ which is clearly the case in your example. ...

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The sample rate needs to be GREATER than (NOT just equal to) twice the highest non-zero frequency content of the signal being sampled. Just a little bit greater might work, but the closer the sample rate is to twice the signal frequency, the longer in time you may need to sample to raise the signal above the noise and complex conjugate image in a DFT/FFT ...

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I have a doubt about (Edit: this was later removed from the question): The distribution of these AM and PM noise components can be reasonably assumed to be uniform as long as the input signal is uncorrelated to the sampling clock Consider the signal: $$\operatorname{signal}(t) = \cos(t) + j\sin(t)$$ and its quantization: $$\operatorname{quantized\_signal}... 4 Depends on assumptions you are willing to make and what type of signals are you trying to sample, but in theory I think that sampling rate equal to the Planck time would be a gold standard for anything... This translates to sampling frequency of 1.855 \times 10 ^ {43} \mathtt{Hz} (18.55 tredecillion hertz). Personally I believe that machines will never ... 4 First, use a timer and an ISR to get accurate timing (don't forget to configure the NVIC so that this timer interrupt takes over any other ISR that would be running). Only this will ensure a consistant sample rate. Little variations in timing would create noticeable degradations of audio quality. In particular, in your current example, unless the "filters" ... 4 There are two aspects to how this works. First, since the signal is oversampled there is a great deal of correlation between samples that we can take advantage of via the low-pass filter. The noise, on the other hand, has no correlation (assuming it is white noise), and thus will often destructively interfere with itself. Your question seems to be more ... 3 Complex sampling does not "break" Nyquist. IQ quadrature sampling produces twice as many bits per second of information (at the same sample rate for real or complex samples), and the 90 degree phase offset between the I and Q channel in those bits provides extra information about the spectrum. One typical example to demonstrate aliasing is that the samples ... 3 No, and the reason is not so much a question of how fast one can sample a continuous-time signal (as the accepted answer and another one says) but rather the impossibility of representing a real number with perfect accuracy via a quantized representation of the real number (as noted in the answer by Marcus Muller). At best, even if we assume an infinite ... 3 Is it theoretically possible to perfectly quantize a continuous signal? No. A quantization has an information content obviously countable as bits. Now, if you have a continuously distributed 1D random variable X, then the event that any of these real numbers x occurred is unbounded ("infinite"):$$I(x) = -\log_2\left(P(X=x)\right)$$So, for any (non-... 3 I'd like to point out Heisenberg Uncertainty principle, based on which theoretical achievable precision is limited. It states that one can not measure two complementary qualities (e.t. here time and voltage) concurrently and there is a trade off between amount of precision you can get from one or another. In ADCs, for example theoretical limit for ... 3 Cross correlation should work. I think the problem is the waveform that you are using. A square wave has bad auto-correlation properties. If it is a periodic square wave it will have multiple peaks. It sounds like you are just using a single pulse which is better, but it will still have a gradual roll-off which is a problem. Instead, use a Barker code, ... 3 By definition, band-limited signals in the sense of the sampling theorem have finite energy. Sine waves are periodic and thus have infinite energy. So any dirac pulse in the Fourier transform is not permissible. To be more precise, the sampling theorems only applies to signals that can be represented as$$x(t)=\int_{-f_s/2}^{f_s/2} X(f)\,e^{2\pi i\,ft}\,df$... 3 again, even with the Addendum, i think Lutz's answer misses the point. the point is (quoting Wikipedia): To illustrate the necessity of$f_s \ > \ 2B$, consider the family of sinusoids (depicted in Fig. 8) ) generated by different values of$\theta\$ in this formula: $$x(t) = \frac{\cos(2 \pi B t + \theta )}{\cos(\theta )}\ = \ \cos(2 \pi B t) - \sin(2 ... 3 Signals can be classified in many ways, and one of them is according to their nature such as being deterministic or random (stochastic). When a signal is deterministic, its spectral content is given by the Fourier transform provided that the signal is absolutely (or square o.w.) integrable; i.e., its Fourier transform exists. On the other hand when an ... 3 First of all, because it's easy to build a 1-bit ADC. It's a comparator. It's literally the easiest ADC you can build. The \Delta\Sigma ADC was invented (or, rather, published) in 1962¹ ! The 2-bit ADC is more than twice as complex as that, you need some window decision: so if you have the choice of making your 1-bit ADC run faster or building a ... 3 I don't quite understand why you feel "harmonics" are relevant to this discussion. A 5 Mhz has sine wave has no harmonics. If the signal has harmonics, it is not a sine wave any more but a different signal (rectangular, triangle, etc.) For ANY signal: you need to determine the highest frequency that's in the signal and then chose the Nyquist frequency to be ... 2 Your observation is correct, and it has been noted before. One example, for instance, is mentioned on pages 160-161 of "Principles of Communication Systems" by H. Taub and D. Schilling, McGraw-Hill Book Company, 1971. After introducing the sampling theorem, the authors state: "An interesting special case is the sampling of a sinusoidal signal having the ... 2 Here is a very simple example demonstrating how we can get increased precision by taking more samples and then filtering (averaging): Consider "truth" to be a constant 8.2, we quantize by rounding to the closest integer, and we oversample and average (filter) in an attempt to improve the precision of the result. If there was no noise, the result of the ... 2 You are right, DC means zero frequency. The term probably originates from electrical engineers' use of the terms DC and AC. 2 The problem with quantizing Gaussian distributed signals (like the real/imaginary part of an OFDM signal) is that they can take any value in theory. It is thus necessary to clip such signals at threshold C prior to quantization. Low C increases the distortion noise in this process, while large C will lead to strong quantization noise. It can be shown, ... 2 A sine wave has infinitesimally little bandwidth. By rotating, filtering appropriately and decimating, you can reduce the sample rate very much. Each of these filtering operations is typically a summing operation, in which you "average" out noise (which isn't your main concern), but also get a more precise estimate for the amplitude. Decimation in DSP is ... 2 To add to Marcus' good answer, the decimation Marcus mentions WILL increase the number of bits, down to the Spurious-Free Dynamic Range (SFDR) of your ADC. Also if you can modify your sampling rate so that it is incommensurate with your signal of interest: 320 MHz is an integer multiple of 1 MHz, so you will ...to the extent your input signal is coherent ... 2 I think that what you are hinting at is the concept of resolution of the Discrete Fourier Transform. This will determine how much SNR will drop in-band. The standard DFT will break some x[n] down to a complex sum of sinusoids. But if you notice, the variable changes from n to k, with k being the index to X. As an index, it cannot be a Real number. ... 2 How do you transform overflowed signal back to normal? Just by adding or subtracting multiplies of 256. So with every new sample you just determine whether overflow occured - you need to define some threshold and compare it with first derivative of your signal, e.g. x[i+1]-x[i] > 200. If this happened, you know that you need to subtract 256 from x[i+1] to ... 2 Stanley is right and what you need is a data acquisition board. Depending on your application concern, you can either buy it from an online store or build your own if you can do so. If you would like to make it on your own, then as MBaz suggested you can select from mini pc boards such as a sufficient Raspperry pi that can either transmit your grasped data ... 2 I would tend to believe that there is an indeed a flaw in the approach as the datasheet SFDR is achievable given the test conditions which are usually specified in the datasheet as well. 23 dB is SIGNIFICANT and I suspect your sine wave is not really a sine wave and you are seeing a harmonic (or an aliased harmonic). Here are some possible culprits, some ... 2 The sampling is indeed analogous to mixing as to my understanding. In the sampling process, we multiply the time domain signal with an impulse train - the impulses in time are represented as impulses in frequency at integer multiples of the sampling rate. So instead of one or two (for a real sine wave) impulses in frequency, we have an infinite number but ... 2 Most audio interface devices are built for the commercial electronics industry which produces devices to be used by people to listen to music, radio, TV etc; for multimedia reproduction purposes. Therefore the commercial electronics standardisation organisations, suggest or enforce the use of a number of frequency weighting filters (A,B,C etc.) for getting ... 1 What is shown in the plot is the PSD (Power Spectral density), which is measured in W/Hz (or dBc). So, in order to get the actual power in a given frequency range, you'd need to multiply with the bandwidth of each bin (which is determined by the length of the DFT window). Let the bandwidth of each bin be B. Then, your formula becomes$$ THD = 10\log\left(...

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