# Tag Info

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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 ...

6

To add here are two diagrams showing common transceiver architectures: (1) a super-heterodyne where the down-conversion is done first to an IF frequency and then to baseband and (2) a zero-IF receiver where RF is translated directly to baseband. Note in both architectures it is arbitrary (technical / technology choice) where the ADC/DAC boundary is as either ...

5

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 ...

5

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}... 5 can we have re-configurable analog filters? Yes. The knob you turn on your grandma's kitchen or living room radio changes the tuning of an oscillator by changing the capacitance of a component. Any stable oscillator is essentially a filter for its resonant frequency. Also, plenty of other examples: Tunable RC filters with adjustable R; mechanically tunable ... 4 No! The ADC (delta sigma or not) can not reduce the uncertainty in the input. It sounds to me that your friend has not made up a real signal flow diagram and then formed equations. The answer to your second question is affirmative. In fact it's generically true that putting as much gain, without introducing extra, in the front end is a good idea. But...... 4 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 ... 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 ... 4 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 charge) 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 resolution ... 4 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 somewhat ... 4 Sampling is the process of making the x-axis (time) discrete and quantization is the process of making the y-axis (magnitude) discrete. You can sample without quantization (such as done with an analog sample and hold circuit). Quantization is introduced through rounding or truncation when the sampled analog signal is mapped to a digital representation. ... 4 First you will need to determine the number of quantization levels. I am going to assume a power of two for digital convenience's sake. nbits = 8 % 256 qantization levels qLevels = 2^nbits The next step will be to scale your signal to have the same magnitude as your number of bits. signalMin = -1 signalMax = 1 scalingFactor = (signalMax-... 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 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)$$So, for ... 3 By definition, band-limited signals in the sense of the sampling theorem have finite energy. Sine waves are periodic and thus have infinite energy. So any dirac pulse in the Fourier transform is not permissible. To be more precise, the sampling theorems only applies to signals that can be represented as$$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 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 ... 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 ... 3 I think the jitter to SNR formula is based on a gaussian jitter. You use "rand" which yields uniformly distributed numbers. You should use randn() instead which yields numbers distributed according to a gaussian distribution. 3 From the paper you link to: Modulators with a higher order loop filter also exhibit a dead-zone which can be calculated in a similar manner. Because of the higher gain in the loop filter, the dead-zone of these modulators will be considerably smaller than the dead-zone of the first order SDM*. Very small signals (including system noise) will disturb ... 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 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, ...

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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 ...

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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 ...

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