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The signal voltage is a sinusoid signal variant from $-V$ to $+V$, thus the RMS value $$V_{rms} = \frac{V}{\sqrt{2}} = \frac{2^N \Delta V / 2}{\sqrt{2}}$$ because $2V = 2^N\Delta V$ where $\Delta V$ is quantization step. Quantization noise is modeled as uniform random variable in $[-\Delta V /2, +\Delta V /2]$ thus its standard deviation is $V_n = \Delta ... 9 For the last week or so I have been trying to understand how quantization error results in the noise floor outside of a mathematical perspective and I haven't really had any luck finding a source that discussed quantization noise without using equations to show where quantization error comes from. That's because it's a purely mathematical effect, there's no ... 7 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 ... 7 This answer discusses the harmonic spectra of the quantized sequence in five cases: limit$f/f_s \to 0$, synchronous sampling of a cosine with rational$f/f_s$, synchronous sampling of a sinusoid of arbitrary phase with rational$f/f_s$, asynchronous sampling of a sinusoid with rational$f/f_s$, and asynchronous sampling of a sinusoid with rational$f/f_s$... 5 any discrete random variable has a p.d.f. that is a summation of dirac delta functions. $$p_\mathrm{y}(\alpha) = \sum\limits_i P_i \ \delta(\alpha - y_i)$$ where$\sum\limits_i P_i = 1 $. if$y[n]$is the quantization of$x[n]$: $$y[n] = \Delta \bigg\lfloor \frac{x[n]}{\Delta} \bigg\rfloor$$ where$ \lfloor \cdot \rfloor$is the floor() function and ... 4 Suppose you are given a data point$D$and want to classify it into one of$N$possible classes,$C_i$for$i=1,2,\ldots, N$. Hard classification chooses one of the$C_i$to represent the classification of$D$. Soft classification can be done in a number of ways, but the usual (for me) is to generate a weight$w_i$for each of the$C_i$which indicates the ... 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 The phase of the sinusoid does not matter: A phase shift of a sinusoid is equivalent to shifting it in time, which results in a time shift of both the quantized sinusoid and the quantization error. The power spectrum is invariant to time shifts. We choose to work with sinusoid$A\cos(x)$. Optimality by def. 1 Equivalently to maximizing signal-to-noise ... 3 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...... 3 Quantization error is usually modeled as additive white noise, uniformly distributed over the interval$[-\delta/2, \delta/2)$, where$\delta$is the step size that the signal is quantized to. As you described, the quantized signal model is described by: $$s_Q[n] = s[n] + w[n]$$ where$s[n]$is the original signal and$w[n]$is a white noise process that ... 3 Your low-pass filter's pass band is smaller than your signal's bandwidth, so it is destroying a significant portion of your signal. Given that, you cannot reliably reproduce the original signal without using sophisticated techniques like error correction codes. I would try increasing$\alpha$to 2400 Hz. Then the rolloff point will be 2400 * .5 = 1200. ... 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 Is this a well-known phenomenon? Yes, of course. You will see harmonics as soon as your clip point is lower than the maximum amplitude in the time domain. The latter is a function of the relative phases between the harmonic components. In your case the max amplitude is indeed 2.5 (plus whatever the noise adds). If you change the phases you will get a ... 3 Say you quantize a signal with$L=2^n$levels. Then, every quantized sample is represented by$n$bits. So, as you increase$L$, you need more and more bits to represent the samples. This means that transmitting the signal will require transmitting more bits. Also, keep in mind that, in order to convey bits at a rate of$R_b$bits per second, you need a ... 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 The harmonics will be in the range of -150dB to -170dB. The exact value will depend on exact frequency, phase relationship to sampling frequency, phase locked or unlocked, rate of phase drift, integration interval, index of harmonics, etc. You need to specify this all in excruciating detail to get anywhere near the resolution that you are asking for. I ... 2 Oftentimes some rounding occurs in storing the coefficients. This is why many image compression algorithms are lossy, i.e. they lose information when converting the floating point coefficients to integer format. The process of rounding is called quantization. See this wikipedia article for an example. http://en.wikipedia.org/wiki/JPEG#Quantization 2 i dunno what a "Lipshitz filter" is but i do know who Stanley Lipshitz is. now, i don't know for sure what Stanley would say to you, but the best advice i got from him is that it does little good to color the additive dither if you are also using error feedback. if there is no noise shaping feedback, probably the simplest way to get triangular PDF dither ... 2 I point you to this https://stackoverflow.com/questions/12723699/changing-image-bit-depth-using-matlab it can be done. This particular post is only for png, but if I'm not mistaken all image types have the depth parameter so they can likely all be changed in a similar manner EDIT For your particular problem, since your images are already grayscale you ... 2 You are correct. When using uniform sampling, the sampling rate is unrelated to the quantization noise. Only the quantization step and the signal's dynamic range are relevant for calculating the quantization SNR. 2 Theoretically, VQ is always better than scalar quantization (SQ). However, as you already mentioned, the complexity of VQ is higher than the complexity of SQ. It increases exponentially together with the dimensionality of vectors. Therefore, the question is how to design a practical coding system which can provide a best rate-distortion performance under a ... 2 From what I read there are certain quantisation matrices for different applications, Adobe Photoshop has like 15 or something. The idea is that although there are error based mathematical calculations for these matrices the best way to deduce a good quantisation matrix is by simply using a human eye (in the case of JPEGs)I haven't seen any quantisation ... 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 Different. Lloyd-Max is a special type of scalar quantizer design which is optimized (in terms of MSE) to source pdf. Hence the quantizer is generally non-uniform. Lloyd's algorithm (and the more generalized LBG algorithm) is a scheme to design vector quantization. More information here and here. That said, one should note that Lloyd's algorithm for VQ is ... 2 To make it more clear, I suppose your question is Why it is said that the compressor gain at low input amplitudes is higher, while the step size of a nonuniform quantizer is small in that region. Similarly, Why it is said that the gain of the compressor is higher for high input amplitudes, while the step size is larger for those inputs. First, notice ... 2 It often doesn't sound like white noise. The quantization noise of a sine wave is fairly harmonic with a clearly detectable pitch. The quantization of a pulse train sounds like a pulse train. If the quantization has enough bits, and if the signal is reasonably stationary in time and reasonably dense in the frequency domain, the quantization noise will be ... 2 No need to do the convolution to find the output. Just use the fact that$z^n$is an eigenfunction of the system with$z=e^{j\omega_0}$. This will lead to the following rule $$x[n]=A\sin(\omega_0 n+\phi) \Rightarrow y[n]=|H(e^{j\omega_0})|A \sin(\omega_0 n+\phi+\angle H(e^{j\omega_0}))$$ That is, the output to a sinusoidal signal with angular frequency$\...