I was watching a video on quantization noise and the instructor kept referring to the "no overload" region of a quantizer? Could anyone please explain what that term means?
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1never heard that term before, but I guess it's the area of the quantizer between its minimum and maximum input value boundary? – Marcus Müller Jul 21 '17 at 16:02
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Was it a sigma-delta modulator? In these one-bit quantizers, if the signal slope is too large, the converter can't keep up (it's overloaded). – MBaz Jul 21 '17 at 16:10
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My guess is that the overload region, is when the input amplitude is larger than the dynamic range of the converter. When this happens, some converters have a saturation recovery time associated with the overload, so there may be a period of time when the signal drops back into the dynamic range but the converter remains overloaded for some time. Since you saw the video, perhaps you could respond if this makes sense. – Jul 21 '17 at 16:36
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@MBaz, delta modulation is not the same thing as sigma-delta. the latter doesn't have a problem with slope. – robert bristow-johnson Jul 21 '17 at 18:34
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@robertbristow-johnson Indeed I got them mixed up -- thanks for the correction. – MBaz Jul 22 '17 at 18:55
1 Answers
It is stated in chapter 5 (Scalar Quantization I: Structure and Performance) of the book "Vector Quantization and Signal Compression" by Gersho-Gray that
Associated with every $N$ point quantizer is a partition of the real line $\mathcal R$ into $N$ cells or atoms $R_i$, for $i = 1,2,\cdots, N$. The $i$th cell is given by $\mathcal R_i=\{x \in \mathcal R: Q(x) = y_i\} \equiv Q^{-1}(y_i)$, the inverse image of $y_i$ under $Q$. It follows from this definition that $\bigcup_i \mathcal R_i = \mathcal R$ and $R_i\bigcap R_j = \emptyset$ for $i \neq j$. A cell that is unbounded is called an overload cell. Each bounded cell is called a granular cell. Together all of the overload (granular) cells are called the overload region (granular region).
So in other words, the union of bounded regions (i.e. granular region) where saturation does not occur is referred to as the "no overload" region and is identified in the following picture
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the width of the "No overload (no saturation) region" is 7 units. i would widen it by $\frac12$ unit on both sides so that there are 8 valid levels. – robert bristow-johnson Jul 22 '17 at 22:16
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actually, for a three-bit word, there should be 4 negative levels and 4 non-negative levels (of which the zero level is one of them). so the nine-levels displayed is one too many. – robert bristow-johnson Jul 22 '17 at 22:17
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The picture shows a mid-tread quantizer with 9 levels (so it is not a 3-bit quantizer). Still, all 9 levels are valid (clipping occurs at the two extreme cells). The 3-bit quantizer you explained is a mid-riser which has four negative and four positive cells. But in that case there would be three positive and three negative "no overload" cells (6 in total). – msm Jul 23 '17 at 03:57
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i don't get "cells". for some value of input, the extreme positive and extreme negative outputs are still value. the overload occurs beyond those limits. – robert bristow-johnson Jul 23 '17 at 05:19
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i know about "mid-tread" and "mid-riser". but the issue is more general than that. for an 8-bit signed quantizer, there are legit inputs that will map to the codes for -128 and +127. but for input beyond those coded values, then the quantizer is overloaded or saturated. – robert bristow-johnson Jul 23 '17 at 05:21
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Yes, the extreme points are still value (cell/atom/quanta/etc.). The point is, given a continuous pdf all values are mapped to a code (this normally includes the overloaded levels). So for example in the 8-bit quantizer example, the two extreme codewords should represent overloaded levels. The codes are just indices that are assigned to the levels. Here for instance, we should have $2^8$ levels overall. – msm Jul 23 '17 at 06:30
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so you're saying that if the code is either -128 or +127, we should infer from that code, that the signal exceeded the input range? – robert bristow-johnson Jul 23 '17 at 06:42
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There is usually no input "range". We normally have an analog source which is modeled by a continuous pdf (such as Gaussian). Although a Gaussian pdf (where the support is infinity) does not make much sense in reality, the analysis is based on such abstract models. – msm Jul 23 '17 at 06:57
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We may call those as tail levels, for instance, (or something like that) rather than "out of range". – msm Jul 23 '17 at 07:03
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in audio there certainly is an input range. and we call that "full-scale". it's about a volt peak-to-peak for a magnetic guitar pickup and a guitar amp. – robert bristow-johnson Jul 23 '17 at 07:31
