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I have an formula for this "Maximum Quantization Error" but i dont know what it is based in. Its just thrown in my study material without further explanation.

It is defined as:

$$Q = \dfrac {\Delta x}{2^{N+1}}$$

where $N$ is the number of bits used for quantization in a analog to digital conversion, and $\Delta x$ is, in portuguese "Faixa de Excursão do Sinal", I don't know what would be the correct translation, but I bet on something like "Signal Excursion Band". I know, its a strange name.

Can someone help me with this? What is this $\Delta x$? Sorry for my bad english, it isnt my native language.

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1 Answer 1

up vote 4 down vote accepted

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 range of the input signal so we could rewrite this as $$Q = \frac{max(x)-min(x)}{2^{N+1}}$$

Let's look at a quick example. Let's assume you have a signal that's uniformly distributed between -1 and +1 and you want to quantize this with 3 bits. You have a total 8 of quantizaton steps which would map to [-1 -.75 -.5 -25 0 .25 .5 .75]. The difference between steps is 0.25. If you round during quantization the maximum error will be half of that (i.e. 0.125).

Now let's try the formula: $$Q = \frac{max(x)-min(x)}{2^{N+1}} = \frac{1-(-1)}{2^{3+1}}= \frac{2}{16} = 0.125$$

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i found the answer before and forget to come check here. Thanks anyway! –  Diedre Jun 8 at 18:43
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