Quantization of Continuous-Amplitude Signals
This point is not clear to me:
"The smallest quantization levels ($\pm\Delta$) correspond to the least significant bit of the binary code word."
Could someone explain it please?
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1$\begingroup$ i retitled the graphic because the previous title was completely wrong. $\endgroup$– robert bristow-johnsonCommented Dec 21, 2020 at 19:08
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2 Answers
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Quantization is a process of representing real values of a signal using integer numbers. That process results in loss of accuracy, which is determined by the selection of the smallest value of quantization level.
If the smallest quantization level is $\Delta$, then every signal level will be rounded to the multiple of that $\Delta$. For example, if a signal level is $x=3,3 V$ and $\Delta = 1V$, then the corresponding quantized representation of the signal level will be $Q(x)=3\Delta$ and the resulting $\hat x = 3_{10} = 11_2$. If $\Delta = 0.5V$, then the signal level will be $Q(x)=7\Delta$ and the resulting $\hat x = 7_{10} = 111_2$.
So, to rephrase the original point: the weight of the least significant bit of the binary code word equals $\Delta$.
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$\begingroup$ This is not correct for negative signal values and moreover the binary number format is two's complement which is different from base 2. $\endgroup$ Commented Dec 21, 2020 at 20:40
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1$\begingroup$ This answer is still correct for negative values/2s-compliment. All they are saying is ‘this is how we encode the output of the quantizer in binary’. $\endgroup$ Commented Dec 21, 2020 at 21:20
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1$\begingroup$ @DSPinfinity To add to what Dan Szabo said, a note regarding negative numbers and different codes: if x is a signed binary, then two's complement code is obtained with this expression: 2^N - x, and offset binary code is obtained with this: x + 2^(N-1). So the code just adds a constant and doesn't change the weight of LSB. $\endgroup$ Commented Dec 22, 2020 at 10:33
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That is well-said. Here, $x_Q$ is a multiple of $\Delta$. A value $x$ can be writen as $x_Q+e_Q$, with remainder $|e_Q|<\Delta$. The last significant bit will be either 0 or 1, depending on the location of $|e_Q|$ with respect to $\Delta/2$: if $|e_Q|<\Delta/2$, the last bit is $0$, else it is $1$.
There is not finer quantization level.
answered Dec 20, 2020 at 23:29
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$\begingroup$ your explanation was harder to understand. $\endgroup$ Commented Dec 20, 2020 at 23:35
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$\begingroup$ I'll try again tomorrow, not on my phone $\endgroup$ Commented Dec 20, 2020 at 23:36
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$\begingroup$ Meanwhile, do not hesitate to add what you {partly} understood so far $\endgroup$ Commented Dec 20, 2020 at 23:42
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$\begingroup$ For example, for $\Delta$, the corresponding code is 001 for which the least significant bit is 1. For $3\Delta$, the corresponding code is 011 for which again the least significant bit is 1. $\endgroup$ Commented Dec 20, 2020 at 23:58