## 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."

• i retitled the graphic because the previous title was completely wrong. – robert bristow-johnson Dec 21 '20 at 19:08

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 $$\Delta$$.

• This is not correct for negative signal values and moreover the binary number format is two's complement which is different from base 2. – DSPinfinity Dec 21 '20 at 20:40
• 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’. – Dan Szabo Dec 21 '20 at 21:20
• @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. – megasplash Dec 22 '20 at 10:33

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.

• your explanation was harder to understand. – DSPinfinity Dec 20 '20 at 23:35
• I'll try again tomorrow, not on my phone – Laurent Duval Dec 20 '20 at 23:36
• thank you Laurent. – DSPinfinity Dec 20 '20 at 23:37
• Meanwhile, do not hesitate to add what you {partly} understood so far – Laurent Duval Dec 20 '20 at 23:42
• 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. – DSPinfinity Dec 20 '20 at 23:58