Quantization confusion in pulse code modulation.

[shivajikobardan]

I have a question-:
This is a quantized version of an information signal. Here number of bits used to encode is 4 so there are 16 levels.

Here is the encoding.
I tried to do this with similar type of questions.

But as you see here is the problem. There are 5 Voltage levels but only 4 possible encodings. There is a solution to this afaik it is to let like-:

4V-2V=00
2V-0V=01
0V-–2V=10
-2V- -4V=11.
But that is making less sense to me at the moment.
Can you shed some light in this issue?

 

[stenzer]

Hi,

I'm not able to follow your thoughts completely. But the quantization step voltage can be determined by

V_FS / (2^n) = (4 V - -4V) / 2^4 = 8 V / 16 = 0.5 V.​

​

So there are 16 quantization steps, each one covers an interval of 0.5 V. Have you choesen a 2 Bit ADC (n = 2 --> L = 4) on purpose for your pen & paper example, I assume so?

4V-2V=00
2V-0V=01
0V-–2V=10
-2V- -4V=11.

Here you are dealing with ranges, as you have correctly shown. So there are no five specific voltages, only ranges, and of course TWO voltage levels define ONE interval/range. Consequently, you have one range less than specific voltage levels. Have a look on the links below, they might help you, especially [3].

[1] https://www.analog.com/en/education/education-library/data-conversion-handbook.html
[2] https://www.analog.com/media/en/tra...ndbooks/Data-Conversion-Handbook/Chapter2.pdf
[3] https://microchipdeveloper.com/adc:adc-quantization-error

BR

 

[KlausST]

Hi,

the first picture shows an "unsigend" behaviour from 0 to 15.

the second one is a signed behaviour (-4 ... 0 ... +3). But a signed behaviour is 16 steps including zero, thus just 15 steps are available for non_zero values. You can´t have symmetric behaviour, unless:
* you define the range to be -3 ... 0 ... +3 (leave the "-4" redundant)
* you define the steps to be -3.5 / -2.5 / -1.5 / -0.5 / +0.5 / ... /+3.5. But you can´t have true zero

Klaus