matlab code about function of quantization

I have been given a quantization function (matlab code) but I'm confused in some places, pls help....(PS: y is a signal(-1 to 1) and 12 bits is used to quantize)

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function [symbols_quant]=quantization(y)

    b=12;
  q=2/(2^b);
  for i =1: length(y)
  symbols_quant(i)= max(min((round(y(i) * 2^(b-1))/(2^(b-1)))-sign(y(i))*q/2 , 1) , -1);

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Question 1: why we should multiply the sample,y(i), with by 2^(b-1) , and then round it to get the quantized value? But not directly round it...What is the principle behind?

Thank you very much if you can give me a hand.

[Azzi Abdelmalek]

What is y?

[fdsa]

the signal, y = wavread('.wav');

Answer 1

[Walter Roberson]

Consider a number such as 3/8. If you round it directly you will get 0; multiply that by 2^2 (4) and you would still get 0. But if you multiply 3/8 by 2^2 (4) and then round you will get round(3/2) = 2. So clearly multiplying before rounding makes a difference.

Keep in mind the only rounding operation available is to round to integer. There is no function to round to nearest 1/4 or 1/10(for example.)

The principle is the same you would use for rounding to a certain number of decimal places. For example, to round 3.141 to 1 decimal place, multiply it by 10 to get 31.41, round to get 31, divide again by 10 to get 3.1

[fdsa]

Great! I get the idea but...

(as to your example) why we should multiply by 2^2 (4) but not 10/100/1000...what I means is we want to round the number to what by multiplying 2^2 (4)...

Thank you very much for your response!!

[Walter Roberson]

You want to quantize to a certain number of bits. bits is binary. binary uses 2^n representation, not 10^n

[fdsa]

Get it!!! One last problem: In the code, why we need the expression: -sign(y(i))*q/2 ? Am I missing any knowledge about quantization?

I wonder it just take the mid-value between two quantization levels but why the sign is "-" but not "+".....?

Thanks!!

[fdsa]

Do not know why???

Thank you.

Answer 2

[Azzi Abdelmalek]

function [symbols_quant]=quantization(y)  
b=12
q=2/(2^b-1)   
symbols_quant=round(y/q)*q

[fdsa]

I've got this expression once... However what is the meaning of the division by q and then ...multiplication by q?

Thanks!

[Azzi Abdelmalek]

Example

y=10
q=0.3
% after quantization y should be 9.9
% round(y/q)=round(10/0.3)=round(33.33)=33
% round(y/q)*q=33*0.3=9.9

[fdsa]

get it!! thz!!

Answer 3

[Azzi Abdelmalek]

One way how a real ADC works

% for values from 0 to 10
n=12
q=10/(2^n-1)
max_val=10/2
u=6.5555   % Example
bit=zeros(1,n)  
for k=1:n
  if u>max_val
      bit(k)=1
      u=u-max_val
  end
  max_val=max_val/2
end
disp(bit)   % The bits that codify your input.
quantized_value=sum(bit.*2.^(n-1:-1:0)*q)

[fdsa]

I've tried my best to understand but... anyway, thanks for your help!!!

[fdsa]

One more question: In the code, why we need the expression: -sign(y(i))*q/2 ? Am I missing any knowledge about quantization?

I wonder it just take the mid-value between two quantization levels but why the sign is "-" but not "+".....?

Thanks!!

Answer 4

[fdsa]

One more question: In the code, why we need the expression: -sign(y(i))*q/2 ? Am I missing any knowledge about quantization?

I wonder it just take the mid-value between two quantization levels but why the sign is "-" but not "+".....?

Thanks!!