What is the reason for adding 1 to magnitude of sine when we want to send samples to DAC (12 bit resolution)?

I'm not understanding reason for using y=y+1; We know that sin(0) is 0 . But getting 2048 as first sample.

What is the reason for adding 1 to Y?

Y = Y + 1;   

This is the code written to generate sine wave for DAC input .

clear all;
close all;
clc;   % Clear The Previous Points
Ns     = 128; % Set The Number of Sample Points
RES    = 12;  % Set The DAC Resolution
OFFSET = 0;   % Set An Offset Value For The DAC Output
%------------[ Calculate The Sample Points ]-------------
T = 0:((2*pi/(Ns-1))):(2*pi);
Y = sin(T);
Y = Y + 1;    
Y = Y*((2^RES-1)-2*OFFSET)/(2+OFFSET); 
Y = round(Y);                  
plot(T, Y); 
grid
%--------------[ Print The Sample Points ]---------------
fprintf('%d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, \n', Y);

Answer 1

[Walter Roberson]

Adding one transforms it from the range [-1 1] to [0 2] . Multiply by 2^12 to get a value in the range [0 4096] . Divide by 2 to get [0 2048]. Which is suitable for transmission over a digital interface using unsigned bits.

(The way they use OFFSET looks a bit dubious to me though. But it doesn't affect anything when OFFSET is 0.)

If 1 were not added to y then you would start having to transmit values in the range [-2048 2048], and then you would start having to argue about which of the three major representations you use for negative values.

[632541]

Thanks Walter Roberson...Your answers are always crystal clear.

Can you please give reference link or any article for further reference..

[Walter Roberson]

https://en.wikipedia.org/wiki/Signed_number_representations

The particular scheme being used there is "Offset binary"

[632541]

Thank you!!!