What is the reason for adding 1 to magnitude of sine when we want to send samples to DAC (12 bit resolution)?
2 views (last 30 days)
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 .
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);
%--------------[ Print The Sample Points ]---------------
fprintf('%d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, %d, \n', Y);
Walter Roberson on 11 Jun 2021
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.