how to covert negative samples of sine wave into positive?

hello

I am having trouble to get unsigned samples of sinewave. For example generated 1MHz sine wave. But samples are with both positive and negative values. But this sinewave LUT not be stored in FPGA. i have to convert negative samples into positive. I tried 2's complement method but it changing shape of sine wave. could some one help on this how to do it in matlab?

%clc;
%clear all;
%close all;
% t=linspace(0,1e-8,10000);
t = 0:0.00406901041666666666666666666667e-6:1e-6;
f=1e6;
x=sin(2*pi*f*t);
plot(t,x);
xlabel('Time (us)');
ylabel('Amplitude');

 a1=x*1000;
 b1=round(a1);
 [c,v]=size(b1); 
    n3 = c*v;
    word_len =11;
    data = reshape(b1',n3,1);
    databin =fi(data,1,10);
   h = bin(databin);
   d = bin2dec(h);

when i plot d vlaues from the code i am getting sine wave as above figure. not exact sine wave. please help on this.

[Star Strider]

Maybe I’m missing something, however if I want a strictly positive sine curve (here ‘x’), I would just add 1.001 (or something similar) to it, then multiply it by 1000.

[Adam Danz]

or abs()? But I'm not sure I understand the goal. maybe

sin(x) - min(sin(x(sin(x)<0)))

[Robert U]

Hi kowsalya ramaraj,

I hope you are aware of that fi() produces a fixed point representation of your values of data. The value produced contains 10 bits. The highest bit is the sign (where '1' means 'negative'). All values are integers. If you convert the values to decimal without "decoding" you'll get values ranging from 0 to 1024 (which is 2^10). All values greater 512 are negative. Thus, values to be displayed are actually -511 till 512.

Kind regards,

Robert

[kowsalya ramaraj]

Thank you so much all for kind response. iI am getting positive samples and sine wave with this expression sin(x) - min(sin(x(sin(x)<0))).....

[Adam Danz]

Great! I'll move that comment to the answers section so the question is marked as answered.

Answer 1

[Adam Danz]

To shift the sin curve upward so that it's minimum value is not negative,

x1 = sin(x) - min(sin(x(sin(x)<0)));

Another way to think about that is

y = sin(x);
y1 = y- min(y(y<0));

[Thorsten]

Instead of y1 = y- min(y(y<0)); you can simply use

y1 = y - min(y);

or

y1 = y + 1;

because you know the minimum value of sin is -1.

[Adam Danz]

In the simple case of sin(x), yes this is true. But with additional parameters such as a*sin(x) you'd have to find the minimum negative value.