How to generate sine wave using Verilog?

[h_upadhyay]

I have to generate a two different sine wave(1st sine wave normal sine wave which is generate from 0 degree and 2nd one phase shifted sine wave ) using Verilog, and on google, I found something related to it but somehow I didn't understand. So, could anyone explain to me the logic behind this code?

module sine_cos(clk, reset, en, sine, cos);
   input clk, reset, en;
   output [7:0] sine,cos;
   reg [7:0] sine_r, cos_r;
   assign      sine = sine_r + {cos_r[7], cos_r[7], cos_r[7], cos_r[7:3]};
   assign      cos  = cos_r - {sine[7], sine[7], sine[7], sine[7:3]};
   always@(posedge clk or negedge reset)
     begin
         if (!reset) begin
             sine_r <= 0;
             cos_r <= 120;
         end else begin
             if (en) begin
                 sine_r <= sine;
                 cos_r <= cos;
             end
         end
     end
endmodule

but here what is the logic for ?

assign sine = sine_r + {cos_r[7], cos_r[7], cos_r[7], cos_r[7:3]};
assign cos = cos_r - {sine[7], sine[7], sine[7], sine[7:3]};

[KlausST]

Hi,

I expect there is some description where you found the code.
--> please post the link to the internet page.

Klaus

[FvM]

The oscillator operation has been discussed in previous threads, e.g. https://www.edaboard.com/thread39599.html

The Verilog implementation is somehow clumsy, the function can be made much clearer by using signed data type.

- - - Updated - - -

Code Verilog - [expand]
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output signed [7:0] sine,cos;
reg signed [7:0] sine_r, cos_r;
assign      sine = sine_r + (cos_r >>> 3);
assign      cos  = cos_r - (sine >>> 3);

[asdf44]

Consider that the derivative (slope) of sin is cos and the derivative of cos is sin. Thus you can increment one by a value proportional to the other and you get a self-oscillating output that's a very good (not perfect) sin wave.

The specific code you're questioning is just a power of 2 divide. FvM posted a better way to do this. The more you divide the longer the period is.

The biggest problem is rounding errors. You need to get lucky and find a set of parameters that returns exactly to where it started or you need to reset every cycle introducing a small amount of distortion. If you don't the amplitude will 'walk away' towards zero or infinity.

Bottom line this is a very 'cheap' way to generate a sin wave in an FPGA. For high precision work you're best off using the IP blocks that all FPGA manufacturers provide.

[h_upadhyay]

The oscillator operation has been discussed in previous threads, e.g. https://www.edaboard.com/thread39599.html

The Verilog implementation is somehow clumsy, the function can be made much clearer by using signed data type.

- - - Updated - - -

Code Verilog - [expand]
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output signed [7:0] sine,cos;
reg signed [7:0] sine_r, cos_r;
assign      sine = sine_r + (cos_r >>> 3);
assign      cos  = cos_r - (sine >>> 3);

Hi, FvM
Heartily thanks for the replay ,
so by using this shift operators we rotate back Or right shift signal by 3 ?
by using sine and cos equations ?

[FvM]

>>> is arithmetic shift, >>> 3 means multiply with 0.125.

For a deeper insight how it works please review the linked old thread.

[h_upadhyay]

Here by using a just normal procedure for generating sine wave , through verilog code. " i have to generate a two different sine wave , 1st normal sine wave (which is start from 0 degree ) and 2nd wave which is start from phase shift(+- 90 degree) sine wave.and that i already done in vivado simulation but Now i want to again generate a square wave from this sine wave, (sine wave --> counter ) so please suggest me the logic for verilog !

[wwfeldman]

generate a square wave from this sine wave, (sine wave --> counter )

same frequency?
sketch the sin wave, then sketch the square wave on top of it. where along the sin wave if the square wave high? when low?
now write the appropriate logic statements

different frequency?
how do you get from the sinwave frequency to the square wave frequency?

[FvM]

but now i want to again generate a square wave from this sine wave

Use the most significant (= sign) bit.

[wwfeldman]

as FvM says, use the sign bit

you specified sine and cos as signed values, so the square waves will be specified by sign + or -
use a statement that checks the sign bit, as FvM suggested