CODEWITHSUNDEEP
bit-manipulationverilogbit
titus.andronicus
titus.andronicus

Reputation: 517

Verilog bit change location

Assuming I have a register reg [15:0] my_reg, which contains a 16-bit signed sample:

How can I find the place where the first bit change is located? Meaning, that if assuming that my_reg = 16'b0001011011010111, how can I know that the first change from 0 to 1 is at my_reg [12]? Same for numbers starting with 1,negative numbers, e.g. my_reg = 16'b1111011011010111 would be interested in the position of the first appearing 0 (which is 11 in this case).

The ultimate goal (to add a little bit of context) is to implement a digital FPGA built-in automatic gain control (AGC).

Upvotes: 1

Views: 4272

Answers (3)

Guy
Guy

Reputation: 177

Same technique as described above but parametrized. Use XOR shifted by one bit to determine where bits change, then use a descending priority encoder to output the first change location. I stuffed with my_reg[0] so the first bit doesn't create a delta.

localparam width=16;

reg  [width-1:0] my_reg;
wire [width:0] delta;
reg  [$clog2(width)-1:0] index; // Note: $clog2 was added in IEEE1364-2005
integer i;

assign delta = my_reg ^ { my_reg, my_reg[0] };

always @* begin
  index = 0;
  for (i=0; i<width; i=i+1)
    if (delta[i])
      index = i;
end

Above code on EDA playground (thanks for heads up on this, BTW) http://www.edaplayground.com/x/3uP

Upvotes: 4

Ari
Ari

Reputation: 7556

A two-liner solution for this is the following:

my_reg1_diff = (my_reg1 ^ (my_reg1 << 1))>>1 ;
my_reg1_diff_pos = $floor($ln(my_reg1_diff)/$ln(2));    //log2 of (my_reg1_diff)

See a working example on edaplayground.com.

It is similar to the idea described in starbox's answer. You can model a synthesizable log2 function using a look-up table or case statement. See edaplayground.com for a sample.

Upvotes: 2

Veridian
Veridian

Reputation: 3667

What you need is a leading signs detector.

You can do this by performing an XOR of the 16-bit my_reg variable, then a case statement that will count the number of repeating bits, then you need to add one to this number.

For example:

if we have a four bit register, we can count the number of leading bits. You can modify this code for your purposes depending on how you want to handle what happens if all the bits are identical.

wire [3:0] my_reg;
wire [2:0] xor_bits;
reg [2:0] count;

// XOR bits
assign xor_bits = {my_reg[3] ^ my_reg[2], my_reg[2] ^ my_reg[1], my_reg[1] ^ my_reg[0];

always @ (*) begin
case (xor_bits)
000: change_location= 4;
00X: change_location= 3;
0XX: change_location= 2;
default: change_location= 1;
endcase

end

Upvotes: 1

Related Questions