fpga/lp20khz_1MSa_iir_filter.v

   1 //-----------------------------------------------------------------------------
   2 // Copyright (C) 2014 iZsh <izsh at fail0verflow.com>
   3 //
   4 // This code is licensed to you under the terms of the GNU GPL, version 2 or,
   5 // at your option, any later version. See the LICENSE.txt file for the text of
   6 // the license.
   7 //-----------------------------------------------------------------------------
   8 // Butterworth low pass IIR filter
   9 // input: 8bit ADC signal, 1MS/s
  10 // output: 8bit value, Fc=20khz
  11 //
  12 // coef: (using http://www-users.cs.york.ac.uk/~fisher/mkfilter/trad.html)
  13 // Recurrence relation:
  14 // y[n] = (  1 * x[n- 2])
  15 //      + (  2 * x[n- 1])
  16 //      + (  1 * x[n- 0])
  17
  18 //      + ( -0.8371816513 * y[n- 2])
  19 //      + (  1.8226949252 * y[n- 1])
  20 //
  21 // therefore:
  22 // a = [1,2,1]
  23 // b = [-0.8371816513, 1.8226949252]
  24 // b is approximated to b = [-0xd6/0x100, 0x1d3 / 0x100] (for optimization)
  25 // gain = 2.761139367e2
  26 //
  27 // See details about its design see
  28 // https://fail0verflow.com/blog/2014/proxmark3-fpga-iir-filter.html
  29 module lp20khz_1MSa_iir_filter(input clk, input [7:0] adc_d, output rdy, output [7:0] out);
  30
  31     // clk is 24MHz, the IIR filter is designed for 1MS/s
  32     // hence we need to divide it by 24
  33     // using a shift register takes less area than a counter
  34     reg [23:0] cnt = 1;
  35     assign rdy = cnt[0];
  36     always @(posedge clk)
  37         cnt <= {cnt[22:0], cnt[23]};
  38
  39     reg [7:0] x0 = 0;
  40     reg [7:0] x1 = 0;
  41     reg [16:0] y0 = 0;
  42     reg [16:0] y1 = 0;
  43
  44     always @(posedge clk)
  45     begin
  46         if (rdy)
  47         begin
  48             x0 <= x1;
  49             x1 <= adc_d;
  50             y0 <= y1;
  51             y1 <=
  52                 // center the signal:
  53                 // input range is [0; 255]
  54                 // We want "128" to be at the center of the 17bit register
  55                 // (128+z)*gain = 17bit center
  56                 // z = (1<<16)/gain - 128 = 109
  57                 // We could use 9bit x registers for that, but that would be
  58                 // a waste, let's just add the constant during the computation
  59                 // (x0+109) + 2*(x1+109) + (x2+109) = x0 + 2*x1 + x2 + 436
  60                 x0 + {x1, 1'b0} + adc_d + 436
  61                 // we want "- y0 * 0xd6 / 0x100" using only shift and add
  62                 // 0xd6 == 0b11010110
  63                 // so *0xd6/0x100 is equivalent to
  64                 // ((x << 1) + (x << 2) + (x << 4) + (x << 6) + (x << 7)) >> 8
  65                 // which is also equivalent to
  66                 // (x >> 7) + (x >> 6) + (x >> 4) + (x >> 2) + (x >> 1)
  67                 - ((y0 >> 7) + (y0 >> 6) + (y0 >> 4) + (y0 >> 2) + (y0 >> 1)) // - y0 * 0xd6 / 0x100
  68                 // we want "+ y1 * 0x1d3 / 0x100"
  69                 // 0x1d3 == 0b111010011
  70                 // so this is equivalent to
  71                 // ((x << 0) + (x << 1) + (x << 4) + (x << 6) + (x << 7) + (x << 8)) >> 8
  72                 // which is also equivalent to
  73                 // (x >> 8) + (x >> 7) + (x >> 4) + (x >> 2) + (x >> 1) + (x >> 0)
  74                 + ((y1 >> 8) + (y1 >> 7) + (y1 >> 4) + (y1 >> 2) + (y1 >> 1) + y1);
  75         end
  76     end
  77
  78     // output: reduce to 8bit
  79     assign out = y1[16:9];
  80
  81 endmodule