1 //-----------------------------------------------------------------------------
2 // Copyright (C) 2014 iZsh <izsh at fail0verflow.com>
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
7 //-----------------------------------------------------------------------------
8 // Butterworth low pass IIR filter
9 // input: 8bit ADC signal, 1MS/s
10 // output: 8bit value, Fc=20khz
12 // coef: (using http://www-users.cs.york.ac.uk/~fisher/mkfilter/trad.html)
13 // Recurrence relation:
14 // y[n] = ( 1 * x[n- 2])
18 // + ( -0.8371816513 * y[n- 2])
19 // + ( 1.8226949252 * y[n- 1])
23 // b = [-0.8371816513, 1.8226949252]
24 // b is approximated to b = [-0xd6/0x100, 0x1d3 / 0x100] (for optimization)
25 // gain = 2.761139367e2
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
);
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
37 cnt
<= {cnt
[22:0], cnt
[23]};
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
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
);
78 // output: reduce to 8bit
79 assign out
= y1
[16:9];