192 lines
4.9 KiB
OCaml
192 lines
4.9 KiB
OCaml
(***********************************************************************)
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(* *)
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(* Objective Caml *)
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(* *)
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(* Xavier Leroy, projet Cristal, INRIA Rocquencourt *)
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(* *)
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(* Copyright 1996 Institut National de Recherche en Informatique et *)
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(* en Automatique. All rights reserved. This file is distributed *)
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(* under the terms of the Q Public License version 1.0. *)
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(* *)
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(***********************************************************************)
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(* $Id$ *)
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open Bigarray
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let pi = 3.14159265358979323846
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let tpi = 2.0 *. pi
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let fft (px : (float, float64_elt, c_layout) Array1.t)
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(py : (float, float64_elt, c_layout) Array1.t) np =
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let i = ref 2 in
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let m = ref 1 in
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while (!i < np) do
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i := !i + !i;
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m := !m + 1
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done;
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let n = !i in
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if n <> np then begin
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for i = np+1 to n do
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px.{i} <- 0.0;
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py.{i} <- 0.0
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done;
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print_string "Use "; print_int n;
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print_string " point fft"; print_newline()
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end;
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let n2 = ref(n+n) in
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for k = 1 to !m-1 do
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n2 := !n2 / 2;
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let n4 = !n2 / 4 in
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let e = tpi /. float !n2 in
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for j = 1 to n4 do
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let a = e *. float(j - 1) in
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let a3 = 3.0 *. a in
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let cc1 = cos(a) in
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let ss1 = sin(a) in
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let cc3 = cos(a3) in
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let ss3 = sin(a3) in
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let is = ref j in
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let id = ref(2 * !n2) in
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while !is < n do
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let i0r = ref !is in
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while !i0r < n do
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let i0 = !i0r in
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let i1 = i0 + n4 in
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let i2 = i1 + n4 in
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let i3 = i2 + n4 in
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let r1 = px.{i0} -. px.{i2} in
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px.{i0} <- px.{i0} +. px.{i2};
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let r2 = px.{i1} -. px.{i3} in
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px.{i1} <- px.{i1} +. px.{i3};
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let s1 = py.{i0} -. py.{i2} in
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py.{i0} <- py.{i0} +. py.{i2};
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let s2 = py.{i1} -. py.{i3} in
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py.{i1} <- py.{i1} +. py.{i3};
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let s3 = r1 -. s2 in
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let r1 = r1 +. s2 in
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let s2 = r2 -. s1 in
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let r2 = r2 +. s1 in
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px.{i2} <- r1*.cc1 -. s2*.ss1;
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py.{i2} <- -.s2*.cc1 -. r1*.ss1;
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px.{i3} <- s3*.cc3 +. r2*.ss3;
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py.{i3} <- r2*.cc3 -. s3*.ss3;
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i0r := i0 + !id
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done;
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is := 2 * !id - !n2 + j;
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id := 4 * !id
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done
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done
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done;
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(************************************)
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(* Last stage, length=2 butterfly *)
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(************************************)
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let is = ref 1 in
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let id = ref 4 in
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while !is < n do
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let i0r = ref !is in
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while !i0r <= n do
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let i0 = !i0r in
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let i1 = i0 + 1 in
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let r1 = px.{i0} in
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px.{i0} <- r1 +. px.{i1};
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px.{i1} <- r1 -. px.{i1};
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let r1 = py.{i0} in
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py.{i0} <- r1 +. py.{i1};
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py.{i1} <- r1 -. py.{i1};
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i0r := i0 + !id
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done;
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is := 2 * !id - 1;
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id := 4 * !id
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done;
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(*************************)
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(* Bit reverse counter *)
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(*************************)
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let j = ref 1 in
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for i = 1 to n - 1 do
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if i < !j then begin
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let xt = px.{!j} in
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px.{!j} <- px.{i};
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px.{i} <- xt;
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let xt = py.{!j} in
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py.{!j} <- py.{i};
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py.{i} <- xt
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end;
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let k = ref(n / 2) in
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while !k < !j do
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j := !j - !k;
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k := !k / 2
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done;
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j := !j + !k
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done;
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n
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let test np =
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print_int np; print_string "... "; flush stdout;
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let enp = float np in
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let npm = np / 2 - 1 in
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let pxr = Array1.create float64 c_layout (np+2)
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and pxi = Array1.create float64 c_layout (np+2) in
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let t = pi /. enp in
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pxr.{1} <- (enp -. 1.0) *. 0.5;
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pxi.{1} <- 0.0;
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let n2 = np / 2 in
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pxr.{n2+1} <- -0.5;
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pxi.{n2+1} <- 0.0;
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for i = 1 to npm do
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let j = np - i in
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pxr.{i+1} <- -0.5;
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pxr.{j+1} <- -0.5;
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let z = t *. float i in
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let y = -0.5 *. (cos(z)/.sin(z)) in
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pxi.{i+1} <- y;
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pxi.{j+1} <- -.y
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done;
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(**
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print_newline();
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for i=0 to 15 do Printf.printf "%d %f %f\n" i pxr.{i+1} pxi.{i+1} done;
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**)
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let _ = fft pxr pxi np in
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(**
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for i=0 to 15 do Printf.printf "%d %f %f\n" i pxr.{i+1} pxi.{i+1} done;
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**)
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let zr = ref 0.0 in
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let zi = ref 0.0 in
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let kr = ref 0 in
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let ki = ref 0 in
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for i = 0 to np-1 do
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let a = abs_float(pxr.{i+1} -. float i) in
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if !zr < a then begin
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zr := a;
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kr := i
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end;
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let a = abs_float(pxi.{i+1}) in
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if !zi < a then begin
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zi := a;
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ki := i
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end
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done;
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let zm = if abs_float !zr < abs_float !zi then !zi else !zr in
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print_float zm; print_newline()
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let _ =
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let np = ref 16 in for i = 1 to 13 do test !np; np := !np*2 done
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