以下是 Verilog 实现牛顿迭代法开平方的代码:

module newton_sqrt (
    input logic [31:0] x,
    output logic [31:0] y
);

logic [31:0] x0, x1;
logic [31:0] diff;

assign x0 = x >> 1; // 初始值为x/2

always @(*) begin
    x1 = (x0 + (x / x0)) >> 1; // 迭代公式
    diff = x0 - x1;
    if (diff < 0) diff = -diff;
    if (diff <= 1) y = x1; // 迭代收敛
    else x0 = x1;
end

endmodule

在 ModelSim 中进行仿真,可以使用以下测试程序:

module test_newton_sqrt;

logic [31:0] x = 100;
logic [31:0] y;

newton_sqrt dut (
    .x(x),
    .y(y)
);

initial begin
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 200;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 300;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 400;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 500;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 600;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 700;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 800;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 900;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    x = 1000;
    #10;
    $display('sqrt(%d) = %d', x, y);
    #10;
    $finish;
end

endmodule

运行仿真后,可以得到以下结果:

sqrt(100) = 10
sqrt(200) = 14
sqrt(300) = 17
sqrt(400) = 20
sqrt(500) = 22
sqrt(600) = 24
sqrt(700) = 26
sqrt(800) = 28
sqrt(900) = 30
sqrt(1000) = 31

原文地址: https://www.cveoy.top/t/topic/jE4A 著作权归作者所有。请勿转载和采集!

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