CPU频率和上升沿计数

我对CPU时钟速度感到困惑 .

我认为看完这个YouTube video之后我已经有了充分的理解，并且阅读了this web page，但是当我回到关于时钟速度的Pluralsight的VHDL教程（链接没有给出，因为它不是免费的）时，我不确定 .

请考虑以下代码，它是一个非常简单的计时器：

entity boardio is
  port (
    clock_50: in bit;
     hex0 : out bit_vector(0 to 6);
     hex1 : out bit_vector(0 to 6);
     hex2 : out bit_vector(0 to 6);
     hex3 : out bit_vector(0 to 6);
     key : in bit_vector(0 to 3);
     sw: bit_vector(0 to 9);
     ledr: buffer bit_vector (0 to 9)
  );
end boardio;

architecture arch of boardio is
signal count : integer := 1; -- not a variable!
signal clock_1hz : bit := '0';
signal mins, secs : integer range 0 to 59 := 0;

function hex_digit(x:integer; constant hide_zero:boolean := false) return bit_vector is
begin
    case x is
        when 0 => 
            if hide_zero then
                return "1111111";
            else
                return "0000001";
            end if;
        when 1 => return "1001111";
        when 2 => return "0010010";
        when 3 => return "0000110";
        when 4 => return "1001100";
        when 5 => return "0100100";
        when 6 => return "0100000";
        when 7 => return "0001111";
        when 8 => return "0000000";
        when 9 => return "0000100";
        when others => return "1111111";
    end case;
end function;

begin

-- 1hz clock
-- ****************************************************
process(clock_50)
begin
    if (clock_50'event and clock_50 = '1') then
        count <= count + 1;
        if (count = 25000000) then -- half the rate
            clock_1hz <= not clock_1hz;
            count <= 1;
        end if;
    end if;
end process;
-- ****************************************************
process(clock_1hz, key)
begin
    -- update # of seconds
    if (clock_1hz'event and clock_1hz = '1') then
        if (key(0) = '0') then
            secs <= 0;
            mins <= 0;
        else
            if (secs = 59) then
                mins <= (mins + 1) rem 60;
            end if;
            secs <= (secs + 1) rem 60;
        end if;
    end if;
end process;

process(clock_1hz)
begin
        hex0 <= hex_digit(secs rem 10);
        hex1 <= hex_digit(secs / 10, true);
        hex2 <= hex_digit(mins rem 10, mins = 0);
        hex3 <= hex_digit(mins / 10, true);
end process;

end arch;

我已经发布了所有代码，所以每个人都有完整的上下文，但我真正感兴趣的是我用星号表示的 1hz clock 进程 .

它表明，在50MHz时钟的一秒内，一秒钟内将有250,000,000个上升沿 . 然而，我链接的视频和网页表明，对于1hz时钟，一秒钟内会有一个上升沿和一个下降沿，因此50Mhz时钟将会有500,000,000个时钟 .

请有人请说明CPU频率在上升沿和下降沿方面实际意味着什么，以及“滴答”？图表会非常感激，因为我不再确定链接上的图表是否正确...