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        EEPW首頁 > 嵌入式系統(tǒng) > 設計應用 > 用VHDL設計的任意頻率分頻器

        用VHDL設計的任意頻率分頻器

        作者: 時間:2017-06-06 來源:網(wǎng)絡 收藏
        Sometimes I need to generate a clock at a lower frequency than the main clock driving the FPGA. If the ratio of the frequencies is a power of 2, the logic is easy. If the ratio is an integer N, then a divide-by-N counter is only a little harder. But if the ratio isn't an integer, a little (and I mean a little) math is required.

        Note that the new clock will have lots of jitter: there's no escaping that. But it will have no drift, and for some applications that's what counts.

        If you have a clock A at frequency a, and want to make a clock B at some lower frequency b (that is, b a), then something like:

        d = 0;
        forever {
        Wait for clock A.
        if (d 1) {
        d += (b/a);
        } else {
        d += (b/a) - 1; /* getting here means tick for clock B */
        }
        }


        but comparison against zero is easier, so subtract 1 from d:

        d = 0;
        forever {
        Wait for clock A.
        if (d 0) {
        d += (b/a);
        } else {
        d += (b/a) - 1; /* getting here means tick for clock B */
        }
        }


        want an integer representation, so multiply everything by a:

        d = 0;
        forever {
        Wait for clock A.
        if (d 0) {
        d += b;
        } else {
        d += b - a; /* getting here means tick for clock B */
        }
        }

        For example. I just bought a bargain batch of 14.1523MHz oscillators from BG but I need to generate a 24Hz clock.

        So a=14152300 and b=24:


        d = 0;
        forever {
        Wait for clock A.
        if (d 0) {
        d += 24;
        } else {
        d += 24 - 14152300; /* getting here means tick for clock B */
        }
        }


        For a hardware implementation I need to know how many bits are needed for d: here it's 24 bits to hold the largest value (-14152300) plus one more bit for the sign. In VHDL this looks like:


        signal d, dInc, dN : std_logic_vector(24 downto 0);
        process (d)
        begin
        if (d(24) = '1') then
        dInc = 0000000000000000000011000; -- (24)
        else
        dInc = 1001010000000110110101100; -- (24 - 14152300)
        end if;
        end process;
        dN = d + dInc;
        process
        begin
        wait until A'event and A = '1';
        d = dN;
        -- clock B tick whenever d(24) is zero
        end process;



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