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It has been established that the FTW sets the step size of the accumulator output. The step size, in turn, establishes the average number of steps required to span the capacity of the accumulator. Specifically, the average number of steps, S, is the capacity divided by the FTW. Therefore, in terms of the bus width of the accumulator N, the average number of steps between 0 and the accumulator capacity is 2^N over FTW. It has also been shown that each step of the accumulator output is separated by one period of the system clock, Ts. Therefore, multiplying the average number of steps, S, necessary to make the accumulator overflow by the time per step, Ts, yields the average overflow period TAVG. By substitution, TAVG is found in terms of the system clock frequency, the accumulator bus width, and the FTW. Specifically, TAVG equals 2^N divided by Fs times FTW. By definition, TAVG is the average period between successive accumulator rollovers. The reciprocal of TAVG is the average rollover frequency and corresponds to the output frequency of the DDS, Fo. This yields the basic DDS equation shown on this slide: Fo equals Fs times FTW divided by 2^N. Recall that two to the N is the capacity of the accumulator C. Furthermore, FTW is always less than C, so the quantity FTW divided by 2^N is always less than one. This implies that Fo is always less than Fs.
