Lets’ assume we have a few buckets A….E filled with zeroes except A, which has sampled a one from input.
in -> A B C D E 1 1 0 0 0 0 BBDs have two copy phases, 1 and 2. In Phase 1, A is transferred to B and D to E: in A -> B C -> D E 0 1 1 0 0 0 Phase 2 does this, assuming A samples a 2 from input; in -> A B -> C D -> E 2 2 1 1 0 0 Back to phase 1, we see: in A -> B C -> D E 3 2 2 1 1 0 Phase 1 & 2 are one clock period T or 1/f. But the signal traversed over 4 stages, from A to D. So, it’s faster than light (no, not yet, but it’s 2 stages per clock cycle). Nonetheless, the actual sampling period is T, so the nyquist frequency is still f/2. Hope this helps a bit, it’s indeed a bit confusing at times. Steffan > On 13. Feb 2024, at 11:41, Victor <[email protected]> wrote: > > This Message Is From an External Sender > This message came from outside your organization. > Hi all, > > a BBD circuit is said to have a delay time equivalent to N/(2f), where N is > the number of stages and > f is the clock frequency (Hz). The 2 in this seems to indicate that each > stage is traversed in T/2 secs, > the clock period being T = 1/f. > > Ok, so this means the clock is running at twice the sampling rate of the > system. In this case, the > discrete-time baseband would extend to f Hz (positive side only), should it > not? > Why I see it quoted everywhere that the BBD bandwidth is f/2 Hz? Where did I > go wrong? > > > Victor > >
