For the path below, determine which latches borrow time and if any setup time violations occur. Repeat for cycle times of 1200, 1000, and 800 ps. Assume there is zero clock skew and that the latch delays are accounted for in the propagation delay, and ?1 = 300 ps; ?2 = 900 ps; ?3 = 200 ps; ?4 = 350 ps.
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This is done by adding up the individual propagation delays: - For the first cycle, the total propagation delay is Δ1 + Δ2 = 300 ps + 900 ps = 1200 ps. - For the second cycle, the total propagation delay is Δ3 + Δ4 = 200 ps + 350 ps = 550 ps. Show more…
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