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Solution
:
a) Maximum bus speed = bus driver delay + propagation delay + address decoder delay + time to
fetch the requested data + set up time
=2 +10 + 6 + 25 +1.5
= 44.5 ns per cycle
Ans: maximum speed of the bus: 44.5 ns
b) Number of clock cycles needed to complete the
input operation is 4.
Because a new transfer is started in clock cycle 4

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Solutiona) Maximum bus speed = bus driver delay + propagation del.pdf

  • 1. Solution : a) Maximum bus speed = bus driver delay + propagation delay + address decoder delay + time to fetch the requested data + set up time =2 +10 + 6 + 25 +1.5 = 44.5 ns per cycle Ans: maximum speed of the bus: 44.5 ns b) Number of clock cycles needed to complete the input operation is 4. Because a new transfer is started in clock cycle 4