Question

Given the read cycle diagram below, if the memory speed is 52.5 ns, how many wait states are needed if the bus clock is 72.1-MHz? ? ADDRESS T Read Cycle with One Wait State T T T Memory Address to be Read Tos Data Read T DATA T MREQ T T RD#/WR WAIT TAD = 6.6 ns max TML = 1.9 ns min TM = 1.1 ns max THL = 3.6 ns max Tos = 9.2 ns min TMH = 0.5 ns max THH = 0.6 ns max ToH = 1.9 ns min T T

          Given the read cycle diagram below, if the memory speed is 52.5 ns, how many wait states are needed if the bus clock is 72.1-MHz?
?
ADDRESS
T
Read Cycle with One Wait State
T
T
T
Memory Address to be Read
Tos
Data Read
T
DATA
T
MREQ
T
T
RD#/WR
WAIT
TAD = 6.6 ns max
TML = 1.9 ns min
TM = 1.1 ns max
THL = 3.6 ns max
Tos = 9.2 ns min
TMH = 0.5 ns max
THH = 0.6 ns max
ToH = 1.9 ns min
T
T
        
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Given the read cycle diagram below, if the memory speed is 52.5 ns, how many wait states are needed if the bus clock is 72.1-MHz?
?
ADDRESS
T
Read Cycle with One Wait State
T
T
T
Memory Address to be Read
Tos
Data Read
T
DATA
T
MREQ
T
T
RD#/WR
WAIT
TAD = 6.6 ns max
TML = 1.9 ns min
TM = 1.1 ns max
THL = 3.6 ns max
Tos = 9.2 ns min
TMH = 0.5 ns max
THH = 0.6 ns max
ToH = 1.9 ns min
T
T

Added by Emilio P.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Read Cycle with One Wait State ADDRESS Memory Address to be Read DATA Data Read MREQ Te RDAWR WITH TAo=6.6ns max Tm=1.9ns min TM=1.1ns max T=3.6ns max Tos=9.2ns min Tm=0.5ns max T=0.6ns max To=1.9ns min
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00:02 We have n -a, so refrat index equals 1, sine theta -a equals nb, sign theta -b, which is a over 2.
00:17 Now, theta -a equals a -over -2 plus alpha.
00:24 So we have sine of a over 2 plus alpha equals nb, which is n, sine a over 2, sine a over 2...
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