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Artificial Intelligence and Soft Computing: Behavioral ... - Arteimi.info

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For the sake of convenience, the pipelining <strong>and</strong> parallelism of different<br />

modulus is presented in fig. 22.18. An estimation of the cycle time for firing a<br />

transition is computed in fig. 22.19 following the pipelining of the stages in<br />

fig. 22.18. It is evident from fig. 22.19 that the TSF works in parallel with the<br />

pipelined stages comprising of THF, PTVVM <strong>and</strong> Matcher or THF, PTVVM<br />

<strong>and</strong> FPS in sequence. The matcher <strong>and</strong> the FPS start working only when the<br />

PTVVM completes the current task. The time required for the matcher being<br />

larger than that of FPS we consider the time required by the matcher only<br />

following THF <strong>and</strong> PTVVM. It is also clear from the figure that the matcher<br />

finishes its tasks at time e, which is larger than the time d required by the<br />

TSF. An additional 2 gate delay followed by a few memory access cycles plus<br />

gate delay makes the total cycle time equal to g, which is the effective cycle<br />

time for firing a transition. The time requirement for each unit is shown in the<br />

fig. 22.19 itself <strong>and</strong> needs no further clarification.<br />

1 Memory<br />

access<br />

THF<br />

TSF<br />

PTVVM Matcher<br />

25 Tc 10 gate delay<br />

FPS<br />

2 gate delay<br />

Delay<br />

Fig. 22.18: Pipelining among the major modules of the architecture for<br />

the execution of FOL programs; each circuit corresponds to<br />

one transition.<br />

Since a number of transitions are concurrently firable, it is expected that<br />

the execution of a complete logic program will require an integer multiple of<br />

this transition firing cycle time. Assuming a clock cycle time equal to Tc <strong>and</strong><br />

ignoring smaller gate delays, it has been found that the total cycle time<br />

required for firing a transition is 25 Tc .<br />

Thus assuming a 100M-Hz clock frequency, the cycle time for one<br />

transition is equal to 25x10 -8 S = 250 nS (nanosecond ). Now for the sample<br />

logic program represented by the Petri net of fig. 22.17, the total time for

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