2019
DOI: 10.15276/aait.02.2019.2
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Power-Oriented Checkabilityand Monitoring of the Currentconsumption in Fpga Projects of the Critical Applications

Abstract: The article is devoted to the problem of checkability of the circuits as an essential element in ensuring the functional safety of informational and control safety-related systems that monitoring objects of increased risk in the energy, transport, military, space and other industries to prevent accidents and reduce their consequences occurrence.The key role of checkabilityin the transformation of fault-tolerant structures used in such systems into fail-safe ones is noted. The problems of logical checkabilityar… Show more

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Cited by 3 publications
(15 citation statements)
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“…Let us consider several variants of extra processors with probabilities of failure-free operation p10 (1) = 0.999, p10 (2) = 0.9995, p10 (3) = 0.9999. In the same way we can calculate probabilities of failure-free operation of the system, extended by the corresponding processors (considering that such a system will be already 3-fault-tolerant): P(S ← p10 (1) ) = 0.9999999999842588; P(S ← p10 (2) ) = 0.9999999999889713; P(S ← p10 (3) ) = 0.999999999992741.…”
Section: Examplesmentioning
confidence: 99%
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“…Let us consider several variants of extra processors with probabilities of failure-free operation p10 (1) = 0.999, p10 (2) = 0.9995, p10 (3) = 0.9999. In the same way we can calculate probabilities of failure-free operation of the system, extended by the corresponding processors (considering that such a system will be already 3-fault-tolerant): P(S ← p10 (1) ) = 0.9999999999842588; P(S ← p10 (2) ) = 0.9999999999889713; P(S ← p10 (3) ) = 0.999999999992741.…”
Section: Examplesmentioning
confidence: 99%
“…Let us consider several variants of extra processors with probabilities of failure-free operation p10 (1) = 0.999, p10 (2) = 0.9995, p10 (3) = 0.9999. In the same way we can calculate probabilities of failure-free operation of the system, extended by the corresponding processors (considering that such a system will be already 3-fault-tolerant): P(S ← p10 (1) ) = 0.9999999999842588; P(S ← p10 (2) ) = 0.9999999999889713; P(S ← p10 (3) ) = 0.999999999992741. Therefore, ΔP(S ← p10 (1) ) = P(S ← p10 (1) ) -P(S) = 9.4153 • 10 -9 ; ΔP(S ← p10 (2) ) = P(S ← p10 (2) ) -P(S) = 9.42 • 10 -9 ; ΔP(S ← p10 (3) ) = P(S ← p10 (3) ) -P(S) = 9.42378 × × 10 -9 .…”
Section: Examplesmentioning
confidence: 99%
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