Aspire Faculty ID #16334 · Topic: NIMCET 2010 · Just now
NIMCET 2010

Access time = $45\text{ ns}$, gap = $5\text{ ns}$ Bandwidth = ?

Solution

Cycle time = $45 + 5 = 50\text{ ns}$ Bandwidth = $\frac{1}{50\times10^{-9}} = 20\text{ MHz}$

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