Consider a 33 MHz CPU based system. What is the number of wait states required if it is interfaced with a 60 ns memory? Assume a maximum of 10 ns delay for additional circuitry like buffering and decoding.
Explanation: A wait state is a delay experienced by a computer processor when accessing external memory or another device that is slow to respond.
Total memory access time = 60 ns + 10 ns = 70 ns.
Given, CPU frequency = 33 MHz
1 clock time = 1 / (33 MHz) = (1/33)*10-6 = 30.30 ns.
Total number of wait states = Total number of cycle needed = 70 ns / (30.30 ns) = 2.31 ≈ 3 cycles.
So, option (D) is correct.
Quiz of this Question
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