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The hexadecimal number (4B)16 is transmitted as an 8-bit word in parallel. What is the time required for this transmission if the clock frequency is 2.25 MHz?(a) 444 ns(b) 444 s(c) 3.55 s(d) 3.55 msI have been asked this question in final exam.This intriguing question comes from Fast Adder & Serial Adder topic in chapter Arithmetic Circuits of Digital Circuits

Answer»

Right answer is (a) 444 NS

The BEST explanation: Because the clock pulse of 4-bit transmits the data of 8-bit WORD in parallel MODE and this transmission is done at 2.25 MHz frequency. We know that: f=1/t and we can find the time required for this transmission by the clock pulse.

Therefore, time = (1/2.25) = 0.4444 US = 444.44 ns ~ 444ns.



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