Question

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 ms

I 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 1

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.