Answer:
0.0000045 s
Explanation:
f = Frequency = 8 MHz
Clock cycle is given by
[tex]\dfrac{1}{f}=\dfrac{1}{8\times 10^6}=1.25\times 10^{-7}\ s[/tex]
Time taken for 12 clock cycles
[tex]12\times 1.25\times 10^{-7}=0.0000015\ s[/tex]
Time taken per instruction is 0.0000015 s
In reading and displaying information it requires 3 processes
1 for reading, 1 for searching and 1 for displaying.
[tex]3\times 0.0000015=0.0000045\ s[/tex]
Time taken is 0.0000045 s
We have that for the Question "How long does it take to scan and service the device, given a clocking rate of 8 MHz?"
Answer:
From the question we are told
A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the processor every 20 ms.
since,
[tex]Frequency = 8Mhz\\\\Time period will be = \frac{1}{8}*10^{-6}\\\\= 1.25*10^{-7}[/tex]
Instruction cycle = 12cycles
Therefore, each instruction will take
[tex]12 * \frac{1}{8}*10^{-6}\\\\= 1.5*10^{-6}[/tex]
Total time
[tex]= 3 * 1.5*10^{-6}\\\\= 4.5*10^{-6}sec[/tex]
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