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digital-logic

How to find period of the clock pulse with frequency.

for example: the period is 10ms and i need to find the frequency it would look like this: f= 1/t = 1/10ms = 100Hz because (10ms=.01 seconds, so its really 1/.01=100)

I understand that but when it changes to a larger unit such as GHz i get confused.

Example: if the frequency of a clock pulse is 2GHz, what is the period of the clock pulse?

My thoughts: F= 2GHz= 2 000 000 000Hz
T= 1/F = 1/2 000 000 000?? i somehow doubt this is even close to the answer. Could someone explain this one to me? Thanks a million.

Solution

  • The period is correct as you calculated it.

    P = 1/F
P = 1/2GHz = 1/(2×10^9)
P = 0.0000000005s (second)
P = 0.0000005ms (millisecond)
P = 0.0005us (microsecond)
P = 0.5ns (nanosecond)
P = 500ps (picosecond)

    Another way to look at it is 2Ghz is 2×10^7 times faster than 100hz so the period should be 2×10^7 times shorter

    2Ghz/100Hz = 2×10^7
10ms/(2×10^7) = 500ps