An input signal of 2MHz is given to the frequency counter. The time base oscillator has the frequency of 1KHz. So, what I understood is (2MHz/1KHz) is shown in the display of 10 digits. ie 0000002000.

But if the measurement time(Tm) is given (ie 0.2sec)than time base osscillator is ignored and we use Tm. So,(2MHz/5Hz) is what is shown in the dispaly. ie. 0000400000

  • \$\begingroup\$ Because it isn't what you say i.e. "0000400000", what do you conclude actually happens? \$\endgroup\$
    – Andy aka
    Jan 13, 2015 at 12:31
  • \$\begingroup\$ @Andyaka this is not my topic and after doing long research I came to the above conclusion. I could not find anywhere on the internet where there is explanation about relation between Input signals, timebase-oscillator and measurement time. You are free to correct me. \$\endgroup\$ Jan 13, 2015 at 12:39

1 Answer 1


The time base oscillator in a frequency counter is used to set the measurement time. For convenience most measurement times are in decade increments ranging from 10 seconds down to 1 microsecond or less depending on the counter. If the measurement time is 1 millisecond, and the input is 2 MHz, then the counter will accumulate 2000 counts. It will display this with the appropriate units to indicate that the input frequency is 2 MHz, for example as 2.000 MHz. If the measurement time is 0.2 seconds, then the counter will accumulate 400,000 counts. However the counter will need to process this result so that it can correctly display it as a frequency, for example as 2000.00 kHz. This is why most modern counters include microprocessors to allow non-decade measuremnt times as well as for other purposes. In any case, the time base oscillator is not ignored, it still provides the signal to generate the 0.2 second measurement time.

  • \$\begingroup\$ you answer helped me a lot. But one thing i don't understand is if the frequency of TBO is given (ie. 1MHz) and measurement time is given(ie. 1 sec) than what measurement time shall I use? \$\endgroup\$ Jan 14, 2015 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.