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I've been reading some guides and papers on data acquisition and often times on the subject of measuring frequency they will simply say

For low-frequency signals, it is sufficient to use one counter, or timebase. The rising edge of the input signal triggers the number of timebase ticks to be counted

They fail to define what constitutes a high frequency, what is the threshold? My online research often takes me into the realm of radio frequency and Rf at which point I wonder if the same applies to electronics in general.

So if I'm looking to measure the frequency of a 4,100Hz signal - is that considered LOW or HIGH frequency?

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    \$\begingroup\$ It's all relative, but in this specific case it comes down to the accuracy and resolution you need. If you can get a sufficient number of timebase ticks in one period of the input signal (or in as many periods as you're willing to wait) then the signal is low enough frequency to use the method described. \$\endgroup\$
    – John D
    Commented Dec 22, 2014 at 1:46

4 Answers 4

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It depends on the application. In this case, it's relative to the frequency of the time base.

As the period of the measured signal is not necessarily a multiple of that of the time base, the ticks counted are only an approximation for the frequency and there will be some error in your measurement. However, the more ticks fit into one period of the measured signal (the higher the frequency difference is between the two signals), the smaller the error.

Knowing the period of your internal time base and the desired precision, you can decide whether the signal is low or high frequency for your measurement.

For example, 4100 Hz seems like low using a 10 MHz time base (2439 or 2440 ticks per period, depends on the starting phase difference). The error can be calculated: as the measuring device can only count integer ticks, exactly 2439 ticks would mean a frequency of 4100.04100... Hz, while exactly 2440 ticks would account for 4098.3607 Hz. In the first case, you have an error of 0.041 Hz (0.001%), in the second case, the error is 1.6393 Hz (0.04%).

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  • \$\begingroup\$ This question has opened up a much larger topic :) I understand now that the high/low determination is relative to the measuring device timing. In the case of the HW I'm considering the datasheet states "Each device supports one 32-bit TTL-level counter that accepts inputs up to 1 MHz" With a timebase of 1MHz I would be looking at 243.902 timebase ticks per period. I've never worked with a timer before so I'm still unclear what the consequences of this new information are. \$\endgroup\$
    – Steve K
    Commented Dec 22, 2014 at 3:13
  • \$\begingroup\$ What exactly do you mean by new information here? \$\endgroup\$
    – hryghr
    Commented Dec 22, 2014 at 23:45
  • \$\begingroup\$ What I meant was that I realize by reading your post that there is more to this than I originally thought. I've never used a timer before and I was thinking I would magically ask it for it's value and it would give me a count. I hadn't though about HOW it would do that or that a timebase was even a factor. After reading about counters along with your reply I understand that the counter counts the ticks for a positive phase of a period and from that count I calculate the frequency. \$\endgroup\$
    – Steve K
    Commented Dec 23, 2014 at 3:13
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It could be either, depending on context, really. 4100 Hz might be high frequency for the bandwidth of a programmable power supply, but low frequency to a broadband amplifier.

In the particular case you mention, it is relative to the clock used to time the period between edges of the incoming waveform.

Imagine you're trying to measure the frequency of a waveform to (say) 4 digits resolution. You want to do it by measuring the period and calculating the reciprocal, which means you need to know the period (time between edges) to within 4 digits resolution. Suppose your timebase is 20MHz. That means you can only time within 50nsec, so the highest frequency you can measure to within 4 digits is f = 20MHz/10,000 = 2kHz. If you don't need 4 digits resolution you can go higher.

Again, in the example shown, you can divide down the input signal so it's lower, and thus get higher resolution at the expense of slower updates and more hardware consumed.

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If I understand this this app note by NI correctly, high frequency or low frequency is relative to the timebase.

enter image description here
Figure 3. Digital Signal with Respect to Internal Timebase (One Counter for Low Frequency)

In this app note, a frequency is considered low if it's much lower than the timebase. In this case, the result is the number of cycles of internal timebase per one period of the input signal.

enter image description here
Figure 4. Digital Signal Frequency Measured with Two Counters (High Frequency)

Otherwise, the frequency is considered high. In this case, the result is the number of cycles of the input signal per one period of the gating signal generated from the internal timebase.

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In the context of the question and the link, low frequency would be when the frequency can be determined with a counter of fixed timebase that yields sufficient precision.

An example will help: if the timebase is 100MHz, the precision required is 0.01%, the fastest period that can be counted that meets the requirement would be 100MHz x 0.01% = 10KHz. So I suppose anything below 10KHz would be considered low frequency in this context.

So for "high frequency", such as Rf frequency, you would not use this scheme to measure the frequency.


3 answers within 2 minutes.

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  • \$\begingroup\$ Thanks for the reply. I'm having a difficult time understanding how I can determine what my required precision is. I know that my application will allow a +- 1Hz tolerance for a measured signal. For example, if the input should be 4,100Hz I will accept 4,099Hz - 4,101Hz as passing. \$\endgroup\$
    – Steve K
    Commented Dec 22, 2014 at 3:26
  • \$\begingroup\$ The precision required would be approx 1 Hz. So in relative term, for 4100 Hz the precision required is 1/4100 = 0.000244. So minimum timebase required to measure this frequency would be 4100/0.000244 = 16.8MHz. \$\endgroup\$
    – rioraxe
    Commented Dec 23, 2014 at 19:55

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