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I'd like to understand how a frequency meter treats a sine wave, i think i got it, but still I'd like some confirmation.

So i was taught with square signals where the meter counts the impulses in a determined gate time, my question is does the meter "see" the sine wave like a square wave? meaning does it see the positive part as the impulse and the negative part as the "zero"?

Does it actually transform the waves into square waves every time?

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  • \$\begingroup\$ Can you be more specific on what frequency meter you are referring to? \$\endgroup\$
    – kjgregory
    Dec 18, 2017 at 19:16
  • \$\begingroup\$ It uses a comparator and possibly a frequency divider to turn the sine wave into the clock signal to a binary counter circuit. A binary counter increments on each clock cycle. \$\endgroup\$
    – Keegan Jay
    Dec 18, 2017 at 19:36

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In general, a digital frequency counter would have a threshold level that it uses to discriminate between zero and one. This could be a set-able threshold or it could be inherent to the hardware, depending on the system. This would convert a sinusoid into a zero or a one (effectively, a square wave), assuming that it is crossing the threshold.

enter image description here

The digital system would have its own oscillator that drives the counter. The counter would count edges from like crossings (rising to rising or falling to falling). The digital counter may do some averaging or some other processing to provide a stable reading or statistics.

The digital frequency counter's accuracy would be dependent on the accuracy of its oscillator and its precision would be a function of the oscillator's frequency.

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  • \$\begingroup\$ Hi! thank you very much for your answer, i am using a Kenwood FC-757A. \$\endgroup\$
    – wrong1man
    Dec 18, 2017 at 19:38
  • \$\begingroup\$ I have one more question, when using this meter for period measurement, it uses the target wave as a the gate time, so again it converts it into a square wave, and uses an internal oscillator to create a signal with a known frequency and counts the amount of pulses it could read. So is counts the number of impulses when the sine wave is positive correct? If so then it needs to do N*internal wave period*2 correct? \$\endgroup\$
    – wrong1man
    Dec 18, 2017 at 19:47
  • \$\begingroup\$ If you are reading time, then you won't be expecting a sine wave input, more likely a pulse edge to pulse edge measurement. \$\endgroup\$ Dec 18, 2017 at 19:51
  • \$\begingroup\$ I'm not sure what you meant by pulse edge to pulse edge, do you mean it uses a whole period as a pulse? \$\endgroup\$
    – wrong1man
    Dec 18, 2017 at 19:58
  • \$\begingroup\$ @wrong1man That Kenwood unit appears to have several modes. In the frequency, period, and period average mode I think it would count between same edges. If it only counted the positive time, then it would have to assume 50% duty cycle when it does the calculation as you say. This operation would almost always introduce some additional error, hence I doubt it would do that. In the time interval mode, it will probably do the measurement as you are saying and report time. \$\endgroup\$
    – kjgregory
    Dec 18, 2017 at 19:59
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That depends on the measurement method.

In a high-end spectrum analyzer that I had the pleasure of working on, frequency counting is coupled to a marker on a peak, so coarse estimation has already happened during the frequency sweep, and it is assumed that there is only a single peak within a small window around it.

Measurement consists of demodulating the area around the marker with a narrow filter to a low-frequency IQ signal, converting to amplitude and phase and then monitoring phase change over the measurement time. The requested number of decimal points determines measurement time, 0.1mHz can be resolved in approx. 10ms of measurement time, while 0.1µHz accuracy requires about five seconds if I remember correctly.

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  • \$\begingroup\$ This answer does not describe a count-based frequency measurement though, which was what the question was about. \$\endgroup\$
    – Joren Vaes
    Dec 19, 2017 at 12:15
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Generally you're going to convert the sine wave into a logic-level square wave.

Here is a very old schematic illustrating one way of doing it- the CA3130 op-amp is used as a comparator with about 50mV of hysteresis which squares up the input waveform and prevents noise from adding counts. The input is AC-coupled through the 220nF capacitor and the other resistors and 1uF capacitor bias the inputs within the common mode range of the amplifier and some measure of input protection is provided by the 33k resistor.

enter image description here

Because it is AC coupled, a waveform of more than 50mV p-p is all that is required, so about 18mV RMS.

The gate time is a fixed time, say 1 second or 0.1 second, and a simple frequency meter counts the edges (either rising or falling, but not both) that occur within that time. For example, with a 1 second gate if you count 50 edges the frequency is 50Hz. The count can be latched at the end of the gate period and sent to the display, and the counter can be reset to 0.

Some more sophisticated frequency meters measure the time between some number of edges and calculate the frequency. For example, if it takes 166.67 ms for 10 edges, the frequency is 60Hz.

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