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I have seen the crystal oscillator output in the oscilloscope and it looks something similar to a sine wave (at least nothing similar to a square wave and with a hard to calculate duty cycle)!

So how does a type 1 phase frequency detector (the xor gate) functions with such a reference signal from the crystal oscillator as one of its input?

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    \$\begingroup\$ Can you tell us what the frequency of the oscillator is, please? And also the bandwidth of the scope, and what type of probe (10x or 1x) that you're using? \$\endgroup\$ – John Honniball Nov 6 '14 at 17:59
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    \$\begingroup\$ A picture of the scope trace would be helpful too. \$\endgroup\$ – tcrosley Nov 6 '14 at 18:06
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    \$\begingroup\$ And the oscillator schematic. \$\endgroup\$ – Leon Heller Nov 6 '14 at 18:19
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    \$\begingroup\$ Are you looking at the output signal from a crystal oscillator or one of the signals on the pins of a crystal itself? \$\endgroup\$ – EM Fields Nov 6 '14 at 22:00
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    \$\begingroup\$ There's no such thing as a period of 12MHz. There's a frequency of 12MHz, and its period is its reciprocal: 83.3ns. But, I was trying to determine whether what you're probing is a crystal or a crystal oscillator. No insult intended, but do you know the difference between them? \$\endgroup\$ – EM Fields Nov 7 '14 at 0:33
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Some oscillators, particularly crystal VCOs, produce a sine wave or clipped sine wave output. For example, these.

You would not likely get good results using a digital phase detector on an analog signal, so you would likely want to use a comparator to turn it into a digital signal.

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