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Considering the following ring oscillator:

oscillator
I know that the Frequency at \$Q\$ is given by $$f=\frac{1}{2*t_{PD}*3}$$
Yet I want to calculate the period of each signal at \$V_{1}\$,\$V_{2}\$, and \$V_{3}\$. Is there a formula to do so or am I just missing something?

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    \$\begingroup\$ They all have the same period, 1/f. The only difference is the phase. Unless I'm not understanding the question ... \$\endgroup\$ – Dave Tweed Feb 21 '19 at 0:36
  • \$\begingroup\$ What I wanted to ask is whether the formula for \$f\$ is valid only for the last stage or for all the stages at \$V_1\$, \$V_2\$, and \$V_3\$. \$\endgroup\$ – billyandriam Feb 21 '19 at 0:40
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In any oscillator, the period or frequency of the signal at all points in the feedback loop will be identical (unless the oscillator is using some multiplication or division technique to create a higher or lower frequency from a simpler oscillator). The signal may have a completely different phase, polarity, or waveform, but the period will be the same throughout.

In the particular case of a ring oscillator, consider this argument: you could tap the outputs of all three gates, producing three outputs instead of one, and then (ignoring any fanout/loading effects) the circuit would be completely symmetrical and therefore each stage is doing the same thing at different times.

If there were some signal in some design of oscillator that was at a different frequency (that's not in a deliberate simple ratio with it) it would either be completely irrelevant or making the oscillator more unstable (causing frequency modulation). At least, that's true as far as I know; I'd be interested to hear of examples to the contrary.

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