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steadyj
Joined: 17 Aug 2008
Posts: 13
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 22 Aug 2008 19:10 "Lead" or "Lag" in phase detector? |
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Hi all, I am confused about the judge of "lead" or "lag" in phase detector.
Figure 1 is a phase detector. Figure 2 is the waveform when A leads B. But the fact "A leads B" seems to depend on the initial time. As shown in Figure 3, if the initial time is the red vertical line, it becomes "A lags B". And the output pulses will be on Qb instead of on Qa.
So what is wrong with my logic? I believe phase detector should work in a reliable way instead of depending on the arbitrary initial time. What is the real way phase detector works?
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FvM
Joined: 22 Jan 2008
Posts: 5161
Helped: 767
Location: Bochum, Germany
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22 Aug 2008 19:25 Re: "Lead" or "Lag" in phase detector? |
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| what is initial time in a real circuit? Only a fiction. Actually, the power supply is turned on at an arbitrary moment and the FFs initial states are unknown anyway. So a edge counting phase detector will have an in initially uncertainty of one cycle. But it's no problem in any real application (typically a PLL). I guess, you'll findout, why.
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steadyj
Joined: 17 Aug 2008
Posts: 13
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22 Aug 2008 19:38 "Lead" or "Lag" in phase detector? |
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I really don't know why. For instance, A is the reference input and B is the VCO output. After power on, B's frequency is lower than A. So we wish the phase detector can output voltage on the "UP" port. Unfortunately the uncertainty after power on maybe make the contrary action, which seems to un-lock for ever.
-- Please correct me. Thank you.
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LvW
Joined: 07 May 2008
Posts: 1466
Helped: 242
Location: Germany
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23 Aug 2008 9:28 Re: "Lead" or "Lag" in phase detector? |
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| steadyj wrote: |
Figure 1 is a phase detector. Figure 2 is the waveform when A leads B. But the fact "A leads B" seems to depend on the initial time. As shown in Figure 3, if the initial time is the red vertical line, it becomes "A lags B". And the output pulses will be on Qb instead of on Qa. |
I am afraid, your understanding of the term "lead" is not correct.
"Lead" does not mean that the pulse train is "more to the right" - just the opposite is true: Take your red line and verify that the train A is already in a high state and train B is not. Therefore, A leads B.
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FvM
Joined: 22 Jan 2008
Posts: 5161
Helped: 767
Location: Bochum, Germany
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23 Aug 2008 10:14 Re: "Lead" or "Lag" in phase detector? |
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| Quote: |
| Unfortunately the uncertainty after power on maybe make the contrary action, which seems to un-lock for ever. |
Comparing your circuit with a 4046 digital phase detector, that is known not to have similar issues, it should work.
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