Timing Diagram

Discussion in 'Homework Help' started by ra1ph, Oct 16, 2010.

  1. ra1ph

    Thread Starter Active Member

    Jan 5, 2010
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    Hi,

    Would like if someone could look at the attached image and tell me how I am doing in attempting to answer the question. I have filled out A and B and started OUT
    [​IMG]

    Thanks
     
    Last edited: Oct 16, 2010
  2. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    You already have done a mistake. The JK FF is positively triggered (as indicated by the lack of a dot in the input of its clock). That means it will change state only when its clock goes from LOW to HIGH. As a result, the JK FF will have half the frequency in its changes than the D FF.
    You have made it change state when the clock goes from HIGH to LOW too, which is wrong.

    When you plot the outputs of the FFs correctly, just combine them with Din as the gates indicate and get the OUT signal.
     
  3. ra1ph

    Thread Starter Active Member

    Jan 5, 2010
    31
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    Thanks, I see where I went wrong for the JK FF. Still a bit unsure for OUT. I have attempted it though,

    [​IMG]
     
  4. Georacer

    Moderator

    Nov 25, 2009
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    You are correct so far. Continue to draw the output.
    I remind you that the formula is (A \cdot D_{in})+(B \cdot \bar{D_{in}}).
     
  5. ra1ph

    Thread Starter Active Member

    Jan 5, 2010
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    So here is my final attempt!

    [​IMG]
     
  6. Georacer

    Moderator

    Nov 25, 2009
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    You nailed it! Good job!
     
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  7. ra1ph

    Thread Starter Active Member

    Jan 5, 2010
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    Thanks for your Help!
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Not quite. U2 toggles off A', so B changes state when A goes low.
     
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  9. Georacer

    Moderator

    Nov 25, 2009
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    Ouch! Thanks for the correction Ron_H.

    What Roh_H says is that since we said that the JK FF is positively triggered, it will change state at a transition from LOW to HIGH. But! Since that signal comes from Q' of the D FF, that transition will happen when A signal goes from HIGH to LOW.

    The correction to be done is do shift the B signal by left by one tick... and do all the calculation for the gates from the begining.

    I appologize again for the trouble.
     
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  10. ra1ph

    Thread Starter Active Member

    Jan 5, 2010
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    Thanks for the intervention; So how is it looking now. Such a simple problem giving me a lot of hassle...

    [​IMG]
     
    Last edited: Oct 17, 2010
  11. Georacer

    Moderator

    Nov 25, 2009
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    Yes, that's it. (I hope)
     
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