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OP image

For this sequential circuit are the outputs "next state function" for q1 and q0?

$$D1 = (Q1)Y' + (Q0)Y' $$ $$J0 = Y'(Q1)' + Y((Q1)'+(Q0)')$$ $$K0 = Y$$ $$Q1 = D1$$ $$Q0 = (J0)(Q0)'+(K0)'(Q0)$$

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closed as not a real question by Brian Carlton, Olin Lathrop, Dave Tweed, clabacchio Mar 26 '13 at 9:18

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Are you sure the circuit is drawn correctly? \$\endgroup\$ – Dave Tweed Mar 25 '13 at 21:20
  • \$\begingroup\$ I think I can see an error too \$\endgroup\$ – Andy aka Mar 25 '13 at 21:30
  • \$\begingroup\$ Have you tried simulating it? \$\endgroup\$ – Leon Heller Mar 25 '13 at 21:42
  • \$\begingroup\$ Sorry, I should have posted this question on my account. The picture is straight out of a textbook, what is the error in the drawing? \$\endgroup\$ – Aaron Mar 25 '13 at 21:44
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    \$\begingroup\$ And what textbook is it from and what does the textbook say it does? Why should we not be priveleged with the information you have when you would like this site to answer a question about its functionality \$\endgroup\$ – Andy aka Mar 25 '13 at 21:50
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Just think about it systematically:

  1. Write the boolean equations for D1, J0, and K0
  2. Simplify those equations to make life easier.
  3. Replace the JK flip flop with an equivalent input to a D flip-flop. Let's call the input of that FF D0.
  4. Simplify the equations a little more.
  5. The next state is just D0 and D1.
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