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I have to design a counter with two inputs: x and y. If y = 0, the counter behaves like a 3-bit ring counter, and if y = 1, it behaves as a 3-bit Johnson counter. If x = 0, it counts up, and if x = 1, it counts down. I may only use D flip flops, and any logic gates I require.

For reference, here are the state tables of a 3-bit ring and Johnson counter (in that order):

Ring table

Johnson table

So naturally, I created this big table of states:

Table of states

Since there are two inputs, and three states, each following state depends on five bits. Therefor the K-maps for Q1+, Q2+ and Q3+ (which are actually D1, D2 and D3 for the flip flops) are maps of five variables, making this somewhat complicated.

The question is: is there a way to do the minimization with k-maps in a simpler manner (perhaps I am missing something)? Or, if there is no way to simplify the minimization, then is it wiser to use k-maps of five variables or perhaps another method (quine-mccluskey maybe, or something completely different)?

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  • \$\begingroup\$ What do you mean by "simpler manner" for the minimization? - Are you asking for a minimized form of your truth table, or for a simpler method to actually do the minimization? - Simpler than the standard method, described e.g. in Wikipedia? \$\endgroup\$ – JimmyB Jun 11 '13 at 10:16
  • \$\begingroup\$ @HannoBinder Is there a simpler method to minimize this truth table than using K-maps of 5 variables? \$\endgroup\$ – user24994 Jun 11 '13 at 10:18
  • \$\begingroup\$ Thinking about it, I believe a truth table is not the right approach for the task, because, as you already noted, the desired output isn't really a direct logic function of the input at any time. \$\endgroup\$ – JimmyB Jun 11 '13 at 10:27
  • \$\begingroup\$ The ring counter has three valid states and the Johnson counter has six. What is supposed to happen to the counting state when Y changes? Also, I would suggest that you work out separate truth tables for your three state bits. That will make it much easier to work out the circuit for each bit separately. \$\endgroup\$ – supercat Jun 11 '13 at 15:31
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A design simplification is to first design the up/down Johnson and ring counters independently. Then just place a mux that selects Q* of either one depending on y.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Undoubtedly that would be a simplification, and thank you for the suggestion. Yet, I am not given freedom in experimenting, and I am stuck with this way of doing things. That's why I need to know if only the minimization can be simplified. \$\endgroup\$ – user24994 Jun 11 '13 at 8:18
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You explicitly stated that you can use any logic gates that you require. If that is the case you do not need to (and should not) do any minimization at all.

Here are two options:

  1. Use the state table to read off a non-minimal sum-of-products form. (Each row is a product of the five input variables, and you sum all the products that result in the output variable being true.)

  2. I understand you can't use @apalopohapa's solution because it has 6 flip-flops instead of 3, but you can do the combinational equivalent: do the next state functions for the ring counter alone, do the next state functions for the johnson counter alone, then use muxes controlled by y to output the final next state values. (Both ring counters and johnson counters are popular precisely because the next state logic is really small, by the way.)

There are also logic minizimization programs available (Google "espresso logic minimizer"), but you really don't need one in this case.

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