boolean-algebradigital-logickarnaugh map
From my understanding, to use karnaugh maps to simplify an expression, you would have an SOP expression that does not have to be canonical and then group the 1s in the table.
So that got me wondering. Does this mean I can then have a POS expression and then group the 0s instead of the 1s?
Just simply make a truth table that shows values of F for each of the combinations of A, B and C. Then transfer this information to the K-Map format table to do the minimization.
Do I combine all of the expressions (as in W+X+Y+Z or WXYZ)
No. What the equations show you are how to derive outputs for four separate pins. If you wanted to combine the four pins you should have done it earlier when coming up with the Karnaugh map (the Karnaugh map would then only have one output instead of four).
First step, treat the four as four separate circuits:
A ─┬───────────────┐
│ │
B ──┬─[NOT]────────[AND]┐
││ │ │
C ────[NOT]──┬─────┘ [OR]── W
││ └[OR]─┐ │
D ────────────┘ │ │
│└──────────────[AND]┘
└──[NOT]────────┘
.... implementation of other circuits left as homework
A ──[NOT]────────┐
│
B ──[NOT]──┐ [OR]─┐
[AND]─┘ │
C ──[NOT]──┘ [AND]── Z
│
D ────────────────────┘
Second step, join A to A, B to B etc.. This is obvious because obviously A is the same as A etc.
A ─┬────┬───────────────┐
│ │ │
B ──┬────┬─[NOT]────────[AND]┐
││ ││ │ │
C ───┬─────[NOT]──┬─────┘ [OR]── W
│││ ││ └[OR]─┐ │
D ────┬────────────┘ │ │
││││ │└──────────────[AND]┘
││││ └──[NOT]────────┘
││││
└───────[NOT]────────┐
│││ │
└──────[NOT]──┐ [OR]─┐
││ [AND]─┘ │
└─────[NOT]──┘ [AND]─ Z
│ │
└──────────────────────┘
.... implementation of other circuits left as homework
The circuit should be working at this point but if you're paying attention you may notice that some parts of the W and Z circuits are sharing the same logic. For example, we're using NOT on B and C twice. The next (optional) step is to refactor the circuit and remove redundant/repeated subcircuits and components:
A ───┬──────────────────┐
│ │
B ────┬─[NOT]──┬────────[AND]┐
││ │ │ │
C ──────[NOT]───┬──┬────┘ [OR]── W
││ ││ │ │
││ ││ [OR]─┐ │
D ─┬───────────────┘ │ │
│ ││ ││ │ │
│ │└─────────────────[AND]┘
│ │ ││ │
│ └──[NOT]─┬─────────┘
│ │││
│ └─────────┐
│ │└─┐ [OR]─┐
│ │ [AND]─┘ │
│ └──┘ [AND]─ Z
└─────────────────────────┘
.... implementation of other circuits left as homework
Best Answer
Given PoS: \$ f=\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}\$
Becomes SoP: \$ f= (\overline B + \overline{D})\cdot (A + \overline C)\cdot (A + \overline D ) \$
Circle 0's. Write OR'ed equations with inverted inputs.
Intersection of B & D have 4 zeros. This becomes \$ \overline B + \overline D\$