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Combinational Logic Part 2:PowerPoint Presentation

Combinational Logic Part 2:

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Combinational Logic Part 2:

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Combinational LogicPart 2:

Karnaugh maps (quick)

- OR all of the minterms of truth table for which the function
value is 1

F = m0 + m2 + m5 + m7

F = X’Y’Z’ + X’YZ’+

XY’Z + XYZ

F = Y’ + X’YZ’ + XY

- Simplifying sum-of-minterms can yield a sum of products
- Difference is that each term need not have all variables
- Resulting gates
- ANDs and one OR

- Sum of products has 2 levels of gates

Fig 2-6

- What’s best?
- Hard to answer
- More gate delays (more on this later)
- But maybe we only have 2-input gates

- Can express F as AND of Maxterms for all rows that should evaluate to 0

- or

This makes one Maxterm fail each time F should be 0

- ORs followed by AND

- Graphical depiction of truth table
- A box for each minterm
- So 2 variables, 4 boxes
- 3 variable, 8 boxes
- And so on

- Useful for simplification
- by inspection
- Algebraic manipulation harder

- There are implied 0s in empty boxes

- Can generate function from K-map

Simplifies to X + Y (in a moment)

- Karnaugh maps were mildly useful when people did simplification
- Computers now do it!
- We’ll cover Karnaugh maps as a way for you to gain insight,
- not as real tool

- Eight minterms
- Look at encoding of columns and rows

- Adjacent squares (horizontally or vertically) are minterms that vary by single variable
- Draw rectangles on map to simplify function
- Illustration next

instead of

- Note that wraps from left edge to right edge.

is

- Help me solve this one

- One box -> 3 literals
- Rectangle of 2 boxes -> 2 literals
- Rectangle of 4 boxes -> 1 literal
- Rectangle of 8 boxes -> Logic 1 (on 3-variable map)
- Covers all minterms

- Overlap is OK.
- No need to use full m5-- waste of input

- At limit of K-map

- A Prime Implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map into a rectangle with the number of squares a power of 2.
- A prime implicant is called an Essential Prime Implicant if it is the only prime implicant that covers (includes) one or more minterms.
- Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map.
- A set of prime implicants "covers all minterms" if, for each minterm of the function, at least one prime implicant in the set of prime implicants includes the minterm.

CD

C

B

B

D

D

B

C

1

1

1

1

1

1

1

1

BD

BD

1

1

B

B

1

1

1

1

A

A

1

1

1

1

1

1

1

1

A

B

D

D

AD

Minterms covered by single prime implicant

- Find ALL Prime Implicants

B’D’ and BD are ESSENTIAL Prime Implicants

C

- Find all prime implicants for:

B

D

B

C

C

1

1

1

B

1

1

A

1

1

1

1

1

1

D

A

- Find all prime implicants for:

- Find all prime implicants.
- Include all essential prime implicants in the solution
- Select a minimum cost set of non-essential prime implicants to cover all minterms not yet covered.
- The solution consists of all essential prime and the selected minimum cost set of non-essential prime implicants

minimum cost

selected minimum cost

- Obtaining a good simplified solution: Use the Selection Rule

- Minimize the overlap among prime implicants as much as possible.
- In the solution, make sure that each prime implicant selected includes at least one minterm not included in any other prime implicant selected.

Essential

Selected

C

C

1

1

1

1

1

1

1

1

1

1

1

1

B

B

1

1

A

A

1

1

1

1

D

D

Minterms covered by essential prime implicants

- Simplify F(A, B, C, D) given on the K-map.

- So far have dealt with functions that were always either 0 or 1
- Sometimes we have some conditions where we don’t care what result is
- Example: dealing with BCD
- Only care about first 10

- In a K-map, mark don’t care with X
- Simpler implementations
- Can select an X either as 1 or 0

or

What would we have if Xs were 0?

Essential

C

C

1

1

x

x

x

x

1

1

x

x

1

1

B

B

x

x

A

A

x

x

1

1

1

1

D

D

Minterms covered by essential prime implicants

- Simplify F(A, B, C, D) given on the K-map.

Selected

F

- Find the optimum POS solution:
- Hint: Use and complement it to get the result.