CDA-4101 Lecture 12 Notes

Boolean Logic Exercises

Karnaugh Map Examples

For the following given boolean functions, use a Karnaugh Map to find a minimal, but equivalent function. Also draw the resulting circuit.

Given:

      F(X,Y,Z) = X'YZ' + X'YZ + XYZ' + XYZ + XY'Z

Answer:

      F(X,Y,Z) = Y + XZ

Given:

      F(W,X,Y,Z) = W'X'Y'Z' + W'X'Y'Z + W'X'YZ + W'XYZ + WX'Y'Z'
                   + WX'Y'Z + WX'YZ' + WXYZ

Answer:

      F(W,X,Y,Z) = X'Y' + W'X'Z + XYZ + WX'Z'

Given:

         F(X,Y,Z) = X'Y'Z + X'YZ + XY'Z + XYZ' + XYZ

Answer:

        F(X,Y,Z) = Z + XY

Given:

       F(A,B,C,D) = A'B'C'D + A'BCD + AB'C'D + ABC'D + ABCD

Answer:

       F(A,B,C,D) = B'C'D + BCD + ABD
          OR
       F(A,B,C,D) = B'C'D + BCD + AC'D

Given:

       F(W,X,Y,Z) = W'X'Y'Z' + W'X'YZ' + W'XY'Z + W'XYZ
                    + WX'Y'Z' + WX'YZ' + WXY'Z + WXYZ

Answer:

      F(W,X,Y,Z) = XZ + X'Z'

Given:

       F(A,B,C,D) = A'B'C'D + A'BC'D' + A'BC'D + A'BCD
                    + ABC'D' + ABCD' + ABCD

Answer:

        F(A,B,C,D) = BC'D + BCD + A'BC' + ABD'
           OR
        F(A,B,C,D) = BC'D + A'BD + BC'D' + ABC
           OR
        F(A,B,C,D) = BC'D + BCD + BC'D' + ABD'
           OR
        F(A,B,C,D) = BC'D + BCD + BC'D' + ABC

Boolean Identity Examples

For the following given boolean expressions, use some boolean identities to covert it to the resulting expression.

Given:
```
      (AB')'
    
```
Result:
```
      A' + B
    
```

Given:

      (X'YZ' + X'Y'Z)'

Result:

      (X + Y' + Z)(X + Y + Z')

Given:

      (XY + Z)(Y + XZ)

Result:

      XY + XYZ + YZ + XZ

Computing Dual of Boolean Expression

Compute the duals of the following given boolean expressions.

Given:

      AB + C'D + E'

Answer:

      (A + B)(C' + D) E'

Given:

      (A + 1)(B0 + C)

Answer:

      (A0) + ((B + 1) C)

Given:

      (A' + B0 + 1)(AB' + C) D + A'B

Answer:

      ((A' (B + 1) 0) + ((A + B') C) + D)(A' + B)

Circuit using only NOR/NAND gates.

Show how to implement the following boolean logic functions using only NOR gates.

Function:
```
      AC' + BC'
    
```
Answer:

Implementing circuits with MUX or Decoder

Using only an 8 to 1 multiplexor, show how to implement the circuit computing:

Function:

      F(X,Y,Z) = X'YZ' + X'YZ + XYZ' + XYZ + XY'Z

Answer:

Function:

      F(X,Y,Z) = X'Y'Z + X'YZ + XY'Z + XYZ' + XYZ

Answer:

Using only an 3 to 8 decoder, show how to implement the circuit computing:

Function:

         F(X,Y,Z) = X'YZ' + X'YZ + XYZ' + XYZ + XY'Z

Answer:

Function:

      F(X,Y,Z) = X'Y'Z + X'YZ + XY'Z + XYZ' + XYZ

Answer: