Questions for Part 1
1) Why does the Multisim circuit Digital_Exp_05_Part_01 use three cascaded OR gates to OR the AND
2) Suppose you wish to modify the circuit in Digital_Exp_05_Part_01 so that the output indicates that a
project is rejected rather than approved. How could you do this?
Part 2: Logic Simplification
5.6 Boolean Algebra
The laws and rules of Boolean algebra shown in Table 5-3 can be used to simplify logic expressions so that circuits require fewer gates with fewer inputs. This reduces circuit complexity and the costs of parts and manufacturing. Advantages of Boolean algebra are that it rigorously proves that two logic expressions are equivalent and it can be used to reduce expressions with (in theory) any number of variables and terms. A disadvantage is that Boolean algebra does not guarantee that an expression is the minimum SOP expression or even that you will find the minimum SOP expression. Another disadvantage is that there is no “standard procedure” on how to proceed, so that some ingenuity is often required in simplifying an expression and more than one process can arrive at the same result.
Table 5-4 shows a sequence that simplifies the 2-of-3 voting to its minimum SOP form, but you must determinethe rule or law that validates each step.
1) Begin with the Boolean expression in the first row of Table 5-4.
2) Indicate the Boolean rule or law in Table 5-3 that is applied to derive the new expression for each row
of Table 5-4 (the first two simplifications are done for you)
3) Continue through the table, indicating the Boolean rule or law that is applied at each step to simplify
the 2-of-3 voting logic to its minimum SOP form.
What does the final logic expression indicate?
5.7 Karnaugh Map
The Karnaugh map is a structured way to simplify a Boolean expression. Its main advantage is that it can
always reduce an expression to its simplest SOP form. Its major drawback is that it is limited to simplifying expressions with five or fewer input variables.
1) Use the data from Table 5-1 to fill each cell of the blank Karnaugh map in Figure 5-1 with the output
corresponding to the combination of input variables represented by that cell.
2) Group the cells containing a “1” according to the following rules:
a. Each cell containing a “1” must be included in at least one group. Cells may, but need not be,
included in more than one group.
b. Cells in each group must be horizontally or vertically adjacent to each other (i.e., no diagonal
groups). Karnaugh maps “wrap around” so that the top and bottom rows are vertically adjacent to each other and the leftmost and rightmost columns are horizontally adjacent to each other.
c. Groups must be essentially rectangular in form, so that every row in a group includes the same number of columns, and every column in a group includes the same number of rows.
d. The number of cells in each group must be a power of 2, so that each group contains 1 cell, 2 cells, 4 cells, etc.
e. Each group must contain as many cells as possible.
3) Each group corresponds to an AND term in the final simplified expression. To determine the variables
AND term, write down the input variables that are common to each cell in the group.
AND Term 1: AB
AND Term 2: BC
AND Term 3: AC
4) OR the AND terms together to derive the final expression.
Final expression: AB + BC + AC
5.8 Verification of Simplified Logic
1) Open the Multisim file Digital_Exp_05_Part_02.
2) For each combination of votes,
a. Use the “A”, “B”, and “C” keys to set the vote for each judge. Alternatively, you can click on each
switch to open or close it.
b. Record the result of the vote Table 5-5.
3) Compare the results of Table 5-5 with the results of Table 5-2. Do the table results match?
(Closed=”0”, Open=”1”_) OUTPUT
ON=1) SWITCH SETTINGS
(Closed=”0”, Open=”1”) OUTPUT
A B C A B C
0 0 0 1 0 0
0 0 1 1 0 1
0 1 0 1 1 0
0 1 1 1 1 1
Questions for Part 2
1) How does simulating the Multisim circuit Digital_Exp_05_Part_02 verify that the simplified logic
expression is functionally equivalent to the original expression?
2) If you wished to ensure that a three-variable Boolean expression was in its minimum SOP form, which of the two simplification methods would you use? Why?
3) Is the Karnaugh map group shown in Figure 5-2 a valid group? Why or why not?